Capacitors and Dielectrics

1 C/1 V

Reason — The capacitance of a conductor is defined by the relation

$\text C = \frac{\text Q}{\text V}$

so that the unit of capacitance is (unit of charge)/(unit of potential difference), that is, coulomb/volt. This unit is called the farad. Hence 1 farad is that capacitance which acquires a charge of 1 coulomb when its potential is raised by 1 volt.

dielectric material

Reason — A conductor cannot maintain an electric field inside it, whereas a dielectric can. The metal plates only act as the means of applying an electric field to the dielectric medium between them, and it is this medium that sustains the field. Hence the real seat of charge in a capacitor is the dielectric material.

The straight line passing through the origin.

Reason — For a given conductor the capacitance C is a constant determined by its size, shape and the surrounding medium. From

$\text q = \text{CV}$

the charge q is directly proportional to the potential V. Hence the graph of q against V is a straight line passing through the origin, its slope being equal to the capacitance C.

constant

Reason — The plates of a parallel-plate capacitor carry equal and opposite charges spread uniformly, with surface charge density σ. The field between two such closely-spaced plates is

$$\begin{aligned}\text E = \frac{\sigma}{\varepsilon_0} \\[1em] \text E = \frac{\text Q}{\varepsilon_0 \text A}\end{aligned}$$

This expression contains no distance term, so the field is uniform everywhere in the space between the plates. It neither increases nor decreases as we move from one plate to the other.

charge on the plates

Reason — For a parallel-plate capacitor,

$\text C = \frac{\text K \varepsilon_0 \text A}{\text d}$

The capacitance is fixed by the geometry of the plates, that is, their area A (which depends on size and shape), the separation d between them, and the nature of the medium between the plates through K. On giving more charge the potential difference rises in the same proportion, so the ratio $\frac{\text Q}{\text V}$, and hence C, remains unchanged.

potential difference between the plates

Reason — Capacitance is a purely geometrical property of the arrangement together with the medium between the plates,

$\text C = \frac{\text K \varepsilon_0 \text A}{\text d}$

The charge and the potential difference always change together in such a way that their ratio $\frac{\text Q}{\text V}$ stays constant. Hence the potential difference does not affect the capacitance, whereas the shape of the plates, their separation and the medium between them do.

II only

Reason — The field between the plates is uniform, and the electron and the proton carry equal magnitudes of charge. Hence the force F = qE has the same magnitude on both, only the directions being opposite, so statement I is wrong.

The potential energy of a charge at a point of potential V is U = qV. Near the positive plate the potential is high. For the proton q is positive, so U is large and positive; for the electron q is negative, so U is large and negative. Therefore the potential energy of the proton is greater than that of the electron, and statement II is correct while statement III is wrong.

1 : K

Reason — For a sphere of radius a, C = 4πε₀a in air and C_K = 4πε₀Ka in the medium. Since the spheres are raised to the same potential V, the charge in the medium becomes K times that in air,

$\text q_\text{air} = \text{CV}, \qquad \text q_\text{med} = \text{KCV}$

The force in a medium of dielectric constant K, for a separation r, is

$\text F = \frac{1}{4\pi\varepsilon_0 \text K}\frac{\text q^2}{\text r^2}$

In air,

$\text F_\text{air} = \frac{1}{4\pi\varepsilon_0}\frac{(\text{CV})^2}{\text r^2}$

and in the medium,

$$\begin{aligned}\text F_\text{med} = \frac{1}{4\pi\varepsilon_0 \text K}\frac{(\text{KCV})^2}{\text r^2} \\[1em] \text F_\text{med} = \text K\thinspace\text F_\text{air}\end{aligned}$$

Hence F_air : F_med = 1 : K.

3 C

Reason — The upper wire joins A to the junction between the second and third capacitors, and the lower wire joins the junction between the first and second capacitors to B. Hence the first capacitor lies between A and B, the second between B and A, and the third between A and B. Thus all the three capacitors have the same potential difference across them, that is, they are connected in parallel. For a parallel combination,

$$\begin{aligned}\text C_\text{eq} = \text C + \text C + \text C \\[1em] \text C_\text{eq} = 3\text C\end{aligned}$$

2.5 µF

Reason — The network is a balanced Wheatstone-type bridge, the 5 µF capacitor forming the bridge arm. Since

$\frac{2}{2} = \frac{3}{3}$

the two ends of the 5 µF capacitor are at the same potential, so it carries no charge and may be removed.

The two 2 µF capacitors are then in series,

$$\begin{aligned}\text C_1 = \frac{2 \times 2}{2 + 2} \\[1em] \text C_1 = 1\ \text{µF}\end{aligned}$$

and the two 3 µF capacitors are in series,

$$\begin{aligned}\text C_2 = \frac{3 \times 3}{3 + 3} \\[1em] \text C_2 = 1.5\ \text{µF}\end{aligned}$$

These two branches are in parallel between A and B, so

$$\begin{aligned}\text C_\text{eq} = 1 + 1.5 \\[1em] \text C_\text{eq} = 2.5\ \text{µF}\end{aligned}$$

gets doubled

Reason — On inserting a slab of thickness t and dielectric constant K, the capacitance becomes

$\text C = \frac{\varepsilon_0 \text A}{\text d - \text t + \frac{\text t}{\text K}}$

For a metal, K = ∞, and here $\text t = \frac{\text d}{2}$. Substituting,

$$\begin{aligned}\text C = \frac{\varepsilon_0 \text A}{\text d - \frac{\text d}{2}} \\[1em] \text C = \frac{2\varepsilon_0 \text A}{\text d} \\[1em] \text C = 2\text C_0\end{aligned}$$

Hence the capacitance gets doubled.

9 km

Reason — For an isolated spherical conductor of radius a,

$$\begin{aligned}\text C = 4\pi\varepsilon_0 \text a \\[1em] \text a = \frac{\text C}{4\pi\varepsilon_0} \\[1em] \text a = \text C \times 9.0 \times 10^9\end{aligned}$$

Substituting C = 1.0 µF = 1.0 × 10^-6 F,

$$\begin{aligned}\text a = (1.0 \times 10^{-6})(9.0 \times 10^9) \\[1em] \text a = 9 \times 10^3\ \text m \\[1em] \text a = 9\ \text{km}\end{aligned}$$

2 times

Reason — The volume of mercury is conserved on combination, so for n = 8 small drops of radius r forming a big drop of radius R,

$$\begin{aligned}\frac{4}{3}\pi \text R^3 = 8 \times \frac{4}{3}\pi \text r^3 \\[1em] \text R = 8^{1/3}\thinspace\text r \\[1em] \text R = 2\text r\end{aligned}$$

Since the capacitance of a sphere is C = 4πε₀a, it is directly proportional to the radius. Hence

$$\begin{aligned}\frac{\text C_\text{big}}{\text C_\text{small}} = \frac{\text R}{\text r} \\[1em] \frac{\text C_\text{big}}{\text C_\text{small}} = 2\end{aligned}$$

the medium between the plates

Reason — The work done by the battery in transferring charge from one plate to the other is stored as electric potential energy in the electric field set up in the space between the plates. The energy stored per unit volume of this medium, that is, the energy density, is

$\text u = \frac{1}{2}\varepsilon_0 \text E^2$

which is non-zero only where the field exists, namely in the medium between the plates.

4 W

Reason — The energy stored in a capacitor in terms of its charge is

$\text W = \frac{\text q^2}{2\text C}$

For a given capacitor C is constant, so W ∝ q². On doubling the charge,

$$\begin{aligned}\text W' = \frac{(2\text q)^2}{2\text C} \\[1em] \text W' = 4 \times \frac{\text q^2}{2\text C} \\[1em] \text W' = 4\text W\end{aligned}$$

4

Reason — For the air-filled capacitor,

$\text C_0 = \frac{\varepsilon_0 \text A}{\text d} = 1\ \text{pF}$

When the separation is doubled to 2d and wax of dielectric constant K fills the gap,

$$\begin{aligned}\text C = \frac{\text K \varepsilon_0 \text A}{2\text d} \\[1em] \text C = \frac{\text K}{2}\text C_0\end{aligned}$$

Substituting C = 2 pF and C₀ = 1 pF,

$$\begin{aligned}2 = \frac{\text K}{2}(1) \\[1em] \text K = 4\end{aligned}$$

U = $\frac{\text U_0}{\text K}$

Reason — The capacitor is charged and then the slab is introduced, the charging battery being no longer connected. Hence the charge Q on the plates remains unchanged, while the capacitance rises to KC. The energy stored is

$$\begin{aligned}\text U_0 = \frac{\text Q^2}{2\text C} \\[1em] \text U = \frac{\text Q^2}{2(\text{KC})} \\[1em] \text U = \frac{\text U_0}{\text K}\end{aligned}$$

Since K > 1, the energy decreases; the deficit appears as the work done by the capacitor in pulling the slab in.

2.5 µF

Reason — In the figure the upper right arm and the lower left arm of the diamond are plain connecting wires. Hence the upper vertex is the same point as B, and the lower vertex is the same point as A.

Between A and B there are then three parallel paths :

(i) the 1 µF in the upper left arm,

(ii) the 1 µF in the lower right arm,

(iii) the two 1 µF capacitors of the central branch, which are in series and give

$\frac{1 \times 1}{1 + 1} = 0.5\ \text{µF}$

As all three are in parallel,

$$\begin{aligned}\text C_\text{eq} = 1 + 1 + 0.5 \\[1em] \text C_\text{eq} = 2.5\ \text{µF}\end{aligned}$$

8 joule

Reason — For five equal capacitors of capacitance C in series,

$$\begin{aligned}\frac{\text C}{5} = 4\ \text{µF} \\[1em] \text C = 20\ \text{µF}\end{aligned}$$

In parallel the equivalent capacitance is

$$\begin{aligned}\text C_\text p = 5\text C \\[1em] \text C_\text p = 100\ \text{µF} \\[1em] \text C_\text p = 100 \times 10^{-6}\ \text F\end{aligned}$$

The energy stored at V = 400 V is

$$\begin{aligned}\text U = \frac{1}{2}\text C_\text p \text V^2 \\[1em] \text U = \frac{1}{2}(100 \times 10^{-6})(400)^2 \\[1em] \text U = \frac{1}{2}(100 \times 10^{-6})(1.6 \times 10^5) \\[1em] \text U = 8\ \text J\end{aligned}$$

9 : 1

Reason — For three equal capacitors,

$\text C_\text p = 3\text C \qquad \text{and} \qquad \text C_\text s = \frac{\text C}{3}$

Therefore

$$\begin{aligned}\frac{\text C_\text p}{\text C_\text s} = \frac{3\text C}{\text C/3} \\[1em] \frac{\text C_\text p}{\text C_\text s} = 9\end{aligned}$$

that is, C_p : C_s = 9 : 1.

15 µF

Reason — The charge on the first capacitor before sharing is

$$\begin{aligned}\text Q = \text C_1 \text V_1 \\[1em] \text Q = 10 \times 100 \\[1em] \text Q = 1000\ \text{µC}\end{aligned}$$

The battery is disconnected, so this charge is merely redistributed. If C₂ is the second capacitance, the common potential is

$\text V = \frac{\text Q}{\text C_1 + \text C_2}$

Substituting V = 40 V,

$$\begin{aligned}40 = \frac{1000}{10 + \text C_2} \\[1em] 10 + \text C_2 = 25 \\[1em] \text C_2 = 15\ \text{µF}\end{aligned}$$

In A : V remains same and hence Q changes

Reason — In situation A the key is closed, so the battery stays connected and the potential difference V across the plates remains equal to the emf of the battery. On increasing the separation the capacitance C decreases, and since Q = CV with V fixed, the charge on the plates decreases. Hence the third option is correct and the first is wrong.

In situation B the battery is disconnected, so the charge Q on the plates cannot change. The capacitance decreases and therefore $\text V = \frac{\text Q}{\text C}$ increases. Hence the second and fourth options are also wrong.

C₁ < C₂

Reason — In a series combination each capacitor carries the same charge Q, and the potential difference across a capacitor is

$\text V = \frac{\text Q}{\text C}$

so that V is inversely proportional to C. The graph shows a larger rise of potential across the first capacitor than across the second, that is, V₁ > V₂. Since Q is the same for both,

$\frac{\text Q}{\text C_1} \gt \frac{\text Q}{\text C_2} \quad \Rightarrow \quad \text C_1 \lt \text C_2$

32

Reason — Each capacitor can withstand only 250 V, so to withstand 1000 V the capacitors must be joined in series in rows of

$$\begin{aligned}\text n = \frac{1000}{250} \\[1em] \text n = 4\end{aligned}$$

The capacitance of one such row is

$\frac{8}{4} = 2\ \text{µF}$

To obtain 16 µF, the number of such rows joined in parallel must be

$$\begin{aligned}\text m = \frac{16}{2} \\[1em] \text m = 8\end{aligned}$$

Hence the total number of capacitors required is

$$\begin{aligned}\text n \times \text m = 4 \times 8 \\[1em] \text n \times \text m = 32\end{aligned}$$

10 V

Reason — The two capacitors are in series, so each carries the same charge q. Moving along the section from A to B, the potential falls across C₁ and across C₂, and rises by E on crossing the cell from its negative to its positive terminal. Hence

$$\begin{aligned}\text V_\text A - \frac{\text q}{\text C_1} + \text E - \frac{\text q}{\text C_2} = \text V_\text B \\[1em] \text V_\text A - \text V_\text B = \text q\left(\frac{1}{\text C_1} + \frac{1}{\text C_2}\right) - \text E\end{aligned}$$

Substituting the values,

$$\begin{aligned}5 = \text q\left(\frac{1}{1} + \frac{1}{2}\right) - 10 \\[1em] \frac{3\text q}{2} = 15 \\[1em] \text q = 10\ \text{µC}\end{aligned}$$

Therefore the voltage across C₁ is

$$\begin{aligned}\text V_1 = \frac{\text q}{\text C_1} \\[1em] \text V_1 = \frac{10}{1.0} \\[1em] \text V_1 = 10\ \text V\end{aligned}$$

4 V

Reason — The initial charges are

$$\begin{aligned}\text Q_1 = 3 \times 12 \\[1em] \text Q_1 = 36\ \text{µC} \\[1em] \text Q_2 = 6 \times 12 \\[1em] \text Q_2 = 72\ \text{µC}\end{aligned}$$

Since the positive plate of one is joined to the negative plate of the other, the charges partly neutralise each other, so the net charge available for sharing is

$$\begin{aligned}\text Q = \text Q_2 - \text Q_1 \\[1em] \text Q = 72 - 36 \\[1em] \text Q = 36\ \text{µC}\end{aligned}$$

The two capacitors are now in parallel, hence the common potential difference is

$$\begin{aligned}\text V = \frac{\text Q}{\text C_1 + \text C_2} \\[1em] \text V = \frac{36}{3 + 6} \\[1em] \text V = 4\ \text V\end{aligned}$$

The SI unit of capacitance is the farad (F), which is one coulomb per volt. The capacitance of a conductor is 1 farad if a charge of 1 coulomb raises its potential by 1 volt.

Its dimensions are [M^-1 L^-2 T⁴ A²].

A conductor has a capacitance of one farad when a charge of one coulomb given to it raises its potential by one volt.

$1\ \text F = 10^{12}\ \text{pF}$

The potential difference between the plates will decrease.

Explanation:
On bringing the plates closer, d decreases and so the capacitance $\text C = \frac{\varepsilon_0 \text A}{\text d}$ increases. The charge Q on the plates is unchanged, hence $\text V = \frac{\text Q}{\text C}$ is lowered.

The second plate lowers the potential of the first plate, so that a much larger charge can be stored on the capacitor for the same potential difference.

A single conductor may be regarded as a capacitor whose second plate is at infinity, that is, an earthed conductor placed infinitely far away.

Zero.

Explanation:
The two plates carry equal and opposite charges, +Q and −Q, so their sum is zero.

No, its capacitance is not zero.

Explanation:
The capacitance is fixed purely by geometry, that is, by the shapes, sizes and relative positions of the plates and by the medium between them. It does not depend on whether the capacitor actually carries a charge.

For three equal capacitors each of capacitance C,

$$\begin{aligned}\text C_1 = \frac{\text C}{3} \qquad \text{and} \qquad \text C_2 = 3\text C \\[1em] \frac{\text C_1}{\text C_2} = \frac{\text C/3}{3\text C} \\[1em] \frac{\text C_1}{\text C_2} = \frac{1}{9}\end{aligned}$$

With n plates there are (n − 1) gaps, and the arrangement is equivalent to (n − 1) capacitors joined in parallel. Hence

$\text C = (\text n - 1)\frac{\varepsilon_0 \text A}{\text d}$

The dielectric constant is the ratio of the capacitance with the dielectric to that with vacuum (or air),

$$\begin{aligned}\text K = \frac{\text C}{\text C_0} \\[1em] \text K = \frac{80}{4} \\[1em] \text K = 20\end{aligned}$$

More charge.

Explanation:
The capacitance of a dielectric-filled capacitor is K times that of an air capacitor, and since Q = CV with V the same in both cases, the charge stored is K times larger.

Lower PD.

Explanation:
For the same charge, $\text V = \frac{\text Q}{\text C}$, and the capacitance of the dielectric-filled capacitor is K times larger. Hence its potential difference is $\frac{1}{\text K}$ times that of the air capacitor.

$\text K = \frac{\text C}{\text C_0}$

where C is the capacitance of a capacitor filled completely with the dielectric and C₀ that of the same capacitor with vacuum (or air) between the plates.

Water molecules are polar, that is, they possess a permanent electric dipole moment. In an external field these permanent dipoles orient themselves along the field and produce a very strong polarisation, giving water a much larger dielectric constant.

The dielectric constant decreases with rise in temperature.

Explanation:
At higher temperatures the increased thermal agitation of the molecules disturbs the alignment of the molecular dipoles along the field, so the polarisation produced is smaller.

The energy of a charged conductor is

$$\begin{aligned}\text U = \frac{\text Q^2}{2\text C} \\[1em] \text Q = \sqrt{2\thinspace\text{UC}}\end{aligned}$$

Both spheres carry equal charges.

Explanation:
The charge given to a conductor resides only on its outer surface, so the capacitance C = 4πε₀a depends only on the radius. The two spheres have equal radii, hence equal capacitances, and being at the same potential they carry equal charges Q = CV.

The electric field will be halved.

Explanation:
On introducing a dielectric of dielectric constant K, the field is reduced by the factor K,

$$\begin{aligned}\text E = \frac{\text E_0}{\text K} \\[1em] \text E = \frac{\text E_0}{2}\end{aligned}$$

For a metal, K = ∞.

When a dielectric is placed in a field E₀, the field inside it falls to $\frac{\text E_0}{\text K}$. In a metal the field inside is zero, which is possible only if K is infinite.

There is no effect on the capacitance.

Explanation:
For a metal plate of thickness t, K = ∞ and

$\text C = \frac{\varepsilon_0 \text A}{\text d - \text t}$

Since the plate is of negligible thickness, t → 0 and $\text C = \frac{\varepsilon_0 \text A}{\text d}$, the same as before.

In parallel the potential difference is common, so Q = CV gives

$\text Q_1 : \text Q_2 = \text C_1 : \text C_2$

In series each capacitor receives the same charge Q, so the charges are in the ratio 1 : 1.

The capacitance of a parallel-plate capacitor is

$\text C = \frac{\text K \varepsilon_0 \text A}{\text d}$

It therefore depends on :

(i) the area A of the plates — C increases with plate area,

(ii) the separation d between the plates — C is inversely proportional to d,

(iii) the nature of the medium between the plates, through its dielectric constant K.

It does not depend on the charge given or the potential difference applied.

Since $\text C = \frac{\text K \varepsilon_0 \text A}{\text d}$, the capacitance can be increased by

(i) increasing the area of the plates,

(ii) decreasing the separation between the plates,

(iii) filling the space between the plates with a dielectric of high dielectric constant K.

No.

Explanation:
As charge is given continuously to a capacitor, the potential difference across its plates, and hence the electric field between them, goes on rising. When the field exceeds the dielectric strength of the medium, the insulation of the medium breaks down. The capacitor is then discharged through the medium with sparking, and it cannot hold any further charge.

The earth is an extremely large conductor, so its capacitance C = 4πε₀a is enormous, and may be taken as infinite. Giving or removing charge therefore produces practically no change in its potential, since $\text V = \frac{\text Q}{\text C}$. Hence the potential of the earth is treated as a constant, and it is conveniently chosen as the zero of potential.

(i) Series combination :

$\frac{1}{\text C_\text{eq}} = \frac{1}{\text C_1} + \frac{1}{\text C_2} + \dots + \frac{1}{\text C_\text n}$

(ii) Parallel combination :

$\text C_\text{eq} = \text C_1 + \text C_2 + \dots + \text C_\text n$

The dielectric constant of a medium is the ratio of the absolute permittivity ε of the medium to the permittivity ε₀ of free space,

$\text K = \frac{\varepsilon}{\varepsilon_0}$

Being a ratio of two permittivities, it is also called the relative permittivity ε_r of the medium. It is a pure number having no unit.

(i) The capacitance increases to K times its original value, C = KC₀.

(ii) For a given charge on the plates, the potential difference falls to $\frac{1}{\text K}$ times its original value, since $\text V = \frac{\text Q}{\text C}$.

(i) Since $\text C \propto \frac{1}{\text d}$, on doubling the separation the capacitance becomes

$\text C' = \frac{\text C}{2}$

(ii) With t = d the dielectric fills the whole space between the plates, so

$\text C' = \text{KC}$

A dielectric is an insulating material in which all the electrons are tightly bound to the nuclei of the atoms, so that there are practically no free electrons to carry a current. When placed in an electric field, induced charges appear on the surfaces of a dielectric due to polarisation.

Out of the three, mica is a dielectric.

Paper, mica, plastic film and wax are commonly used as dielectrics in capacitors.

On placing a dielectric slab of dielectric constant K in the field between the plates of a charged capacitor, the slab gets electrically polarised. The face of the slab nearer the positive plate acquires an induced negative charge, and the face nearer the negative plate an equal induced positive charge.

Since each induced charge is opposite in sign to the charge on the plate facing it, the induced charges produce a field E' directed against the original field E₀. The effective electric field between the plates, and hence the potential difference, is therefore reduced by the factor K,

$\text E = \frac{\text E_0}{\text K}, \qquad \text V = \frac{\text V_0}{\text K}$

Since the charge Q on the plates is unaltered and $\text C = \frac{\text Q}{\text V}$, the capacitance increases by the factor K, that is, C = KC₀.

Our body is a conductor and it remains in contact with the earth, so the body and the earth together form a single equipotential surface. Since there is no potential difference between our head and the earth, no electric field exists along the body and hence no current flows through it. Therefore we do not get an electric shock.

Yes, he may get a shock.

Explanation:
The aluminium sheet and the earth act as the two plates of a large capacitor with the insulating slab as the dielectric between them. The atmospheric electric field near the surface of the earth gradually charges this capacitor overnight. By the next morning the sheet may be raised to a high potential, and on touching it the man may receive a shock, depending on the potential attained.

Starting from C² N^-1 m^-2 and using $\text F = \frac{\text C}{\text V}$ and $\text N = \frac{\text J}{\text m}$,

$\text C^2\thinspace\text N^{-1}\thinspace\text m^{-2} = \text C\thinspace(\text{F V})(\text{m J}^{-1})\thinspace\text m^{-2}$

Now using $\text V = \frac{\text J}{\text C}$,

$$\begin{aligned}= \text C\thinspace(\text{F J C}^{-1})(\text{m J}^{-1})\thinspace\text m^{-2} \\[1em] \text C^2\thinspace\text N^{-1}\thinspace\text m^{-2} = \text F\thinspace\text m^{-1}\end{aligned}$$

Hence the two units are identical. They are the units of permittivity.

The total charge on the two conductors before connecting is

$\text Q = \text C_1 \text V_1 + \text C_2 \text V_2$

On connecting them by a wire, charge flows until both are at the same potential; the potential of one rises by ΔV₁ and that of the other falls by ΔV₂. The total charge after connecting is

$\text Q' = \text C_1(\text V_1 + \Delta \text V_1) + \text C_2(\text V_2 - \Delta \text V_2)$

Charge is conserved, so Q' = Q,

$$\begin{aligned}\text C_1 \text V_1 + \text C_1 \Delta \text V_1 + \text C_2 \text V_2 - \text C_2 \Delta \text V_2 = \text C_1 \text V_1 + \text C_2 \text V_2 \\[1em] \text C_1 \Delta \text V_1 = \text C_2 \Delta \text V_2 \\[1em] \frac{\Delta \text V_1}{\Delta \text V_2} = \frac{\text C_2}{\text C_1}\end{aligned}$$

(i) Being joined by a wire, both spheres are at the same potential V. Since Q = CV and C = 4πε₀a, the sphere of larger radius has the greater capacitance. Hence the bigger sphere takes more charge.

(ii) For the two spheres,

$$\begin{aligned}\frac{\text Q_1}{\text Q_2} = \frac{\text C_1}{\text C_2} \\[1em] \frac{\text Q_1}{\text Q_2} = \frac{\text a_1}{\text a_2}\end{aligned}$$

and in terms of the surface charge densities,

$\frac{\text Q_1}{\text Q_2} = \frac{4\pi \text a_1^2 \sigma_1}{4\pi \text a_2^2 \sigma_2}$

Equating the two expressions,

$\frac{\sigma_1}{\sigma_2} = \frac{\text a_2}{\text a_1}$

Thus σ is inversely proportional to the radius, so the smaller sphere has the higher surface density of charge.

Both spheres are at the same potential V. Their capacitances are 4πε₀a and 4πε₀b, so if Q_a and Q_b are their charges,

$$\begin{aligned}\text V = \frac{\text Q_\text a}{4\pi\varepsilon_0 \text a} = \frac{\text Q_\text b}{4\pi\varepsilon_0 \text b} \\[1em] \frac{\text Q_\text a}{\text Q_\text b} = \frac{\text a}{\text b}\end{aligned}$$

The electric field at the surface of a sphere of radius r carrying charge Q is $\text E = \frac{\text Q}{4\pi\varepsilon_0 \text r^2}$. Hence

$$\begin{aligned}\frac{\text E_\text a}{\text E_\text b} = \frac{\text Q_\text a/\text a^2}{\text Q_\text b/\text b^2} \\[1em] \frac{\text E_\text a}{\text E_\text b} = \frac{\text Q_\text a}{\text Q_\text b}\cdot\frac{\text b^2}{\text a^2} \\[1em] \frac{\text E_\text a}{\text E_\text b} = \frac{\text a}{\text b}\cdot\frac{\text b^2}{\text a^2} \\[1em] \frac{\text E_\text a}{\text E_\text b} = \frac{\text b}{\text a}\end{aligned}$$

An electric shock is felt on account of a heavy current passing through the body, not merely on account of a high voltage. The balloon has a very small capacitance, so even at a high potential the charge Q = CV stored on it is extremely small. The current that it can drive through the body to the earth is therefore very feeble, and no shock is felt.

At the same potential the charge stored is Q = CV, so the 2 µF capacitor holds a hundred times the charge of the 0.02 µF capacitor. On holding its wires, this much larger charge flows through the body in a very short time, producing a much larger current. Hence the shock felt is more intense.

When the capacitor is connected across a battery, electrons are transferred from one plate to the other until the potential difference between the plates becomes equal to the emf of the battery. Whatever negative charge is removed from one plate appears on the other, so one plate acquires +Q and the other an exactly equal −Q.

Yes, this remains true even when the plates are of different sizes, since the charge is merely transferred from one plate to the other. However, if the sizes are different, the electric field in the space between the plates will no longer be uniform.

On pressing the key, the capacitor C begins to charge. During the charging a current flows in the circuit, and the needle of the ammeter is deflected.

As the capacitor gets fully charged, the potential difference between its plates becomes equal to the emf of the cell, and no further charge can flow. The current therefore stops and the needle returns to its original position.

The reading of the voltmeter will increase.

Explanation:
On increasing the separation, the capacitance $\text C = \frac{\varepsilon_0 \text A}{\text d}$ decreases. The charge Q on the plates is unchanged, hence the potential difference $\text V = \frac{\text Q}{\text C}$ between the plates increases.

(i) The charge Q on the plates remains unchanged, since the battery has been disconnected and no extra charge can be supplied.

(ii) The capacitance increases, as $\text C = \frac{\varepsilon_0 \text A}{\text d}$ and d has been decreased.

(iii) The potential difference falls, since $\text V = \frac{\text Q}{\text C}$ with Q constant.

(iv) The electric field remains unchanged, because $\text E = \frac{\text Q}{\varepsilon_0 \text A}$ does not contain d.

(v) The stored energy decreases, since $\text U = \frac{\text Q^2}{2\text C}$ and C has increased.

Note: The textbook prints $\text U = \frac{\text Q^2}{2\text Q}$ here, a misprint for $\text U = \frac{\text Q^2}{2\text C}$.

The initial capacitance is $\text C_0 = \frac{\varepsilon_0 \text A}{\text d}$ and the charge Q = C₀V remains constant.

(i) On doubling the separation the capacitance becomes $\text C = \frac{\varepsilon_0 \text A}{2\text d} = \frac{\text C_0}{2}$, so the new potential difference is

$$\begin{aligned}\text V' = \frac{\text Q}{\text C} \\[1em] \text V' = \frac{\text C_0 \text V}{\text C_0/2} \\[1em] \text V' = 2\text V\end{aligned}$$

(ii) The initial and final energies are

$$\begin{aligned}\text U_\text i = \frac{1}{2}\text C_0 \text V^2 \\[1em] \text U_\text i = \frac{\varepsilon_0 \text A \text V^2}{2\text d} \\[1em] \text U_\text f = \frac{\text Q^2}{2\text C} \\[1em] \text U_\text f = \frac{1}{2}\left(\frac{\text C_0}{2}\right)(2\text V)^2 \\[1em] \text U_\text f = \text C_0 \text V^2 \\[1em] \text U_\text f = \frac{\varepsilon_0 \text A \text V^2}{\text d}\end{aligned}$$

(iii) The work done in pulling the plates apart is the increase in the stored energy,

$$\begin{aligned}\text W = \text U_\text f - \text U_\text i \\[1em] \text W = \frac{\varepsilon_0 \text A \text V^2}{\text d} - \frac{\varepsilon_0 \text A \text V^2}{2\text d} \\[1em] \text W = \frac{\varepsilon_0 \text A \text V^2}{2\text d}\end{aligned}$$

Note: The textbook prints the initial energy and the work as $\varepsilon_0 \text A/\text V^2/2\text d$, a misprint for $\frac{\varepsilon_0 \text A \text V^2}{2\text d}$.

The battery remains connected, so the potential difference V between the plates stays constant. On increasing the separation the capacitance C decreases, and hence the charge Q = CV on the plates also decreases.

The charge released from the plates passes back through the connecting wires into the battery, so a part of the energy of the capacitor is transferred to the battery. The work done in separating the plates is used up in this transference and is dissipated as heat in the connecting wires. Since $\text U = \frac{1}{2}\text{CV}^2$ with V constant and C decreased, the energy of the capacitor decreases.

More work is done in the second case.

On separating the plates the capacitance goes on decreasing in both cases.

(i) With the battery connected, V remains constant, so the charge Q = CV decreases and hence the electric field $\text E = \frac{\text Q}{\varepsilon_0 \text A}$ between the plates also decreases. The force of attraction between the plates therefore becomes weaker as they are separated.

(ii) With the battery removed, the charge Q remains constant, so the electric field $\text E = \frac{\text Q}{\varepsilon_0 \text A}$ between the plates remains unchanged in spite of the increase in distance. The force of attraction stays the same throughout.

Since the opposing force is larger in the second case, more work has to be done there.

The electric field will be weakened.

On inserting the slab, the dielectric becomes polarised and induced charges appear on its faces. The field due to these polarisation charges is opposite in direction to the field due to the charges on the capacitor plates, so the resultant field between the plates is reduced to $\frac{\text E_0}{\text K}$.

Let C₀ be the capacitance with air and Q₀ the initial charge, so that the initial potential difference is

$\text V_0 = \frac{\text Q_0}{\text C_0}$

After inserting the dielectric plate the capacitance becomes KC₀. With the charge raised to 3Q₀, the potential difference is again V₀,

$$\begin{aligned}\frac{3\text Q_0}{\text{KC}_0} = \text V_0 \\[1em] \frac{3\text Q_0}{\text{KC}_0} = \frac{\text Q_0}{\text C_0} \\[1em] \frac{3}{\text K} = 1 \\[1em] \text K = 3\end{aligned}$$

The capacitance before inserting the slab is

$\text C_0 = \frac{\varepsilon_0 \text A}{\text d}$

On inserting a slab of thickness $\text t = \frac{3\text d}{4}$ and dielectric constant K,

$$\begin{aligned}\text C = \frac{\varepsilon_0 \text A}{\text d - \text t + \frac{\text t}{\text K}} \\[1em] \text C = \frac{\varepsilon_0 \text A}{\text d - \frac{3\text d}{4} + \frac{3\text d}{4\text K}} \\[1em] \text C = \frac{\varepsilon_0 \text A}{\text d\left[\frac{1}{4} + \frac{3}{4\text K}\right]} \\[1em] \text C = \frac{\varepsilon_0 \text A}{\text d}\left(\frac{4\text K}{\text K + 3}\right) \\[1em] \text C = \text C_0\left(\frac{4\text K}{\text K + 3}\right)\end{aligned}$$

Hence the capacitance increases by the factor $\frac{4\text K}{\text K + 3}$.

The capacitance will be doubled.

On placing the metal plate, equal and opposite charges are induced on its two surfaces facing the capacitor plates. The arrangement is then equivalent to two capacitors joined in series, the separation of the plates of each being $\frac{\text d}{4}$.

Using the general formula with $\text t = \frac{\text d}{2}$ and K = ∞ for a metal,

$$\begin{aligned}\text C = \frac{\varepsilon_0 \text A}{\text d - \text t + \frac{\text t}{\text K}} \\[1em] \text C = \frac{\varepsilon_0 \text A}{\text d - \frac{\text d}{2}} \\[1em] \text C = \frac{2\varepsilon_0 \text A}{\text d} \\[1em] \text C = 2\text C_0\end{aligned}$$

(i) The plate A is joined to the cell and D is earthed, so their potentials are V and zero respectively. The plates B and C are unconnected and equidistant, so the potential falls uniformly across the three equal gaps. Hence

$\text V_\text B = \frac{2\text V}{3}, \qquad \text V_\text C = \frac{\text V}{3}$

(ii) On joining B and C by a wire they come to a common potential, which by symmetry is $\frac{\text V}{2}$ each. The plates A and D remain at V and zero respectively.

(iii) The field between A and B rises from $\frac{\text V}{3\text d}$ to $\frac{\text V}{2\text d}$. Between B and C the field becomes zero, since they are now at the same potential. Between C and D the field again rises from $\frac{\text V}{3\text d}$ to $\frac{\text V}{2\text d}$.

(iv) The capacitance of the system was initially $\frac{\varepsilon_0 \text A}{3\text d}$, where A is the area of each plate, and it has now increased to $\frac{\varepsilon_0 \text A}{2\text d}$. Since the potential difference V is unchanged, the charges on A and D will increase.

The charge on the first capacitor is Q = CV, and it stays on the system because the battery has been removed.

(i) In parallel : The charge is shared equally between the two identical capacitors, so the first capacitor is left with $\frac{\text Q}{2}$. Hence its potential difference falls to

$$\begin{aligned}\text V' = \frac{\text Q/2}{\text C} \\[1em] \text V' = \frac{\text V}{2}\end{aligned}$$

(ii) In series : No path exists for the charge on the first capacitor to redistribute, so it retains its charge Q. Hence its potential difference remains V.

In a parallel combination the equivalent capacitance is C₁ + C₂, and the potential difference is common to both,

$\text V = \frac{\text Q}{\text C_1 + \text C_2}$

The charges on the two capacitors are therefore

$$\begin{aligned}\text Q_1 = \text C_1 \text V \\[1em] \text Q_1 = \frac{\text C_1 \text Q}{\text C_1 + \text C_2} \\[1em] \text Q_2 = \text C_2 \text V \\[1em] \text Q_2 = \frac{\text C_2 \text Q}{\text C_1 + \text C_2}\end{aligned}$$

Let each capacitor have capacitance C. In parallel, each is charged to a potential V and carries a charge CV.

On separating them and joining them in series, each capacitor keeps its own charge CV. Hence :

Charge : The charge of the combination in series is CV, whereas in parallel it was 2CV. So the charge is halved.

Potential difference : The potential differences of the two capacitors now add, giving 2V. So the PD is doubled.

Energy : The energy of each capacitor is unaltered, hence the total energy $2 \times \frac{1}{2}\text{CV}^2$ remains unchanged.

In parallel : the equivalent capacitance is nC, so

$$\begin{aligned}\text{total charge} = (\text{nC})\text V, \\[1em] \text{total PD} = \text V, \\[1em] \text{total energy} = \frac{1}{2}(\text{nC})\text V^2\end{aligned}$$

In series : each capacitor retains its own charge CV, which is also the charge of the combination. The equivalent capacitance is $\frac{\text C}{\text n}$ and the potential differences add, so

$$\begin{aligned}\text{total charge} = \text{CV}, \qquad \text{total PD} = \text{nV} \\[1em] \text{total energy} = \frac{1}{2}\left(\frac{\text C}{\text n}\right)(\text{nV})^2 \\[1em] \text{total energy} = \frac{1}{2}\text{nCV}^2\end{aligned}$$

Let C be the capacitance of the air capacitor and V₂ the potential to which it is charged, so its charge is CV₂. The dielectric-filled capacitor has capacitance KC and is initially uncharged.

On joining them in parallel the common potential difference is V, so the charges become CV and (KC)V. Since charge is conserved,

$$\begin{aligned}\text{CV} + (\text{KC})\text V = \text C \text V_2 \\[1em] \text V(1 + \text K) = \text V_2 \\[1em] \text K = \frac{\text V_2 - \text V}{\text V}\end{aligned}$$

Note: The textbook's printed answer writes the initial potential as V₁, giving $\text K = \frac{\text V_1 - \text V}{\text V}$; the question calls it V₂.

On slipping the slab, the capacitance of C₂ increases, and hence the capacitance of the whole series combination also increases. The battery remains connected, so the total potential difference is constant and the charge on the combination increases. As C₁ and C₂ are in series, they carry equal charges, so the charge on each capacitor increases.

C₁ : its capacitance is unchanged but its charge has increased, so the potential difference across it increases, and its stored energy increases.

C₂ : its capacitance increases. Since the total potential difference is fixed and that across C₁ has risen, the potential difference across C₂ decreases, and its stored energy decreases.

There is a net increase in the energy of the combination.

(i) C₁ will remain the same, while C₂ will increase to K times its value.

(ii) The total charge Q₁ + Q₂ is conserved and is shared in proportion to the capacitances. Since C₂ has increased, Q₂ will increase and Q₁ will decrease.

(iii) The common potential difference $\text V = \frac{\text Q_1 + \text Q_2}{\text C_1 + \text C_2}$ will decrease, since the total capacitance has increased while the total charge is unchanged.

(a) Q represents C₂ and P represents C₁.

The slope of a q–V graph is

$\frac{\text q}{\text V} = \text C$

so the steeper line corresponds to the larger capacitance. Since

$\text C = \frac{\varepsilon_0 \text A}{\text d}$

and the plate separation d is the same for both while the plate area of C₂ is double that of C₁, the capacitance of C₂ is larger. The graph Q, having the greater slope, therefore represents C₂.

(b) The graph B corresponds to C₁.

Here the plate areas are equal but the separation of C₁ is twice that of C₂. Since C is inversely proportional to d, C₁ is the smaller capacitance and must be represented by the line of smaller slope, that is, by B.

Electrical capacitance : When a charge Q is given to a conductor, its potential V rises in direct proportion to the charge, that is, Q ∝ V, or

$\text Q = \text{CV}$

The constant C is called the capacitance of the conductor. It is the charge required to raise the potential of the conductor by unity. Its SI unit is the farad.

Capacitance of an isolated spherical conductor :

Let an isolated spherical conductor of radius a be placed in vacuum and be given a charge +Q. The charge spreads uniformly over the outer surface, so the surface density of charge is the same at every point and the lines of force appear to diverge radially from the centre. For finding the potential at the surface, the whole charge may be assumed to be concentrated at the centre O.

The potential of the sphere is therefore

$\text V = \frac{1}{4\pi\varepsilon_0}\frac{\text Q}{\text a}$

By definition, the capacitance is

$\text C = \frac{\text Q}{\text V}$

Substituting the value of V,

$$\begin{aligned}\text C = \frac{\text Q}{\frac{\text Q}{4\pi\varepsilon_0 \text a}} \\[1em] \text C = 4\pi\varepsilon_0 \text a\ \text{farad}\end{aligned}$$

Thus the capacitance of a spherical conductor is directly proportional to its radius. The larger the radius, the greater is the charge it can hold without running up to too high a potential.

A parallel-plate capacitor consists of two plane metallic plates, each of area A, held parallel to each other at a small distance d apart, with vacuum between them.

Let a charge +Q be given to one plate, so that a charge −Q appears on the other. If σ is the surface density of charge, then

$\sigma = \frac{\text Q}{\text A}$

The electric field in the space between the plates is uniform and is given by

$$\begin{aligned}\text E = \frac{\sigma}{\varepsilon_0} \\[1em] \text E = \frac{\text Q}{\varepsilon_0 \text A}\end{aligned}$$

Since the field is uniform, the potential difference between the plates is

$$\begin{aligned}\text V = \text{Ed} \\[1em] \text V = \frac{\text{Qd}}{\varepsilon_0 \text A}\end{aligned}$$

By definition the capacitance is $\text C = \frac{\text Q}{\text V}$, so

$$\begin{aligned}\text C = \frac{\text Q}{\frac{\text{Qd}}{\varepsilon_0 \text A}} \\[1em] \text C = \frac{\varepsilon_0 \text A}{\text d}\end{aligned}$$

Thus the capacitance of a parallel-plate capacitor is directly proportional to the area of the plates and inversely proportional to the distance between them.

In a parallel combination the left plates of all the capacitors are joined to one point A and the right plates to another point B. Hence each capacitor has the same potential difference V across it, but the charges on them are different.

If Q₁, Q₂ and Q₃ are the charges on the three capacitors, then

$\text Q_1 = \text C_1 \text V, \qquad \text Q_2 = \text C_2 \text V, \qquad \text Q_3 = \text C_3 \text V$

The total charge drawn from the source is

$$\begin{aligned}\text Q = \text Q_1 + \text Q_2 + \text Q_3 \\[1em] \text Q = (\text C_1 + \text C_2 + \text C_3)\text V\end{aligned}$$

If a single capacitor of capacitance C placed between A and B takes the same charge Q at the same potential difference V, then Q = CV. Comparing,

$$\begin{aligned}\text{CV} = (\text C_1 + \text C_2 + \text C_3)\text V \\[1em] \text C = \text C_1 + \text C_2 + \text C_3\end{aligned}$$

Thus, in a parallel combination the equivalent capacitance is the sum of the individual capacitances, and it is greater than even the highest of the capacitances connected.

In a series combination the capacitors are joined end to end, the right plate of one being connected to the left plate of the next. When a charge +Q is given to the first plate, equal and opposite charges are induced on the plates of each capacitor by induction. Hence each capacitor carries the same charge Q, while the potential differences across them are different.

The potential differences are

$\text V_1 = \frac{\text Q}{\text C_1}, \qquad \text V_2 = \frac{\text Q}{\text C_2}, \qquad \text V_3 = \frac{\text Q}{\text C_3}$

If V is the total potential difference between the end points A and B, then

$$\begin{aligned}\text V = \text V_1 + \text V_2 + \text V_3 \\[1em] \text V = \text Q\left(\frac{1}{\text C_1} + \frac{1}{\text C_2} + \frac{1}{\text C_3}\right)\end{aligned}$$

If a single capacitor of capacitance C placed between A and B acquires the same charge Q for the same potential difference V, then $\text V = \frac{\text Q}{\text C}$. Comparing,

$$\begin{aligned}\frac{\text Q}{\text C} = \text Q\left(\frac{1}{\text C_1} + \frac{1}{\text C_2} + \frac{1}{\text C_3}\right) \\[1em] \frac{1}{\text C} = \frac{1}{\text C_1} + \frac{1}{\text C_2} + \frac{1}{\text C_3}\end{aligned}$$

Thus, in a series combination the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances, and the equivalent capacitance is smaller than even the smallest capacitance connected.

When a capacitor is charged, work has to be done in transferring charge from one plate to the other against the potential difference already existing between them. This work is stored as electric potential energy of the capacitor.

Let at some instant during charging the charge on the capacitor be q, so that the potential difference between the plates is

$\text V' = \frac{\text q}{\text C}$

The small work done in transferring a further charge dq is

$$\begin{aligned}\text{dW} = \text V'\thinspace\text{dq} \\[1em] \text{dW} = \frac{\text q}{\text C}\thinspace\text{dq}\end{aligned}$$

The total work done in charging the capacitor from zero charge to the final charge Q is obtained by integrating,

$$\begin{aligned}\text W = \int_0^{\text Q} \frac{\text q}{\text C}\thinspace\text{dq} \\[1em] \text W = \frac{1}{\text C}\left[\frac{\text q^2}{2}\right]_0^{\text Q} \\[1em] \text W = \frac{\text Q^2}{2\text C}\end{aligned}$$

This work is stored as the energy U of the capacitor. Using Q = CV, it may be written in the equivalent forms

$\text U = \frac{\text Q^2}{2\text C} = \frac{1}{2}\text{CV}^2 = \frac{1}{2}\text{QV}$

Dielectric : A dielectric is an insulating material in which the electrons are tightly bound to the nuclei of the atoms, so that it cannot conduct electricity. On applying an electric field, induced charges appear on its surfaces due to polarisation.

Let a slab of thickness t and dielectric constant K be placed between the plates of a parallel-plate capacitor of plate area A and separation d, where t < d.

If σ is the surface density of charge on the plates, the field in the remaining space, which contains vacuum, is

$$\begin{aligned}\text E_0 = \frac{\sigma}{\varepsilon_0} \\[1em] \text E_0 = \frac{\text Q}{\varepsilon_0 \text A}\end{aligned}$$

and the field inside the dielectric slab is reduced to

$\text E = \frac{\text E_0}{\text K}$

The potential difference between the plates is the sum of the contributions from the two regions,

$$\begin{aligned}\text V = \text E_0(\text d - \text t) + \frac{\text E_0}{\text K}\thinspace\text t \\[1em] \text V = \text E_0\left(\text d - \text t + \frac{\text t}{\text K}\right)\end{aligned}$$

Substituting $\text E_0 = \frac{\text Q}{\varepsilon_0 \text A}$,

$\text V = \frac{\text Q}{\varepsilon_0 \text A}\left(\text d - \text t + \frac{\text t}{\text K}\right)$

Hence the capacitance is

$$\begin{aligned}\text C = \frac{\text Q}{\text V} \\[1em] \text C = \frac{\varepsilon_0 \text A}{\text d - \text t + \frac{\text t}{\text K}}\end{aligned}$$

Since K > 1, the denominator is smaller than d, so the capacitance is increased by introducing the slab.

Let the entire space between the plates, of area A and separation d, be filled with a dielectric of dielectric constant K. If σ is the surface density of charge on the plates, the field between the plates without the dielectric would be $\frac{\sigma}{\varepsilon_0}$. On account of the polarisation of the dielectric, the field is reduced to

$$\begin{aligned}\text E = \frac{\sigma}{\text K \varepsilon_0} \\[1em] \text E = \frac{\text Q}{\text K \varepsilon_0 \text A}\end{aligned}$$

The potential difference between the plates is

$$\begin{aligned}\text V = \text{Ed} \\[1em] \text V = \frac{\text{Qd}}{\text K \varepsilon_0 \text A}\end{aligned}$$

Hence the capacitance is

$$\begin{aligned}\text C = \frac{\text Q}{\text V} \\[1em] \text C = \frac{\text K \varepsilon_0 \text A}{\text d}\end{aligned}$$

that is, K times the capacitance $\text C_0 = \frac{\varepsilon_0 \text A}{\text d}$ of the air capacitor.

Energy stored : The energy of a charged capacitor is $\text U = \frac{1}{2}\text{CV}^2$. Substituting the above value of C,

$\text U = \frac{1}{2}\left(\frac{\text K \varepsilon_0 \text A}{\text d}\right)\text V^2$

Energy stored in a charged air capacitor : Let at some instant the charge on the capacitor be q, so that the potential difference is $\frac{\text q}{\text C_0}$. The work done in transferring a further charge dq is

$\text{dW} = \frac{\text q}{\text C_0}\thinspace\text{dq}$

Integrating from 0 to the final charge Q,

$$\begin{aligned}\text W = \int_0^{\text Q}\frac{\text q}{\text C_0}\thinspace\text{dq} \\[1em] \text W = \frac{1}{\text C_0}\left[\frac{\text q^2}{2}\right]_0^{\text Q} \\[1em] \text W = \frac{\text Q^2}{2\text C_0}\end{aligned}$$

This work is stored as the energy of the capacitor,

$\text U_0 = \frac{\text Q^2}{2\text C_0} = \frac{1}{2}\text C_0 \text V^2$

where $\text C_0 = \frac{\varepsilon_0 \text A}{\text d}$.

Effect of introducing a dielectric : On filling the space with a dielectric of dielectric constant K, the capacitance becomes C = KC₀. Two cases arise.

(i) Charging battery disconnected : The charge Q remains constant, so

$$\begin{aligned}\text U = \frac{\text Q^2}{2(\text{KC}_0)} \\[1em] \text U = \frac{\text U_0}{\text K}\end{aligned}$$

Since K > 1, the energy decreases by the factor $\frac{1}{\text K}$. The deficit is accounted for by the work done by the capacitor in drawing the slab in.

(ii) Charging battery remaining connected : The potential difference V remains constant, so

$$\begin{aligned}\text U = \frac{1}{2}(\text{KC}_0)\text V^2 \\[1em] \text U = \text K\thinspace\text U_0\end{aligned}$$

and the energy increases K times, the extra energy being supplied by the battery.

(i) The response of the student is incorrect.

In a series combination the equivalent capacitance is smaller than even the smallest capacitance connected. To increase the capacitance, the second capacitor should have been connected in parallel, for which

$\text C_\text{eq} = \text C_1 + \text C_2$

(ii) The response is correct.

For a parallel-plate capacitor

$$\begin{aligned}\text C = \frac{\varepsilon_0 \text A}{\text d} \\[1em] \text C \propto \frac{1}{\text d}\end{aligned}$$

so decreasing the distance between the plates increases the capacitance.

(iii) The response is correct.

When the dielectric completely fills the space between the plates, the capacitance becomes K times its original value,

$$\begin{aligned}\text C' = \text{KC} \\[1em] \text C' = 5 \times 20 \\[1em] \text C' = 100\ \text{µF}\end{aligned}$$

Factors : The capacitance of a parallel-plate capacitor depends on

(i) the area A of the plates,

(ii) the separation d between the plates,

(iii) the dielectric constant K of the medium filling the space between the plates.

Expression :

Let the plates each of area A be held a distance d apart, with vacuum between them, and let σ be the surface density of charge on them. The uniform field between the plates is

$$\begin{aligned}\text E = \frac{\sigma}{\varepsilon_0} \\[1em] \text E = \frac{\text Q}{\varepsilon_0 \text A}\end{aligned}$$

so the potential difference is

$$\begin{aligned}\text V = \text{Ed} \\[1em] \text V = \frac{\text{Qd}}{\varepsilon_0 \text A}\end{aligned}$$

Hence

$$\begin{aligned}\text C = \frac{\text Q}{\text V} \\[1em] \text C = \frac{\varepsilon_0 \text A}{\text d}\end{aligned}$$

and with a dielectric of dielectric constant K filling the space, $\text C = \frac{\text K \varepsilon_0 \text A}{\text d}$.

Interrelationship when the separation decreases : Since C is inversely proportional to d, decreasing the separation increases the capacitance. If the charge Q on the plates is kept constant, then from

$\text V = \frac{\text Q}{\text C}$

the potential difference between the plates decreases in the same proportion. Thus capacitance and potential difference vary inversely with each other for a given charge.

Expression for the energy stored : Let q be the charge on the capacitor at some instant, so that the potential difference between the plates is $\frac{\text q}{\text C}$. The work done in transferring a further small charge dq from one plate to the other is

$\text{dW} = \frac{\text q}{\text C}\thinspace\text{dq}$

Integrating from q = 0 to q = Q,

$$\begin{aligned}\text W = \int_0^{\text Q}\frac{\text q}{\text C}\thinspace\text{dq} \\[1em] \text W = \frac{1}{\text C}\left[\frac{\text q^2}{2}\right]_0^{\text Q} \\[1em] \text W = \frac{\text Q^2}{2\text C}\end{aligned}$$

This work is stored as electric potential energy,

$\text U = \frac{\text Q^2}{2\text C} = \frac{1}{2}\text{CV}^2 = \frac{1}{2}\text{QV}$

(a) Battery remains connected : The potential difference V is fixed by the battery, and the capacitance becomes KC. Hence

$$\begin{aligned}\text U' = \frac{1}{2}(\text{KC})\text V^2 \\[1em] \text U' = \text{KU}\end{aligned}$$

The energy increases K times, the additional energy being supplied by the battery.

(b) Battery is removed : The charge Q on the plates is now fixed, and the capacitance becomes KC. Hence

$$\begin{aligned}\text U' = \frac{\text Q^2}{2(\text{KC})} \\[1em] \text U' = \frac{\text U}{\text K}\end{aligned}$$

The energy decreases to $\frac{1}{\text K}$ times its value, the loss appearing as the work done by the capacitor in pulling the dielectric slab into the gap.

Sharing of charges : When two charged capacitors are joined together by a conducting wire, charge flows from the capacitor at the higher potential to the one at the lower potential until both attain a common potential. This redistribution of charge is called sharing of charges. The total charge is conserved, but a part of the total electric potential energy is always lost, appearing as heat in the connecting wire.

Let two capacitors of capacitances C₁ and C₂ be charged to potentials V₁ and V₂ respectively.

(i) Joined in series : In a series combination each capacitor carries the same charge and no redistribution of charge takes place between them. Hence for two capacitors already charged and then joined in series with their unlike plates together, there is no sharing and no loss of energy; the equivalent capacitance is simply

$\text C = \frac{\text C_1 \text C_2}{\text C_1 + \text C_2}$

and the energy of the combination is the sum of the energies of the two capacitors.

(ii) Joined in parallel : The total charge is conserved,

$\text C_1 \text V_1 + \text C_2 \text V_2 = (\text C_1 + \text C_2)\text V$

so the common potential is

$\text V = \frac{\text C_1 \text V_1 + \text C_2 \text V_2}{\text C_1 + \text C_2}$

The initial and final energies are

$\text U_\text i = \frac{1}{2}\text C_1 \text V_1^2 + \frac{1}{2}\text C_2 \text V_2^2, \qquad \text U_\text f = \frac{1}{2}(\text C_1 + \text C_2)\text V^2$

Substituting the value of V in U_f,

$$\begin{aligned}\text U_\text f = \frac{1}{2}(\text C_1 + \text C_2)\left(\frac{\text C_1 \text V_1 + \text C_2 \text V_2}{\text C_1 + \text C_2}\right)^2 \\[1em] \text U_\text f = \frac{(\text C_1 \text V_1 + \text C_2 \text V_2)^2}{2(\text C_1 + \text C_2)}\end{aligned}$$

Hence the loss of energy is

$$\begin{aligned}\Delta \text U = \text U_\text i - \text U_\text f \\[1em] \Delta \text U = \frac{1}{2}(\text C_1 \text V_1^2 + \text C_2 \text V_2^2) - \frac{(\text C_1 \text V_1 + \text C_2 \text V_2)^2}{2(\text C_1 + \text C_2)} \\[1em] \Delta \text U = \frac{(\text C_1 + \text C_2)(\text C_1 \text V_1^2 + \text C_2 \text V_2^2) - (\text C_1 \text V_1 + \text C_2 \text V_2)^2}{2(\text C_1 + \text C_2)} \\[1em] \Delta \text U = \frac{\text C_1^2 \text V_1^2 + \text C_1 \text C_2 \text V_2^2 + \text C_1 \text C_2 \text V_1^2 + \text C_2^2 \text V_2^2 - \text C_1^2 \text V_1^2 - \text C_2^2 \text V_2^2 - 2\text C_1 \text C_2 \text V_1 \text V_2}{2(\text C_1 + \text C_2)} \\[1em] \Delta \text U = \frac{\text C_1 \text C_2(\text V_1^2 + \text V_2^2 - 2\text V_1 \text V_2)}{2(\text C_1 + \text C_2)} \\[1em] \Delta \text U = \frac{\text C_1 \text C_2}{2(\text C_1 + \text C_2)}(\text V_1 - \text V_2)^2\end{aligned}$$

Since this expression is always positive, there is always a loss of energy on sharing, and it is zero only when V₁ = V₂.

Dielectric constant : The dielectric constant K of a medium is the ratio of the absolute permittivity ε of the medium to the permittivity ε₀ of free space,

$\text K = \frac{\varepsilon}{\varepsilon_0}$

Equivalently, it is the ratio of the capacitance C of a capacitor completely filled with the dielectric to the capacitance C₀ of the same capacitor with vacuum between its plates, $\text K = \frac{\text C}{\text C_0}$. It is a pure number and is also called the relative permittivity ε_r.

Energy density : For a parallel-plate capacitor of plate area A and separation d with vacuum between the plates, the energy stored is

$$\begin{aligned}\text U = \frac{1}{2}\text{CV}^2 \\[1em] \text U = \frac{1}{2}\left(\frac{\varepsilon_0 \text A}{\text d}\right)\text V^2\end{aligned}$$

The field between the plates is uniform, so V = Ed. Substituting,

$$\begin{aligned}\text U = \frac{1}{2}\frac{\varepsilon_0 \text A}{\text d}(\text{Ed})^2 \\[1em] \text U = \frac{1}{2}\varepsilon_0 \text E^2 (\text{Ad})\end{aligned}$$

The quantity Ad is the volume of the space between the plates, in which the electric field exists. Hence the energy stored per unit volume, called the energy density, is

$$\begin{aligned}\text u = \frac{\text U}{\text{Ad}} \\[1em] \text u = \frac{1}{2}\varepsilon_0 \text E^2\end{aligned}$$

If a dielectric of dielectric constant K fills the space, this becomes

$\text u = \frac{1}{2}\text K \varepsilon_0 \text E^2$

Terms involved :

• u is the energy density, that is, the electric potential energy stored per unit volume of the medium.
• ε₀ is the permittivity of free space, equal to 8.85 × 10^-12 C² N^-1 m^-2.
• E is the electric field intensity in the region between the plates.
• K is the dielectric constant of the medium filling the space between the plates.

(i) The battery maintains a constant potential difference V while a total charge Q = CV is transferred. The work done by the battery is

$$\begin{aligned}\text W = \text{QV} \\[1em] \text W = \text{CV}^2\end{aligned}$$

while the energy stored in the capacitor is

$\text U = \frac{1}{2}\text{CV}^2$

Hence

$$\begin{aligned}\frac{\text W}{\text U} = \frac{\text{CV}^2}{\frac{1}{2}\text{CV}^2} \\[1em] \frac{\text W}{\text U} = \frac{2}{1}\end{aligned}$$

The ratio is 2 : 1.

(ii) The battery remains connected, so V is constant while the capacitance rises to KC. Therefore

$$\begin{aligned}\text U' = \frac{1}{2}(\text{KC})\text V^2 \\[1em] \text U' = \text K\left(\frac{1}{2}\text{CV}^2\right) \\[1em] \text U' = \text{KU}\end{aligned}$$

The energy stored becomes KU.

(iii) The battery has been removed, so the charge Q = CV is constant while the capacitance becomes KC. Hence

$$\begin{aligned}\text V' = \frac{\text Q}{\text{KC}} \\[1em] \text V' = \frac{\text{CV}}{\text{KC}} \\[1em] \text V' = \frac{\text V}{\text K}\end{aligned}$$

(iv) Let C₁ = C and C₂ = 2C. Then

$$\begin{aligned}\text C_\text p = \text C_1 + \text C_2 \\[1em] \text C_\text p = 3\text C \\[1em] \text C_\text s = \frac{\text C_1 \text C_2}{\text C_1 + \text C_2} \\[1em] \text C_\text s = \frac{2\text C}{3}\end{aligned}$$

For the energies to be equal,

$$\begin{aligned}\frac{1}{2}\text C_\text p \text V_\text p^2 = \frac{1}{2}\text C_\text s \text V_\text s^2 \\[1em] \frac{\text V_\text p^2}{\text V_\text s^2} = \frac{\text C_\text s}{\text C_\text p} \\[1em] \frac{\text V_\text p^2}{\text V_\text s^2} = \frac{2\text C/3}{3\text C} \\[1em] \frac{\text V_\text p^2}{\text V_\text s^2} = \frac{2}{9} \\[1em] \frac{\text V_\text p}{\text V_\text s} = \sqrt{\frac{2}{9}} \\[1em] \frac{\text V_\text p}{\text V_\text s} = \frac{\sqrt{2}}{3}\end{aligned}$$

Given,

• Potential of the conductor, V = 1000 V
• Charge taken, Q = 2 µC = 2 × 10^-6 C

The capacitance of a conductor is defined as

$\text C = \frac{\text Q}{\text V}$

Substituting the values,

$$\begin{aligned}\text C = \frac{2 \times 10^{-6}}{1000} \\[1em] \text C = 2 \times 10^{-9}\ \text F \\[1em] \text C = 2 \times 10^{-3}\ \mu\text F\end{aligned}$$

Hence, the capacity of the conductor is 2 × 10^-3 µF.

Given,

• Capacitance of the spherical conductor, C = 1 µF = 1 × 10^-6 F

For an isolated spherical conductor of radius a,

$$\begin{aligned}\text C = 4\pi\epsilon_0 \text a \\[1em] \text a = \frac{\text C}{4\pi\epsilon_0} \\[1em] \text a = \text C \times 9.0 \times 10^9\end{aligned}$$

Substituting the value of C,

$$\begin{aligned}\text a = (1 \times 10^{-6})(9.0 \times 10^9) \\[1em] \text a = 9 \times 10^3\ \text m\end{aligned}$$

The diameter is twice the radius,

$$\begin{aligned}\text D = 2\text a \\[1em] \text D = 2 \times 9 \times 10^3 \\[1em] \text D = 1.8 \times 10^4\ \text m\end{aligned}$$

Hence, the diameter of the spherical conductor is 1.8 × 10⁴ m.

Given,

• Capacitance, C = 25 µF = 25 × 10^-6 F
• Potential difference, V = 250 V

The charge on the plates of a capacitor is

$\text Q = \text{CV}$

Substituting the values,

$$\begin{aligned}\text Q = (25 \times 10^{-6})(250) \\[1em] \text Q = 6.25 \times 10^{-3}\ \text C\end{aligned}$$

The two plates carry equal and opposite charges.

Hence, the charges on the plates are + 6.25 × 10^-3 C and − 6.25 × 10^-3 C.

Given,

• Area of each plate, A = 0.0314 m²
• Radius of the conducting sphere, a = 0.5 m

The capacitance of the parallel-plate air capacitor is ε₀A/d and that of the conducting sphere is 4πε₀a. Equating the two,

$$\begin{aligned}\frac{\epsilon_0 \text A}{\text d} = 4\pi\epsilon_0 \text a \\[1em] \text d = \frac{\text A}{4\pi \text a}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text d = \frac{0.0314}{4 \times 3.14 \times 0.5} \\[1em] \text d = \frac{0.0314}{6.28} \\[1em] \text d = 5.0 \times 10^{-3}\ \text m\end{aligned}$$

Hence, the plates must be kept 5.0 mm apart.

Given,

• Area of each plate, A = 6 × 10^-3 m²
• Separation between the plates, d = 3 mm = 3 × 10^-3 m
• Supply voltage, V = 100 V
• ε₀ = 8.85 × 10^-12 C² N^-1 m^-2

The capacitance of a parallel-plate air capacitor is

$\text C = \frac{\epsilon_0 \text A}{\text d}$

Substituting the values,

$$\begin{aligned}\text C = \frac{(8.85 \times 10^{-12})(6 \times 10^{-3})}{3 \times 10^{-3}} \\[1em] \text C = 1.77 \times 10^{-11}\ \text F \approx 18\ \text{pF}\end{aligned}$$

The charge on each plate is

$$\begin{aligned}\text Q = \text{CV} \\[1em] \text Q = (1.8 \times 10^{-11})(100) \\[1em] \text Q = 1.8 \times 10^{-9}\ \text C\end{aligned}$$

Hence, the capacitance is about 18 pF and the charge on each plate is 1.8 × 10^-9 C.

Given,

• Area of each plate, A = 100 cm² = 1 × 10^-2 m²
• Distance between the plates, d = 0.05 cm = 5 × 10^-4 m
• Capacitance with the dielectric, C = 3.54 × 10^-10 F
• ε₀ = 8.85 × 10^-12 C² N^-1 m^-2

The capacitance with air between the plates is

$\text C_0 = \frac{\epsilon_0 \text A}{\text d}$

Substituting the values,

$$\begin{aligned}\text C_0 = \frac{(8.85 \times 10^{-12})(1 \times 10^{-2})}{5 \times 10^{-4}} \\[1em] \text C_0 = 1.77 \times 10^{-10}\ \text F\end{aligned}$$

The dielectric constant is the ratio of the two capacitances,

$$\begin{aligned}\text K = \frac{\text C}{\text C_0} \\[1em] \text K = \frac{3.54 \times 10^{-10}}{1.77 \times 10^{-10}} \\[1em] \text K = 2.0\end{aligned}$$

Hence, the dielectric constant of the medium is 2.0.

Given,

• Capacitance required, C = 0.004 µF = 4 × 10^-9 F
• Dielectric constant of the paper, K = 2.5
• Thickness of the paper, d = 0.025 mm = 2.5 × 10^-5 m

For a capacitor filled with a dielectric,

$$\begin{aligned}\text C = \frac{\text K \epsilon_0 \text A}{\text d} \\[1em] \text A = \frac{\text{Cd}}{\text K \epsilon_0}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text A = \frac{(4 \times 10^{-9})(2.5 \times 10^{-5})}{(2.5)(8.85 \times 10^{-12})} \\[1em] \text A = \frac{1 \times 10^{-13}}{2.2125 \times 10^{-11}} \\[1em] \text A = 4.52 \times 10^{-3}\ \text m^2\end{aligned}$$

Hence, an area of 4.52 × 10^-3 m² of paper will be required.

Given,

• Glass slab : K = 7, d = 4.0 mm
• Mica foil : K = 6, d = 0.20 mm
• Amber plate : K = 2, d = 2.0 cm = 20 mm

When a slab completely fills the space between the plates, the separation of the plates is equal to the thickness of the slab, so

$\text C = \frac{\text K \epsilon_0 \text A}{\text d}$

For a given plate area A, the capacitance is greatest for the material having the largest value of K/d.

$$\begin{aligned}\text{Glass} : \frac{7}{4.0} = 1.75\ \text{mm}^{-1} \\[1em] \text{Mica} : \frac{6}{0.20} = 30\ \text{mm}^{-1} \\[1em] \text{Amber} : \frac{2}{20} = 0.1\ \text{mm}^{-1}\end{aligned}$$

Hence, the mica foil should be placed between the plates, since K/d is maximum for it.

Given,

• Potential difference, V = 200 V
• Area of each plate, A = 100 cm² = 1 × 10^-2 m²
• Distance between the plates, d = 1 mm = 1 × 10^-3 m
• Dielectric constant of the medium, K = 2.5

The capacitance with air between the plates is

$$\begin{aligned}\text C_0 = \frac{\epsilon_0 \text A}{\text d} \\[1em] \text C_0 = \frac{(8.85 \times 10^{-12})(1 \times 10^{-2})}{1 \times 10^{-3}} \\[1em] \text C_0 = 8.85 \times 10^{-11}\ \text F\end{aligned}$$

Hence the charge taken by the capacitor is

$$\begin{aligned}\text Q = \text C_0 \text V \\[1em] \text Q = (8.85 \times 10^{-11})(200) \\[1em] \text Q = 1.76 \times 10^{-8}\ \text C\end{aligned}$$

On introducing the medium the capacitance becomes KC₀, while the charge is the same. Therefore

$$\begin{aligned}\text V' = \frac{\text Q}{\text K \text C_0} \\[1em] \text V' = \frac{\text V}{\text K} \\[1em] \text V' = \frac{200}{2.5} \\[1em] \text V' = 80\ \text V\end{aligned}$$

Hence, the charge taken is 1.76 × 10^-8 C and the new potential difference is 80 V.

Given,

• Area of each plate, A = 100 cm² = 1 × 10^-2 m²
• Thickness of the paper, d = 0.005 cm = 5 × 10^-5 m
• Dielectric constant, K = 2.5
• Maximum field the paper can bear, E_max = 5 × 10⁵ V cm^-1 = 5 × 10⁷ V m^-1

The maximum voltage is

$$\begin{aligned}\text V_\text{max} = \text E_\text{max}\thinspace\text d \\[1em] \text V_\text{max} = (5 \times 10^7)(5 \times 10^{-5}) \\[1em] \text V_\text{max} = 2500\ \text V\end{aligned}$$

The capacitance of the arrangement is

$$\begin{aligned}\text C = \frac{\text K \epsilon_0 \text A}{\text d} \\[1em] \text C = \frac{(2.5)(8.85 \times 10^{-12})(1 \times 10^{-2})}{5 \times 10^{-5}} \\[1em] \text C = 4.425 \times 10^{-9}\ \text F\end{aligned}$$

Hence the charge stored at this voltage is

$$\begin{aligned}\text Q = \text C\thinspace\text V_\text{max} \\[1em] \text Q = (4.425 \times 10^{-9})(2500) \\[1em] \text Q = 1.1 \times 10^{-5}\ \text C\end{aligned}$$

Hence, the capacitor can be charged up to 2500 V and the charge stored is 1.1 × 10^-5 C.

Given,

• EMF of the battery, V = 20 V
• Charge with air, Q₀ = 30 µC
• Charge with oil, Q = 75 µC

(i) The capacitances in the two cases are

$$\begin{aligned}\text C_0 = \frac{\text Q_0}{\text V} \\[1em] \text C_0 = \frac{30}{20} \\[1em] \text C_0 = 1.5\ \mu\text F \\[1em] \text C = \frac{\text Q}{\text V} \\[1em] \text C = \frac{75}{20} \\[1em] \text C = 3.75\ \mu\text F\end{aligned}$$

Hence the dielectric constant of the oil is

$$\begin{aligned}\text K = \frac{\text C}{\text C_0} \\[1em] \text K = \frac{3.75}{1.5} \\[1em] \text K = 2.5\end{aligned}$$

(ii) The energy contained in the oil-capacitor is

$$\begin{aligned}\text U = \frac{1}{2}\text{CV}^2 \\[1em] \text U = \frac{1}{2}(3.75 \times 10^{-6})(20)^2 \\[1em] \text U = \frac{1}{2}(3.75 \times 10^{-6})(400) \\[1em] \text U = 7.5 \times 10^{-4}\ \text J\end{aligned}$$

Hence, the dielectric constant of the oil is 2.5 and the energy contained is 7.5 × 10^-4 J.

Given,

• Plate area, A = 6 × 10^-3 m²
• Plate separation, d = 3 mm = 3 × 10^-3 m
• Potential difference, V = 100 V
• Dielectric constant of mica, K = 6

The capacitance with air is

$$\begin{aligned}\text C_0 = \frac{\epsilon_0 \text A}{\text d} \\[1em] \text C_0 = \frac{(8.85 \times 10^{-12})(6 \times 10^{-3})}{3 \times 10^{-3}} \\[1em] \text C_0 = 1.77 \times 10^{-11}\ \text F \approx 18\ \text{pF}\end{aligned}$$

and the initial charge is

$$\begin{aligned}\text Q_0 = \text C_0 \text V \\[1em] \text Q_0 = (18 \times 10^{-12})(100) \\[1em] \text Q_0 = 1.8 \times 10^{-9}\ \text C\end{aligned}$$

Since the mica sheet is 3 mm thick, it completely fills the space between the plates, so the final capacitance in both the cases is

$$\begin{aligned}\text C = \text{KC}_0 \\[1em] \text C = 6 \times 18 \\[1em] \text C = 108\ \text{pF}\end{aligned}$$

(a) Battery disconnected : the charge on the plates cannot change.

(i) The new potential difference is

$$\begin{aligned}\text V' = \frac{\text V}{\text K} \\[1em] \text V' = \frac{100}{6}\ \text V\end{aligned}$$

(ii) The final capacitance is 108 pF.

(iii) The final charge is the same as before, that is, 1.8 × 10^-9 C.

(b) Battery remaining connected : the potential difference is fixed by the battery.

(i) The potential difference remains 100 V.

(ii) The final capacitance is again 108 pF.

(iii) The final charge is

$$\begin{aligned}\text Q = \text{CV} \\[1em] \text Q = (108 \times 10^{-12})(100) \\[1em] \text Q = 1.08 \times 10^{-8}\ \text C\end{aligned}$$

Given,

• Plate area, A = 90 cm² = 9 × 10^-3 m²
• Separation, d = 2.5 mm = 2.5 × 10^-3 m
• Potential difference, V = 400 V

(a) The capacitance is

$$\begin{aligned}\text C = \frac{\epsilon_0 \text A}{\text d} \\[1em] \text C = \frac{(8.85 \times 10^{-12})(9 \times 10^{-3})}{2.5 \times 10^{-3}} \\[1em] \text C = 3.186 \times 10^{-11}\ \text F\end{aligned}$$

Hence the stored electrostatic energy is

$$\begin{aligned}\text U = \frac{1}{2}\text{CV}^2 \\[1em] \text U = \frac{1}{2}(3.186 \times 10^{-11})(400)^2 \\[1em] \text U = \frac{1}{2}(3.186 \times 10^{-11})(1.6 \times 10^5) \\[1em] \text U = 2.55 \times 10^{-6}\ \text J\end{aligned}$$

(b) The field exists only in the space between the plates, whose volume is

$$\begin{aligned}\text{Ad} = (9 \times 10^{-3})(2.5 \times 10^{-3}) \\[1em] \text{Ad} = 2.25 \times 10^{-5}\ \text m^3\end{aligned}$$

Hence the energy density is

$$\begin{aligned}\text u = \frac{\text U}{\text{Ad}} \\[1em] \text u = \frac{2.55 \times 10^{-6}}{2.25 \times 10^{-5}} \\[1em] \text u = 0.113\ \text J\ \text m^{-3}\end{aligned}$$

Hence, the stored energy is 2.55 × 10^-6 J and the energy density is 0.113 J m^-3.

Given,

• Capacitance, C = 100 µF
• Potential difference, V = 500 V
• The charging battery is removed, so the charge Q remains constant

Since C = ε₀A/d, on halving the separation the capacitance becomes

$\text C' = 2\text C$

The new potential difference is

$$\begin{aligned}\text V' = \frac{\text Q}{\text C'} \\[1em] \text V' = \frac{\text{CV}}{2\text C} \\[1em] \text V' = \frac{\text V}{2} \\[1em] \text V' = \frac{500}{2} \\[1em] \text V' = 250\ \text V\end{aligned}$$

The stored energy in terms of the constant charge is U = Q²/2C, so

$$\begin{aligned}\frac{\text U'}{\text U} = \frac{\text C}{\text C'} \\[1em] \frac{\text U'}{\text U} = \frac{1}{2}\end{aligned}$$

Hence, the new potential difference is 250 V and the stored energy is halved.

Given,

• Capacitance, C = 16 µF = 16 × 10^-6 F
• Charge, Q = 3.2 × 10^-4 C
• The battery is disconnected, so Q remains constant

On doubling the separation the capacitance is halved,

$$\begin{aligned}\text C' = \frac{\text C}{2} \\[1em] \text C' = 8 \times 10^{-6}\ \text F\end{aligned}$$

The work required is equal to the increase in the stored energy,

$$\begin{aligned}\text W = \frac{\text Q^2}{2\text C'} - \frac{\text Q^2}{2\text C} \\[1em] \text W = \frac{\text Q^2}{2}\left(\frac{1}{\text C'} - \frac{1}{\text C}\right)\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text W = \frac{(3.2 \times 10^{-4})^2}{2}\left(\frac{1}{8 \times 10^{-6}} - \frac{1}{16 \times 10^{-6}}\right) \\[1em] = \frac{1.024 \times 10^{-7}}{2}\left(1.25 \times 10^5 - 0.625 \times 10^5\right) \\[1em] = (5.12 \times 10^{-8})(6.25 \times 10^4) \\[1em] \text W = 3.2 \times 10^{-3}\ \text J\end{aligned}$$

Hence, the work required is 3.2 × 10^-3 J.

Given,

• Capacitance, C = 6 µF = 6 × 10^-6 F
• Initial potential difference, V₁ = 10 V
• Final potential difference, V₂ = 20 V

The energy stored is U = ½CV², so the increase in the potential energy is

$\Delta \text U = \frac{1}{2}\text C\left(\text V_2^2 - \text V_1^2\right)$

Substituting the values,

$$\begin{aligned}\Delta \text U = \frac{1}{2}(6 \times 10^{-6})\left[(20)^2 - (10)^2\right] \\[1em] \Delta \text U = \frac{1}{2}(6 \times 10^{-6})(400 - 100) \\[1em] \Delta \text U = \frac{1}{2}(6 \times 10^{-6})(300) \\[1em] \Delta \text U = 9 \times 10^{-4}\ \text J\end{aligned}$$

Hence, the increase in electric potential energy is 9 × 10^-4 J.

Given,

• Initial capacitance, C₁ = 7 µF = 7 × 10^-6 F
• Final capacitance, C₂ = 3 µF = 3 × 10^-6 F
• Potential difference, V = 10 V

The battery remains connected, so V is constant and U = ½CV². The change in energy is

$\Delta \text U = \frac{1}{2}\left(\text C_2 - \text C_1\right)\text V^2$

Substituting the values,

$$\begin{aligned}\Delta \text U = \frac{1}{2}(3 \times 10^{-6} - 7 \times 10^{-6})(10)^2 \\[1em] \Delta \text U = \frac{1}{2}(-4 \times 10^{-6})(100) \\[1em] \Delta \text U = -2 \times 10^{-4}\ \text J\end{aligned}$$

The negative sign shows that the energy of the capacitor decreases.

Hence, 2 × 10^-4 J of energy will be transferred to the battery.

Given,

• Capacitance at 180°, C₁ = 950 pF = 9.5 × 10^-10 F
• Capacitance at 0, C₂ = 50 pF = 5 × 10^-11 F
• Charging voltage, V₁ = 400 V

The charge given to the capacitor is

$$\begin{aligned}\text Q = \text C_1 \text V_1 \\[1em] \text Q = (9.5 \times 10^{-10})(400) \\[1em] \text Q = 3.8 \times 10^{-7}\ \text C\end{aligned}$$

(a) The capacitor is disconnected from the battery, so this charge remains constant. When the dial reads 0,

$$\begin{aligned}\text V_2 = \frac{\text Q}{\text C_2} \\[1em] \text V_2 = \frac{3.8 \times 10^{-7}}{5 \times 10^{-11}} \\[1em] \text V_2 = 7600\ \text V\end{aligned}$$

(b) The work required is equal to the increase in the stored energy,

$$\begin{aligned}\text W = \frac{\text Q^2}{2\text C_2} - \frac{\text Q^2}{2\text C_1} \\[1em] \text W = \frac{1}{2}\text Q\left(\text V_2 - \text V_1\right)\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text W = \frac{1}{2}(3.8 \times 10^{-7})(7600 - 400) \\[1em] \text W = \frac{1}{2}(3.8 \times 10^{-7})(7200) \\[1em] \text W = 1.368 \times 10^{-3}\ \text J\end{aligned}$$

Hence, the potential difference is 7600 V and the work required is 1.368 × 10^-3 J.

Given,

• Capacitance, C = 2000 µF = 2 × 10^-3 F
• EMF of the cell, V = 1.5 V
• Time of discharge, t = 0.1 millisecond = 1 × 10^-4 s

The energy stored in the capacitor is

$\text U = \frac{1}{2}\text{CV}^2$

Substituting the values,

$$\begin{aligned}\text U = \frac{1}{2}(2 \times 10^{-3})(1.5)^2 \\[1em] \text U = \frac{1}{2}(2 \times 10^{-3})(2.25) \\[1em] \text U = 2.25 \times 10^{-3}\ \text J\end{aligned}$$

The whole of this energy is released in the time t, so the power of the flash is

$$\begin{aligned}\text P = \frac{\text U}{\text t} \\[1em] \text P = \frac{2.25 \times 10^{-3}}{1 \times 10^{-4}} \\[1em] \text P = 22.5\ \text W\end{aligned}$$

Hence, the energy stored is 2.25 × 10^-3 J and the power of the flash is 22.5 W.

Given,

• Three equal capacitors, each of capacitance C = 9 pF
• Supply voltage, V = 120 V

(a) For three equal capacitors in series,

$$\begin{aligned}\frac{1}{\text C_s} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} \\[1em] \frac{1}{\text C_s} = \frac{3}{9} \\[1em] \text C_s = 3\ \text{pF}\end{aligned}$$

(b) The capacitors are identical and carry the same charge, so the applied potential difference divides equally between them,

$$\begin{aligned}\text V_1 = \text V_2 = \text V_3 = \frac{120}{3} \\[1em] \text V_1 = \text V_2 = \text V_3 = 40\ \text V\end{aligned}$$

Hence, the total capacitance is 3 pF and the PD across each capacitor is 40 V.

Given,

• C₁ = 2 pF, C₂ = 3 pF, C₃ = 4 pF, connected in parallel
• Supply voltage, V = 100 V

(a) For a parallel combination,

$$\begin{aligned}\text C_p = \text C_1 + \text C_2 + \text C_3 \\[1em] \text C_p = 2 + 3 + 4 \\[1em] \text C_p = 9\ \text{pF}\end{aligned}$$

(b) In parallel each capacitor has the full supply voltage across it, so

$$\begin{aligned}\text Q_1 = \text C_1 \text V \\[1em] \text Q_1 = (2 \times 10^{-12})(100) \\[1em] \text Q_1 = 2 \times 10^{-10}\ \text C \\[1em] \text Q_2 = \text C_2 \text V \\[1em] \text Q_2 = (3 \times 10^{-12})(100) \\[1em] \text Q_2 = 3 \times 10^{-10}\ \text C \\[1em] \text Q_3 = \text C_3 \text V \\[1em] \text Q_3 = (4 \times 10^{-12})(100) \\[1em] \text Q_3 = 4 \times 10^{-10}\ \text C\end{aligned}$$

Hence, the total capacitance is 9 pF and the charges are 2 × 10^-10 C, 3 × 10^-10 C and 4 × 10^-10 C respectively.

Given,

• Capacitance of each capacitor, C = 2.0 µF
• Total potential difference, V = 300 V
• Charge given to the arrangement, Q = 75 µC

The equivalent capacitance of the arrangement must be

$$\begin{aligned}\text C_\text{eq} = \frac{\text Q}{\text V} \\[1em] \text C_\text{eq} = \frac{75}{300} \\[1em] \text C_\text{eq} = 0.25\ \mu\text F\end{aligned}$$

Since the equivalent capacitance is smaller than the capacitance of a single capacitor, the capacitors must be joined in series. For n capacitors each of capacitance C in series,

$$\begin{aligned}\text C_\text{eq} = \frac{\text C}{\text n} \\[1em] 0.25 = \frac{2.0}{\text n} \\[1em] \text n = 8\end{aligned}$$

Hence, 8 capacitors should be connected in series.

Given,

• Each capacitor can be varied from 10 pF to 500 pF

Minimum capacitance : This is obtained when both are set to their least value and joined in series,

$$\begin{aligned}\text C_\text{min} = \frac{\text C_1 \text C_2}{\text C_1 + \text C_2} \\[1em] \text C_\text{min} = \frac{10 \times 10}{10 + 10} \\[1em] \text C_\text{min} = 5\ \text{pF}\end{aligned}$$

Maximum capacitance : This is obtained when both are set to their greatest value and joined in parallel,

$$\begin{aligned}\text C_\text{max} = \text C_1 + \text C_2 \\[1em] \text C_\text{max} = 500 + 500 \\[1em] \text C_\text{max} = 1000\ \text{pF}\end{aligned}$$

Hence, the minimum and maximum capacitances obtainable are 5 pF and 1000 pF.

Given,

• Total charge taken in parallel, Q_p = 1.8 µC
• EMF of the cell, V = 2 V
• Number of equal capacitors = 3

In parallel the equivalent capacitance is 3C, so

$$\begin{aligned}\text Q_p = (3\text C)\text V \\[1em] 1.8 = (3\text C)(2) \\[1em] \text C = 0.3\ \mu\text F\end{aligned}$$

In series the equivalent capacitance is

$$\begin{aligned}\text C_s = \frac{\text C}{3} \\[1em] \text C_s = \frac{0.3}{3} \\[1em] \text C_s = 0.1\ \mu\text F\end{aligned}$$

The charge taken from the same cell is

$$\begin{aligned}\text Q_s = \text C_s \text V \\[1em] \text Q_s = (0.1)(2) \\[1em] \text Q_s = 0.2\ \mu\text C\end{aligned}$$

Hence, in series the combination will take a charge of 0.2 µC.

Given,

• C₁ = 20 µF charged to V₁ = 30 V
• C₂ = 30 µF charged to V₂ = 20 V

Since plates carrying the same type of charge are joined together, the two capacitors are connected in parallel and the charges add. By the conservation of charge, the common potential difference is

$\text V = \frac{\text C_1 \text V_1 + \text C_2 \text V_2}{\text C_1 + \text C_2}$

Substituting the values,

$$\begin{aligned}\text V = \frac{(20)(30) + (30)(20)}{20 + 30} \\[1em] \text V = \frac{600 + 600}{50} \\[1em] \text V = \frac{1200}{50} \\[1em] \text V = 24\ \text V\end{aligned}$$

Hence, the common potential difference is 24 V.

Given,

• C₁ = 0.2 µF charged to V₁ = 600 V
• C₂ = 1.0 µF, uncharged

The charge on the first capacitor is

$$\begin{aligned}\text Q = \text C_1 \text V_1 \\[1em] \text Q = (0.2)(600) \\[1em] \text Q = 120\ \mu\text C\end{aligned}$$

The battery is disconnected, so this charge is merely shared between the two capacitors joined in parallel. The common potential is

$$\begin{aligned}\text V = \frac{\text Q}{\text C_1 + \text C_2} \\[1em] \text V = \frac{120}{0.2 + 1.0} \\[1em] \text V = \frac{120}{1.2} \\[1em] \text V = 100\ \text V\end{aligned}$$

Hence, the new potential is 100 V.

Given,

• C₁ = 80 µF charged to V₁ = 50 V
• C₂ = 320 µF, uncharged

The charge on the first capacitor is

$$\begin{aligned}\text Q = \text C_1 \text V_1 \\[1em] \text Q = (80)(50) \\[1em] \text Q = 4000\ \mu\text C\end{aligned}$$

On joining the two capacitors in parallel, this charge is shared and the common potential is

$$\begin{aligned}\text V = \frac{\text Q}{\text C_1 + \text C_2} \\[1em] \text V = \frac{4000}{80 + 320} \\[1em] \text V = \frac{4000}{400} \\[1em] \text V = 10\ \text V\end{aligned}$$

The charge on the second capacitor is

$$\begin{aligned}\text Q_2 = \text C_2 \text V \\[1em] \text Q_2 = (320)(10) \\[1em] \text Q_2 = 3200\ \mu\text C\end{aligned}$$

Hence, the charge on the second capacitor is 3200 µC.

Given,

• C₁ = 4 µF, C₂ = 2 µF, connected in parallel
• EMF of the battery, V = 6 V

(i) In a parallel combination each capacitor has the same potential difference, which equals the emf of the battery.

Hence, the PD across each capacitor is 6 V.

(ii) The charges are

$$\begin{aligned}\text Q_1 = \text C_1 \text V \\[1em] \text Q_1 = (4)(6) \\[1em] \text Q_1 = 24\ \mu\text C \\[1em] \text Q_2 = \text C_2 \text V \\[1em] \text Q_2 = (2)(6) \\[1em] \text Q_2 = 12\ \mu\text C\end{aligned}$$

Hence, the charges on the 4 µF and 2 µF capacitors are 24 µC and 12 µC respectively.

Given,

• Capacitance of each capacitor, C = 0.50 µF
• Potential difference applied, V = 20 V

The two capacitors in series give

$$\begin{aligned}\text C_s = \frac{\text C}{2} \\[1em] \text C_s = \frac{0.50}{2} \\[1em] \text C_s = 0.25\ \mu\text F\end{aligned}$$

The third capacitor is in parallel with this combination, so

$$\begin{aligned}\text C_\text{eq} = \text C_s + \text C \\[1em] \text C_\text{eq} = 0.25 + 0.50 \\[1em] \text C_\text{eq} = 0.75\ \mu\text F\end{aligned}$$

The total charge stored is

$$\begin{aligned}\text Q = \text C_\text{eq}\text V \\[1em] \text Q = (0.75)(20) \\[1em] \text Q = 15\ \mu\text C\end{aligned}$$

Hence, the capacitance of the combination is 0.75 µF and the total charge stored is 15 µC.

Given,

• C₁ = 4 µF, C₂ = 12 µF, connected in series
• Applied potential difference, V = 120 V

(a) For a series combination,

$$\begin{aligned}\text C_\text{eq} = \frac{\text C_1 \text C_2}{\text C_1 + \text C_2} \\[1em] \text C_\text{eq} = \frac{4 \times 12}{4 + 12} \\[1em] \text C_\text{eq} = \frac{48}{16} \\[1em] \text C_\text{eq} = 3\ \mu\text F\end{aligned}$$

(b) In series each capacitor carries the same charge,

$$\begin{aligned}\text Q = \text C_\text{eq}\text V \\[1em] \text Q = (3)(120) \\[1em] \text Q = 360\ \mu\text C\end{aligned}$$

The potential differences are

$$\begin{aligned}\text V_1 = \frac{\text Q}{\text C_1} \\[1em] \text V_1 = \frac{360}{4} \\[1em] \text V_1 = 90\ \text V \\[1em] \text V_2 = \frac{\text Q}{\text C_2} \\[1em] \text V_2 = \frac{360}{12} \\[1em] \text V_2 = 30\ \text V\end{aligned}$$

Hence, the effective capacitance is 3 µF; the potential differences are 90 V and 30 V, and the charge on each capacitor is 360 µC.

Given,

• C₁ = 2 µF, C₂ = 3 µF, joined in series
• Potential of the outer plate of C₁ = 1000 V
• Outer plate of C₂ is earthed, that is, at zero potential

The equivalent capacitance is

$$\begin{aligned}\text C_\text{eq} = \frac{\text C_1 \text C_2}{\text C_1 + \text C_2} \\[1em] \text C_\text{eq} = \frac{2 \times 3}{2 + 3} \\[1em] \text C_\text{eq} = 1.2\ \mu\text F\end{aligned}$$

Since the total potential difference across the combination is 1000 V, the charge on each capacitor is

$$\begin{aligned}\text Q = \text C_\text{eq}\text V \\[1em] \text Q = (1.2 \times 10^{-6})(1000) \\[1em] \text Q = 1.2 \times 10^{-3}\ \text C\end{aligned}$$

The potential difference across C₁ is

$$\begin{aligned}\text V_1 = \frac{\text Q}{\text C_1} \\[1em] \text V_1 = \frac{1.2 \times 10^{-3}}{2 \times 10^{-6}} \\[1em] \text V_1 = 600\ \text V\end{aligned}$$

Hence the potential of the inner plate of the first capacitor is

$1000 - 600 = 400\ \text V$

The two inner plates are joined together, so the inner plate of the second capacitor is also at 400 V. This is consistent, since the potential difference across C₂ is

$$\begin{aligned}\text V_2 = \frac{\text Q}{\text C_2} \\[1em] \text V_2 = \frac{1.2 \times 10^{-3}}{3 \times 10^{-6}} \\[1em] \text V_2 = 400\ \text V\end{aligned}$$

Hence, both the inner plates are at 400 V, and each carries a charge of 1.2 × 10^-3 C.

Given,

• C₁ = 100 pF = 1 × 10^-10 F charged to V₁ = 100 V
• Common potential after joining, V = 30 V

The charge on the first capacitor is

$$\begin{aligned}\text Q = \text C_1 \text V_1 \\[1em] \text Q = (1 \times 10^{-10})(100) \\[1em] \text Q = 1 \times 10^{-8}\ \text C\end{aligned}$$

On joining the second capacitor in parallel, this charge is shared, so

$$\begin{aligned}\text C_1 + \text C_2 = \frac{\text Q}{\text V} \\[1em] \text C_1 + \text C_2 = \frac{1 \times 10^{-8}}{30} \\[1em] \text C_1 + \text C_2 = \frac{1}{3} \times 10^{-9}\ \text F \\[1em] \text C_2 = \frac{10}{3} \times 10^{-10} - 1 \times 10^{-10} \\[1em] \text C_2 = \frac{7}{3} \times 10^{-10}\ \text F\end{aligned}$$

Energy wasted :

$$\begin{aligned}\text U_i = \frac{1}{2}\text C_1 \text V_1^2 \\[1em] \text U_i = \frac{1}{2}(1 \times 10^{-10})(100)^2 \\[1em] \text U_i = 5 \times 10^{-7}\ \text J \\[1em] \text U_f = \frac{1}{2}(\text C_1 + \text C_2)\text V^2 \\[1em] \text U_f = \frac{1}{2}\left(\frac{1}{3} \times 10^{-9}\right)(30)^2 \\[1em] \text U_f = 1.5 \times 10^{-7}\ \text J \\[1em] \Delta \text U = \text U_i - \text U_f \\[1em] \Delta \text U = 5 \times 10^{-7} - 1.5 \times 10^{-7} \\[1em] \Delta \text U = 3.5 \times 10^{-7}\ \text J\end{aligned}$$

Hence, the capacitance of the second capacitor is $\frac{7}{3} \times 10^{-10}$ F and the energy wasted is 3.5 × 10^-7 J. This energy is dissipated as heat in the connecting wires and as electromagnetic radiation.

Given,

• Number of capacitors, n = 20, each of capacitance C = 5 µF
• Operating voltage, V = 100 V

For a parallel combination,

$$\begin{aligned}\text C_\text{eq} = \text{nC} \\[1em] \text C_\text{eq} = 20 \times 5 \\[1em] \text C_\text{eq} = 100\ \mu\text F \\[1em] \text C_\text{eq} = 100 \times 10^{-6}\ \text F\end{aligned}$$

The energy stored, which is radiated in one flash, is

$\text U = \frac{1}{2}\text C_\text{eq}\text V^2$

Substituting the values,

$$\begin{aligned}\text U = \frac{1}{2}(100 \times 10^{-6})(100)^2 \\[1em] \text U = \frac{1}{2}(100 \times 10^{-6})(10^4) \\[1em] \text U = 0.5\ \text J\end{aligned}$$

Hence, the energy radiated in a flash is 0.5 J.

Given,

• C₁ = 3 µF, C₂ = 6 µF, C₃ = 10 µF
• EMF of the battery, V = 10 V

(a) From the figure, C₁ and C₂ are in series and this combination is in parallel with C₃.

For the series part,

$$\begin{aligned}\text C' = \frac{\text C_1 \text C_2}{\text C_1 + \text C_2} \\[1em] \text C' = \frac{3 \times 6}{3 + 6} \\[1em] \text C' = 2\ \mu\text F\end{aligned}$$

For the parallel combination with C₃,

$$\begin{aligned}\text C_\text{eq} = \text C' + \text C_3 \\[1em] \text C_\text{eq} = 2 + 10 \\[1em] \text C_\text{eq} = 12\ \mu\text F\end{aligned}$$

(b) The energy stored is

$$\begin{aligned}\text U = \frac{1}{2}\text C_\text{eq}\text V^2 \\[1em] \text U = \frac{1}{2}(12 \times 10^{-6})(10)^2 \\[1em] \text U = 6 \times 10^{-4}\ \text J\end{aligned}$$

Hence, the equivalent capacitance is 12 µF and the energy stored is 6 × 10^-4 J.

Given,

• Capacitances 2 µF, 3 µF and 1 µF
• Potential difference between A and B, V = 40 V

From the figure the 2 µF and 3 µF capacitors are in series, and this series branch is in parallel with the 1 µF capacitor.

For the series branch,

$$\begin{aligned}\text C' = \frac{2 \times 3}{2 + 3} \\[1em] \text C' = 1.2\ \mu\text F\end{aligned}$$

The effective capacitance between A and B is

$$\begin{aligned}\text C_\text{eq} = \text C' + 1 \\[1em] \text C_\text{eq} = 1.2 + 1 \\[1em] \text C_\text{eq} = 2.2\ \mu\text F\end{aligned}$$

Charge on the 1 µF capacitor : it has the full 40 V across it,

$$\begin{aligned}\text Q = (1)(40) \\[1em] \text Q = 40\ \mu\text C\end{aligned}$$

Charge on the series branch : the same charge flows through both capacitors of this branch,

$$\begin{aligned}\text Q' = \text C'\text V \\[1em] \text Q' = (1.2)(40) \\[1em] \text Q' = 48\ \mu\text C\end{aligned}$$

Hence, the effective capacitance is 2.2 µF; the charges on the 2 µF, 3 µF and 1 µF capacitors are 48 µC, 48 µC and 40 µC respectively.

Given,

• Three capacitors of 20 µF each, joined in parallel
• Equivalent capacitance between P and Q, C_eq = 30 µF

The three 20 µF capacitors are in parallel, so

$$\begin{aligned}\text C' = 20 + 20 + 20 \\[1em] \text C' = 60\ \mu\text F\end{aligned}$$

The capacitance C is in series with this combination, hence

$\frac{1}{\text C_\text{eq}} = \frac{1}{\text C} + \frac{1}{\text C'}$

Substituting the values,

$$\begin{aligned}\frac{1}{30} = \frac{1}{\text C} + \frac{1}{60} \\[1em] \frac{1}{\text C} = \frac{1}{30} - \frac{1}{60} \\[1em] \frac{1}{\text C} = \frac{2 - 1}{60} \\[1em] \frac{1}{\text C} = \frac{1}{60} \\[1em] \text C = 60\ \mu\text F\end{aligned}$$

Hence, the capacitance C is 60 µF.

Given,

• Capacitances 5 µF, 10 µF, 15 µF and 30 µF

Let the junctions along the row be numbered 1, 2 and 3, the 5 µF lying between A and node 1, the 10 µF between nodes 1 and 2, the 15 µF between nodes 2 and 3, and the 30 µF between node 3 and B.

From the figure, the upper connecting wire joins A to node 2, and the lower connecting wire joins node 1 to node 3. Hence

$\text A \equiv \text{node 2}, \qquad \text{node 1} \equiv \text{node 3}$

Therefore the 5 µF, the 10 µF and the 15 µF capacitors all lie between the same pair of points, that is, they are in parallel :

$$\begin{aligned}\text C' = 5 + 10 + 15 \\[1em] \text C' = 30\ \mu\text F\end{aligned}$$

This combination is in series with the 30 µF capacitor,

$$\begin{aligned}\text C_\text{eq} = \frac{30 \times 30}{30 + 30} \\[1em] \text C_\text{eq} = 15\ \mu\text F\end{aligned}$$

Hence, the equivalent capacitance between A and B is 15 µF.

Given,

• Capacitances 2 µF, 3 µF, 1.8 µF and 6 µF
• Applied voltage, V = 150 V

(i) The 2 µF and 3 µF capacitors are in series,

$$\begin{aligned}\text C' = \frac{2 \times 3}{2 + 3} \\[1em] \text C' = 1.2\ \mu\text F\end{aligned}$$

This is in parallel with the 1.8 µF capacitor,

$$\begin{aligned}\text C'' = 1.2 + 1.8 \\[1em] \text C'' = 3\ \mu\text F\end{aligned}$$

This combination is in series with the 6 µF capacitor,

$$\begin{aligned}\text C_\text{eq} = \frac{3 \times 6}{3 + 6} \\[1em] \text C_\text{eq} = 2\ \mu\text F\end{aligned}$$

(ii) The total charge drawn is

$$\begin{aligned}\text Q = \text C_\text{eq}\text V \\[1em] \text Q = (2)(150) \\[1em] \text Q = 300\ \mu\text C\end{aligned}$$

Since the 6 µF capacitor is in series with the rest, it carries this whole charge. Hence

$$\begin{aligned}\text V_6 = \frac{\text Q}{6} \\[1em] \text V_6 = \frac{300}{6} \\[1em] \text V_6 = 50\ \text V\end{aligned}$$

Hence, the equivalent capacitance is 2 µF and the PD across the 6 µF capacitor is 50 V.

Given,

• Six capacitors of 3 µF in the two horizontal rows, two capacitors of 2 µF as the cross links and one 3 µF at the far end
• Potential difference between A and B, V = 100 V

The network is a ladder, and it is reduced from the far end towards A and B.

Last section : the end 3 µF together with the last 3 µF of the upper row and the last 3 µF of the lower row form a series combination,

$$\begin{aligned}\frac{1}{\text C} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \\[1em] \frac{1}{\text C} = 1 \\[1em] \text C = 1\ \mu\text F\end{aligned}$$

This is in parallel with the second 2 µF cross link,

$$\begin{aligned}\text C = 1 + 2 \\[1em] \text C = 3\ \mu\text F\end{aligned}$$

Middle section : this 3 µF is in series with the middle 3 µF of each row,

$$\begin{aligned}\frac{1}{\text C} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \\[1em] \frac{1}{\text C} = 1 \\[1em] \text C = 1\ \mu\text F\end{aligned}$$

and in parallel with the first 2 µF cross link,

$$\begin{aligned}\text C = 1 + 2 \\[1em] \text C = 3\ \mu\text F\end{aligned}$$

First section : this 3 µF is in series with the first 3 µF of each row,

$$\begin{aligned}\frac{1}{\text C_\text{eq}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \\[1em] \frac{1}{\text C_\text{eq}} = 1 \\[1em] \text C_\text{eq} = 1\ \mu\text F\end{aligned}$$

The charge drawn from the supply is

$$\begin{aligned}\text Q = \text C_\text{eq}\text V \\[1em] \text Q = (1)(100) \\[1em] \text Q = 100\ \mu\text C\end{aligned}$$

Since the capacitors nearest to A and B are in series with the whole of the remaining network, this entire charge resides on them.

Hence, the equivalent capacitance is 1 µF and the charge on the capacitor closest to A (or to B) is 100 µC.

Note: The last sentence of the question is incomplete in the textbook; it is taken to mean the capacitor closest to the point A.

Given,

• Four capacitors of equal capacitance joined in series
• EMF of the battery, V_A − V_C = 10 V
• The middle point B is earthed, so V_B = 0

Two equal capacitors lie between A and B, and two equal capacitors lie between B and C. Since the capacitances on either side of B are equal and each carries the same charge, the potential differences on the two sides are equal in magnitude,

$$\begin{aligned}\text V_A - \text V_B = \text V_B - \text V_C \\[1em] \text V_A + \text V_C = 2\text V_B = 0\end{aligned}$$

Also, from the battery,

$\text V_A - \text V_C = 10\ \text V$

Solving the two equations,

$\text V_A = + 5\ \text V \qquad \text{and} \qquad \text V_C = - 5\ \text V$

Hence, the potentials of A and C are + 5 volt and − 5 volt respectively.

Given,

• Capacitance without dielectric, C₀ = ε₀A/d = 0.5 µF
• Dielectric constants, K₁ = 2 and K₂ = 3

Arrangement I : The two dielectrics occupy the two halves of the plate area, each of area A/2 with the full separation d. The two portions behave as two capacitors in parallel,

$$\begin{aligned}\text C_I = \frac{\text K_1 \epsilon_0 (\text A/2)}{\text d} + \frac{\text K_2 \epsilon_0 (\text A/2)}{\text d} \\[1em] \text C_I = \frac{\epsilon_0 \text A}{2\text d}(\text K_1 + \text K_2) \\[1em] \text C_I = \frac{\text C_0}{2}(2 + 3) \\[1em] \text C_I = 2.5\thinspace\text C_0\end{aligned}$$

Arrangement II : The two dielectrics are stacked one above the other, each of thickness d/2 with the full area A. They behave as two capacitors in series,

$$\begin{aligned}\text C_1 = \frac{2\text K_1 \epsilon_0 \text A}{\text d}, \qquad \text C_2 = \frac{2\text K_2 \epsilon_0 \text A}{\text d} \\[1em] \text C_\text{II} = \frac{\text C_1 \text C_2}{\text C_1 + \text C_2} \\[1em] \text C_\text{II} = \frac{2\epsilon_0 \text A}{\text d}\left(\frac{\text K_1 \text K_2}{\text K_1 + \text K_2}\right) \\[1em] \text C_\text{II} = 2\text C_0\left(\frac{2 \times 3}{2 + 3}\right) \\[1em] \text C_\text{II} = 2.4\thinspace\text C_0\end{aligned}$$

Therefore

$$\begin{aligned}\frac{\text C_I}{\text C_\text{II}} = \frac{2.5\thinspace\text C_0}{2.4\thinspace\text C_0} \\[1em] \frac{\text C_I}{\text C_\text{II}} = \frac{25}{24}\end{aligned}$$

Hence, C_I : C_II = 25 : 24.

Given,

• Initial capacitance of the air capacitor, C = ε₀A/d
• The dielectric of constant K fills half the plate area

The arrangement is equivalent to two capacitors joined in parallel, each of plate area A/2 and separation d, one with air and the other with the dielectric.

$$\begin{aligned}\text C_1 = \frac{\epsilon_0 (\text A/2)}{\text d} \\[1em] \text C_1 = \frac{\text C}{2} \\[1em] \text C_2 = \frac{\text K \epsilon_0 (\text A/2)}{\text d} \\[1em] \text C_2 = \frac{\text{KC}}{2}\end{aligned}$$

Hence the new capacitance is

$$\begin{aligned}\text C' = \text C_1 + \text C_2 \\[1em] \text C' = \frac{\text C}{2} + \frac{\text{KC}}{2} \\[1em] \text C' = \frac{1}{2}\text C(1 + \text K)\end{aligned}$$

Hence, the new capacitance is $\frac{1}{2}\text C(1 + \text K)$.

Given, from the graph,

• For the line A : q = 500 µC at V = 10 V
• For the line B : q = 80 µC at V = 10 V

The slope of a q–V line gives the capacitance, C = q/V. The line of greater slope corresponds to the parallel combination and the other to the series combination. Hence

$$\begin{aligned}\text C_p = \frac{500}{10} \\[1em] \text C_p = 50\ \mu\text F \\[1em] \text C_s = \frac{80}{10} \\[1em] \text C_s = 8\ \mu\text F\end{aligned}$$

Let the two capacitances be C₁ and C₂. Then

$$\begin{aligned}\text C_1 + \text C_2 = 50 \qquad \dots(\text i) \\[1em] \frac{\text C_1 \text C_2}{\text C_1 + \text C_2} = 8 \\[1em] \text C_1 \text C_2 = 8 \times 50 \\[1em] \text C_1 \text C_2 = 400 \qquad \dots(\text{ii})\end{aligned}$$

C₁ and C₂ are therefore the roots of

$$\begin{aligned}\text x^2 - 50\text x + 400 = 0 \\[1em] \text x = \frac{50 \pm \sqrt{2500 - 1600}}{2} \\[1em] \text x = \frac{50 \pm 30}{2} \\[1em] \text x = 40\ \text{or}\ 10\end{aligned}$$

Hence, the capacitances of the two given capacitors are 40 µF and 10 µF.

Given,

• Original separation, d = 4 mm
• Thickness of the slab, t = 3 mm
• Dielectric constant, K = 3
• New capacitance, C = $\frac{2}{3}$ C₀

The original capacitance is

$\text C_0 = \frac{\epsilon_0 \text A}{4}$

Let d' be the new separation. With the slab introduced,

$$\begin{aligned}\text C = \frac{\epsilon_0 \text A}{\text d' - \text t + \frac{\text t}{\text K}} \\[1em] \text C = \frac{\epsilon_0 \text A}{\text d' - 3 + \frac{3}{3}} \\[1em] \text C = \frac{\epsilon_0 \text A}{\text d' - 2}\end{aligned}$$

Since C = $\frac{2}{3}$ C₀,

$$\begin{aligned}\frac{\epsilon_0 \text A}{\text d' - 2} = \frac{2}{3}\left(\frac{\epsilon_0 \text A}{4}\right) \\[1em] \frac{\epsilon_0 \text A}{\text d' - 2} = \frac{\epsilon_0 \text A}{6} \\[1em] \text d' - 2 = 6 \\[1em] \text d' = 8\ \text{mm}\end{aligned}$$

Hence, the new distance between the plates is 8 mm.

Given,

• X is an air capacitor, Y contains a dielectric of ε_r = 4
• The two are joined in series across a 12 V battery
• Equivalent capacitance, C_eq = 4 µF

(i) Since the plate areas and separations are equal,

$\text C_Y = 4\thinspace\text C_X$

For the series combination,

$$\begin{aligned}\text C_\text{eq} = \frac{\text C_X \text C_Y}{\text C_X + \text C_Y} \\[1em] \text C_\text{eq} = \frac{\text C_X(4\text C_X)}{5\text C_X} \\[1em] \text C_\text{eq} = \frac{4\text C_X}{5} \\[1em] 4 = \frac{4\text C_X}{5} \\[1em] \text C_X = 5\ \mu\text F \\[1em] \text C_Y = 20\ \mu\text F\end{aligned}$$

(ii) In series both capacitors carry the same charge,

$$\begin{aligned}\text Q = \text C_\text{eq}\text V \\[1em] \text Q = (4)(12) \\[1em] \text Q = 48\ \mu\text C \\[1em] \text V_X = \frac{\text Q}{\text C_X} \\[1em] \text V_X = \frac{48}{5} \\[1em] \text V_X = 9.6\ \text V \\[1em] \text V_Y = \frac{\text Q}{\text C_Y} \\[1em] \text V_Y = \frac{48}{20} \\[1em] \text V_Y = 2.4\ \text V\end{aligned}$$

(iii) For the same charge, U = Q²/2C, so

$$\begin{aligned}\frac{\text U_X}{\text U_Y} = \frac{\text C_Y}{\text C_X} \\[1em] \frac{\text U_X}{\text U_Y} = \frac{20}{5} \\[1em] \frac{\text U_X}{\text U_Y} = 4\end{aligned}$$

Hence, C_X = 5 µF and C_Y = 20 µF; V_X = 9.6 V and V_Y = 2.4 V; and U_X : U_Y = 4 : 1.

Given,

• Capacitors of 10 µF, 7 µF, 3 µF, 15 µF and Z µF
• Applied voltage, V = 200 V
• Total energy stored, U = 160 mJ = 0.16 J

Equivalent capacitance : From the energy stored,

$$\begin{aligned}\text U = \frac{1}{2}\text C_\text{eq}\text V^2 \\[1em] 0.16 = \frac{1}{2}\text C_\text{eq}(200)^2 \\[1em] 0.16 = \frac{1}{2}\text C_\text{eq}(4 \times 10^4) \\[1em] \text C_\text{eq} = \frac{2 \times 0.16}{4 \times 10^4} \\[1em] \text C_\text{eq} = 8 \times 10^{-6}\ \text F \\[1em] \text C_\text{eq} = 8\ \mu\text F\end{aligned}$$

Value of Z : In the upper branch, the 7 µF and 3 µF capacitors are in parallel,

$7 + 3 = 10\ \mu\text F$

This is in series with the 10 µF and the 15 µF capacitors,

$$\begin{aligned}\frac{1}{\text C'} = \frac{1}{10} + \frac{1}{10} + \frac{1}{15} \\[1em] \frac{1}{\text C'} = \frac{3 + 3 + 2}{30} \\[1em] \frac{1}{\text C'} = \frac{8}{30} \\[1em] \text C' = \frac{30}{8} \\[1em] \text C' = 3.75\ \mu\text F\end{aligned}$$

The capacitor Z is in parallel with this branch, so

$$\begin{aligned}\text C_\text{eq} = \text C' + \text Z \\[1em] 8 = 3.75 + \text Z \\[1em] \text Z = 4.25\ \mu\text F\end{aligned}$$

Hence, the equivalent capacitance is 8 µF and Z = 4.25 µF.

Given,

• Each capacitor of the repeating link has capacitance C

Let the capacitance of the whole infinite network between A and B be x. Since the network is infinite, removing the first link leaves a network which is again identical to the original one, of capacitance x.

Hence the arrangement is : one capacitor C in series with the parallel combination of the shunt capacitor C and the remaining network x. Therefore

$$\begin{aligned}\frac{1}{\text x} = \frac{1}{\text C} + \frac{1}{\text C + \text x} \\[1em] \frac{1}{\text x} = \frac{(\text C + \text x) + \text C}{\text C(\text C + \text x)} \\[1em] \frac{1}{\text x} = \frac{2\text C + \text x}{\text C(\text C + \text x)}\end{aligned}$$

Cross-multiplying,

$$\begin{aligned}\text C(\text C + \text x) = \text x(2\text C + \text x) \\[1em] \text C^2 + \text{Cx} = 2\text{Cx} + \text x^2 \\[1em] \text x^2 + \text{Cx} - \text C^2 = 0\end{aligned}$$

Solving this quadratic and retaining the positive root,

$$\begin{aligned}\text x = \frac{-\text C + \sqrt{\text C^2 + 4\text C^2}}{2} \\[1em] \text x = \frac{(\sqrt{5} - 1)\text C}{2}\end{aligned}$$

Hence, the capacitance of the infinite circuit is $\frac{(\sqrt{5} - 1)}{2}\text C$.

Given,

• Area of each plate = A, distance between neighbouring plates = d
• Capacitance of each gap, C₀ = ε₀A/d

(i) From the figure the plates 2 and 4 are joined to P, while the plates 1 and 3 are joined to Q. The three gaps 1–2, 2–3 and 3–4 therefore all lie between P and Q, that is, they are in parallel. Hence

$$\begin{aligned}\text C_\text{PQ} = 3\text C_0 \\[1em] \text C_\text{PQ} = \frac{3\epsilon_0 \text A}{\text d}\end{aligned}$$

(ii) Here there are five plates and four gaps. From the figure the plates 1 and 3 are joined to P, the plate 4 is joined to Q, and the plates 2 and 5 are joined together to form a common node, say D.

The gaps then give :

• gap 1–2 : between P and D, capacitance C₀
• gap 2–3 : between D and P, capacitance C₀
• gap 3–4 : between P and Q, capacitance C₀
• gap 4–5 : between Q and D, capacitance C₀

The first two are in parallel between P and D, giving 2C₀. This is in series with the gap 4–5 between D and Q,

$\frac{2\text C_0 \times \text C_0}{2\text C_0 + \text C_0} = \frac{2\text C_0}{3}$

This path is in parallel with the gap 3–4, which lies directly between P and Q,

$$\begin{aligned}\text C_\text{PQ} = \text C_0 + \frac{2\text C_0}{3} \\[1em] \text C_\text{PQ} = \frac{5\text C_0}{3} \\[1em] \text C_\text{PQ} = \frac{5}{3}\frac{\epsilon_0 \text A}{\text d}\end{aligned}$$

Hence, (i) $\frac{3\epsilon_0 \text A}{\text d}$ and (ii) $\frac{5}{3}\frac{\epsilon_0 \text A}{\text d}$.

(i)

Given,

• Capacitances 1 F, 2 F, 3 F, 4 F, 5 F and 6 F joined as shown

Let T be the top node, at which the 1 F capacitor, the 5 F capacitor and the upper branch meet, and J the node on the lower right wire, at which the 2 F, the 5 F and the 4 F capacitors meet.

Simplifying the upper branch : the 3 F and 6 F capacitors are in series,

$$\begin{aligned}\text C_\text{TB} = \frac{3 \times 6}{3 + 6} \\[1em] \text C_\text{TB} = \frac{18}{9} \\[1em] \text C_\text{TB} = 2\ \text F\end{aligned}$$

Testing the balance : the network is now a Wheatstone bridge with the 5 F capacitor as the bridge arm. Comparing the ratios of the arms,

$$\begin{aligned}\frac{\text C_\text{AT}}{\text C_\text{TB}} = \frac{1}{2} \\[1em] \frac{\text C_\text{AJ}}{\text C_\text{JB}} = \frac{2}{4} \\[1em] \frac{\text C_\text{AJ}}{\text C_\text{JB}} = \frac{1}{2}\end{aligned}$$

The two ratios are equal, so the bridge is balanced. Hence the nodes T and J are at the same potential, no charge flows through the 5 F capacitor, and it may be removed.

Reducing the remaining branches : two series branches are left between A and B,

$$\begin{aligned}\text C_\text{top} = \frac{1 \times 2}{1 + 2} \\[1em] \text C_\text{top} = \frac{2}{3}\ \text F \\[1em] \text C_\text{bottom} = \frac{2 \times 4}{2 + 4} \\[1em] \text C_\text{bottom} = \frac{8}{6} \\[1em] \text C_\text{bottom} = \frac{4}{3}\ \text F\end{aligned}$$

These are in parallel, hence

$$\begin{aligned}\text C_\text{AB} = \frac{2}{3} + \frac{4}{3} \\[1em] \text C_\text{AB} = \frac{6}{3} \\[1em] \text C_\text{AB} = 2\ \text F\end{aligned}$$

Hence, the equivalent capacitance between A and B is 2 F.

(ii)

Given,

• Capacitances 23 µF, 12 µF, 7 µF, 13 µF, 10 µF and two of 1 µF joined as shown

Combining the parallel capacitors : the 23 µF and 12 µF capacitors lie between the same pair of nodes, and so do the two 1 µF capacitors. Hence

$$\begin{aligned}\text C_\text{left} = 23 + 12 \\[1em] \text C_\text{left} = 35\ \mu\text F \\[1em] \text C_\text{right} = 1 + 1 \\[1em] \text C_\text{right} = 2\ \mu\text F\end{aligned}$$

Redrawing as a bridge : the network becomes a Wheatstone bridge with the 13 µF capacitor in the central branch, the four arms being

$35\ \mu\text F, \qquad 7\ \mu\text F, \qquad 10\ \mu\text F, \qquad 2\ \mu\text F$

Testing the balance :

$$\begin{aligned}\frac{35}{7} = 5 \\[1em] \frac{10}{2} = 5\end{aligned}$$

The two ratios are equal, so the bridge is balanced. The nodes on either side of the 13 µF capacitor are at the same potential, no charge flows through it, and it may be removed.

Reducing the remaining branches :

$$\begin{aligned}\text C_\text{top} = \frac{35 \times 7}{35 + 7} \\[1em] \text C_\text{top} = \frac{245}{42} \\[1em] \text C_\text{top} = \frac{35}{6}\ \mu\text F \\[1em] \text C_\text{bottom} = \frac{10 \times 2}{10 + 2} \\[1em] \text C_\text{bottom} = \frac{20}{12} \\[1em] \text C_\text{bottom} = \frac{10}{6}\ \mu\text F\end{aligned}$$

These are in parallel, hence

$$\begin{aligned}\text C_\text{AB} = \frac{35}{6} + \frac{10}{6} \\[1em] \text C_\text{AB} = \frac{45}{6} \\[1em] \text C_\text{AB} = 7.5\ \mu\text F\end{aligned}$$

Hence, the equivalent capacitance between A and B is 7.5 µF.

(iii)

Given,

• Seven identical capacitors, each of capacitance C, joined as shown

The capacitor joining the two upper nodes links the two halves of the network, so it cannot be reduced by series and parallel steps alone. It is solved by the potential method.

Setting up the node potentials : let a potential difference V be applied, so that

$\text V_A = \text V, \qquad \text V_B = 0$

Let V₁ be the potential of the top-left node, V₂ that of the bottom-middle node and V₃ that of the top-right node. The network is symmetrical about the vertical through the middle node, so

$\text V_2 = \frac{\text V}{2}, \qquad \text V_3 = \text V - \text V_1$

Applying conservation of charge at the node V₁ : this node is isolated, so the algebraic sum of the charges on the plates meeting there is zero. Its neighbours are A, the middle node and the top-right node, hence

$(\text V_1 - \text V)\text C + (\text V_1 - \text V_2)\text C + (\text V_1 - \text V_3)\text C = 0$

Dividing by C and substituting the values of V₂ and V₃,

$$\begin{aligned}(\text V_1 - \text V) + \left(\text V_1 - \frac{\text V}{2}\right) + \left[\text V_1 - (\text V - \text V_1)\right] = 0 \\[1em] 4\text V_1 - \frac{5\text V}{2} = 0 \\[1em] \text V_1 = \frac{5}{8}\text V\end{aligned}$$

Finding the total charge : the charge drawn at the terminal A divides between the two capacitors joined to it, one going to the top-left node and the other to the middle node,

$\text Q = (\text V - \text V_1)\text C + (\text V - \text V_2)\text C$

Substituting the potentials,

$$\begin{aligned}\text Q = \left(\text V - \frac{5}{8}\text V\right)\text C + \left(\text V - \frac{1}{2}\text V\right)\text C \\[1em] \text Q = \frac{3}{8}\text{CV} + \frac{4}{8}\text{CV} \\[1em] \text Q = \frac{7}{8}\text{CV}\end{aligned}$$

By definition the equivalent capacitance is

$$\begin{aligned}\text C_\text{AB} = \frac{\text Q}{\text V} \\[1em] \text C_\text{AB} = \frac{7\text C}{8}\end{aligned}$$

Hence, the equivalent capacitance between A and B is $\frac{7\text C}{8}$.

Given,

• Area of each plate = A
• Mass hanging from the string = m
• The system is in equilibrium and the surface is smooth

The two plates carry equal and opposite charges Q, so they attract each other. The force of attraction between the plates of a charged parallel-plate capacitor is

$\text F = \frac{\text Q^2}{2\epsilon_0 \text A}$

Since the strings and the spring are massless and the surface is smooth, the tension in the string equals the weight of the hanging block,

$\text T = \text{mg}$

For the plate assembly to be in equilibrium, the electrostatic force of attraction between the plates must be balanced by this tension,

$$\begin{aligned}\frac{\text Q^2}{2\epsilon_0 \text A} = \text{mg} \\[1em] \text Q^2 = 2\thinspace\text{mg}\thinspace\text A\thinspace\epsilon_0 \\[1em] \text Q = \sqrt{2\thinspace\text{mg}\thinspace\text A\thinspace\epsilon_0}\end{aligned}$$

Hence, the magnitude of the charge on one of the plates is $\sqrt{2\thinspace\text{mg}\thinspace\text A\thinspace\epsilon_0}$.

Given,

• Radius of the earth, a = 6400 km = 6.4 × 10⁶ m
• $\frac{1}{4\pi\epsilon_0} = 9.0 \times 10^9$ N-m² C^-2

For an isolated spherical conductor,

$$\begin{aligned}\text C = 4\pi\epsilon_0 \text a \\[1em] \text C = \frac{\text a}{9.0 \times 10^9}\end{aligned}$$

Substituting the value of a,

$$\begin{aligned}\text C = \frac{6.4 \times 10^6}{9.0 \times 10^9} \\[1em] \text C = 7.11 \times 10^{-4}\ \text F \\[1em] \text C = 711\ \mu\text F\end{aligned}$$

Hence, the capacitance of the earth is 711 µF.

Given,

• Area of each plate, A = 600 cm² = 6 × 10^-2 m²
• Separation between the plates, d = 2.0 mm = 2.0 × 10^-3 m
• Potential difference, V = 200 V
• ε₀ = 8.85 × 10^-12 C² N^-1 m^-2

(i) The field between the plates is uniform, so

$$\begin{aligned}\text E = \frac{\text V}{\text d} \\[1em] = \frac{200}{2.0 \times 10^{-3}} \\[1em] \text E = 1 \times 10^5\ \text V\ \text m^{-1}\end{aligned}$$

(ii) The field between the plates in terms of the surface charge density is E = σ/ε₀, so

$$\begin{aligned}\sigma = \epsilon_0 \text E \\[1em] \sigma = (8.85 \times 10^{-12})(1 \times 10^5) \\[1em] \sigma = 8.85 \times 10^{-7}\ \text C\ \text m^{-2}\end{aligned}$$

Hence, E = 10⁵ V m^-1 and σ = 8.85 × 10^-7 C m^-2.

Given,

• Capacitance with air, C
• Dielectric constant of the sheet, K = 5
• The battery of emf E remains connected in both cases

Potential differences : In both the cases the battery remains connected across the plates, so the potential difference is fixed by the battery,

$\text V_1 = \text V_2 = \text E$

Charges : In case (a) the capacitance is C, hence

$\text Q_1 = \text{CE}$

In case (b) the dielectric fills the whole space, so the capacitance becomes KC and

$$\begin{aligned}\text Q_2 = (\text{KC})\text E \\[1em] \text Q_2 = 5\thinspace\text{CE}\end{aligned}$$

Hence, V₁ = V₂ = E, while Q₁ = CE and Q₂ = 5CE.

Given,

• Capacitance, C = 900 pF = 900 × 10^-12 F
• Potential difference, V = 100 V

The energy stored in a capacitor is

$\text U = \frac{1}{2}\text{CV}^2$

Substituting the values,

$$\begin{aligned}\text U = \frac{1}{2}(900 \times 10^{-12})(100)^2 \\[1em] \text U = \frac{1}{2}(900 \times 10^{-12})(10^4) \\[1em] \text U = 4.5 \times 10^{-6}\ \text J\end{aligned}$$

Hence, the energy stored by the capacitor is 4.5 × 10^-6 J.

While a capacitor is being charged, work has to be done in carrying charge from one plate to the other against the potential difference that already exists between them. This work is stored as the electric potential energy of the capacitor.

Let q be the charge on the capacitor at some instant, so that the potential difference between the plates is q/C. The work done in transferring a further small charge dq is

$\text{dW} = \left(\frac{\text q}{\text C}\right)\text{dq}$

Integrating from q = 0 to the final charge Q,

$$\begin{aligned}\text W = \int_0^{\text Q}\frac{\text q}{\text C}\thinspace\text{dq} \\[1em] = \frac{1}{\text C}\left[\frac{\text q^2}{2}\right]_0^{\text Q} \\[1em] \text W = \frac{\text Q^2}{2\text C}\end{aligned}$$

This work is stored as the energy of the capacitor. Using Q = CV,

$$\begin{aligned}\text U = \frac{\text Q^2}{2\text C} \\[1em] = \frac{1}{2}\text{CV}^2 \\[1em] \text U = \frac{1}{2}\text{QV}\end{aligned}$$

Consider a parallel-plate capacitor whose plates each have area A and are held a distance d apart, with vacuum between them. Let the plates carry charges +Q and −Q, so that the surface density of charge is

$\sigma = \frac{\text Q}{\text A}$

The electric field in the space between the plates is uniform and is given by

$$\begin{aligned}\text E = \frac{\sigma}{\epsilon_0} \\[1em] \text E = \frac{\text Q}{\epsilon_0 \text A}\end{aligned}$$

Since the field is uniform, the potential difference between the plates is

$$\begin{aligned}\text V = \text{Ed} \\[1em] \text V = \frac{\text{Qd}}{\epsilon_0 \text A}\end{aligned}$$

By definition the capacitance is

$$\begin{aligned}\text C = \frac{\text Q}{\text V} \\[1em] \text C = \frac{\text Q\thinspace\epsilon_0 \text A}{\text{Qd}} \\[1em] \text C = \frac{\epsilon_0 \text A}{\text d}\end{aligned}$$

Thus the capacitance is directly proportional to the area of the plates and inversely proportional to the distance between them.

Given,

• C₁ = C₂ = C₃ = C₄ = 10 µF
• C₅ = 5 µF

The network is a Wheatstone-type bridge in which C₅ forms the bridge arm. The balance condition is

$$\begin{aligned}\frac{\text C_1}{\text C_2} = \frac{\text C_3}{\text C_4} \\[1em] \frac{10}{10} = \frac{10}{10}\end{aligned}$$

The bridge is balanced, so the two ends of C₅ are at the same potential and it carries no charge. It may therefore be removed.

The remaining arms give two series branches,

$$\begin{aligned}\text C' = \frac{\text C_1 \text C_2}{\text C_1 + \text C_2} \\[1em] = \frac{10 \times 10}{20} \\[1em] \text C' = 5\ \mu\text F \\[1em] \text C'' = \frac{\text C_3 \text C_4}{\text C_3 + \text C_4} \\[1em] = \frac{10 \times 10}{20} \\[1em] \text C'' = 5\ \mu\text F\end{aligned}$$

These two branches are in parallel between A and B,

$$\begin{aligned}\text C_\text{AB} = 5 + 5 \\[1em] \text C_\text{AB} = 10\ \mu\text F\end{aligned}$$

Hence, the equivalent capacitance between A and B is 10 µF.

Note: The answer is printed in the textbook as "10 mF". The correct unit is the microfarad, since all the given capacitances are in microfarads.

Given,

• C₁ = 8 µF, C₂ = 16 µF, C₃ = 24 µF, joined in series
• Applied potential difference, V = 110 V

For a series combination,

$$\begin{aligned}\frac{1}{\text C_\text{eq}} = \frac{1}{8} + \frac{1}{16} + \frac{1}{24} \\[1em] = \frac{6 + 3 + 2}{48} \\[1em] \frac{1}{\text C_\text{eq}} = \frac{11}{48} \\[1em] \text C_\text{eq} = \frac{48}{11}\ \mu\text F\end{aligned}$$

In a series combination every capacitor carries the same charge, which is the charge drawn from the source,

$$\begin{aligned}\text Q_1 = \text C_\text{eq}\text V \\[1em] \text Q_1 = \left(\frac{48}{11} \times 10^{-6}\right)(110) \\[1em] \text Q_1 = 480 \times 10^{-6}\ \text C \\[1em] \text Q_1 = 4.8 \times 10^{-4}\ \text C\end{aligned}$$

Hence, the charge on the plates of C₁ is 4.8 × 10^-4 C.

Given,

• Capacitance with air, C₀ = 16 µF
• Initial potential difference, V₀ = 100 V
• Dielectric constant, K = 5

(i) When the dielectric fills the space between the plates completely,

$$\begin{aligned}\text C = \text{KC}_0 \\[1em] = 5 \times 16 \\[1em] \text C = 80\ \mu\text F\end{aligned}$$

(ii) The capacitor is isolated, so the charge on the plates remains unchanged,

$$\begin{aligned}\text Q = \text C_0 \text V_0 \\[1em] = (16)(100) \\[1em] \text Q = 1600\ \mu\text C\end{aligned}$$

Hence the new potential difference is

$$\begin{aligned}\text V = \frac{\text Q}{\text C} \\[1em] = \frac{1600}{80} \\[1em] \text V = 20\ \text V\end{aligned}$$

Hence, the new capacitance is 80 µF and the new potential difference is 20 V.

Given,

• Three capacitors each of capacitance C

For the series combination,

$$\begin{aligned}\frac{1}{\text C_s} = \frac{1}{\text C} + \frac{1}{\text C} + \frac{1}{\text C} \\[1em] \frac{1}{\text C_s} = \frac{3}{\text C} \\[1em] \text C_s = \frac{\text C}{3}\end{aligned}$$

For the parallel combination,

$$\begin{aligned}\text C_p = \text C + \text C + \text C \\[1em] \text C_p = 3\text C\end{aligned}$$

Therefore

$$\begin{aligned}\frac{\text C_p}{\text C_s} = \frac{3\text C}{\text C/3} \\[1em] \frac{\text C_p}{\text C_s} = 9\end{aligned}$$

Hence, C_p : C_s = 9 : 1.

Given,

• C₁ = 6 µF, C₂ = 12 µF, C₃ = 20 µF
• EMF of the battery, V = 100 V

From the figure, C₁ and C₂ are in series and this branch is in parallel with C₃ across the battery.

(i) For the series branch,

$$\begin{aligned}\text C' = \frac{\text C_1 \text C_2}{\text C_1 + \text C_2} \\[1em] = \frac{6 \times 12}{6 + 12} \\[1em] \text C' = 4\ \mu\text F\end{aligned}$$

The full 100 V acts across this branch, and in series both capacitors carry the same charge,

$$\begin{aligned}\text Q_1 = \text C'\text V \\[1em] = (4)(100) \\[1em] \text Q_1 = 400\ \mu\text C\end{aligned}$$

(ii) The capacitor C₃ is directly across the battery, so it has 100 V across it,

$$\begin{aligned}\text U_3 = \frac{1}{2}\text C_3 \text V^2 \\[1em] \text U_3 = \frac{1}{2}(20 \times 10^{-6})(100)^2 \\[1em] \text U_3 = 0.1\ \text J\end{aligned}$$

Hence, the charge on each plate of C₁ is 400 µC and the energy stored in C₃ is 0.1 J.

Given,

• Capacitance with air, C₀ = 5 µF
• Capacitance with the dielectric, C = 50 µF

The dielectric constant is the ratio of the two capacitances,

$$\begin{aligned}\text K = \frac{\text C}{\text C_0} \\[1em] \text K = \frac{50}{5} \\[1em] \text K = 10\end{aligned}$$

Hence, the dielectric constant of the medium is 10.

Given,

• Capacitance of each capacitor = 8 µF
• Required resultant capacitance = 20 µF

Two capacitors of 8 µF joined in series give

$\frac{8 \times 8}{8 + 8} = 4\ \mu\text F$

If this series pair is joined in parallel with two more single capacitors of 8 µF each, the resultant capacitance is

$$\begin{aligned}\text C = 8 + 8 + 4 \\[1em] \text C = 20\ \mu\text F\end{aligned}$$

Hence, a minimum of 4 capacitors is required — two of them joined in series, and this pair connected in parallel with the other two.

In a parallel combination, one plate of each capacitor is connected to the point A and the other plate to the point B. Hence each capacitor has the same potential difference V across it, while the charges on them are different.

The charges on the three capacitors are

$\text Q_1 = \text C_1 \text V, \qquad \text Q_2 = \text C_2 \text V, \qquad \text Q_3 = \text C_3 \text V$

The total charge drawn from the source is

$$\begin{aligned}\text Q = \text Q_1 + \text Q_2 + \text Q_3 \\[1em] \text Q = (\text C_1 + \text C_2 + \text C_3)\text V\end{aligned}$$

If a single capacitor of capacitance C between A and B takes the same charge Q at the same potential difference V, then Q = CV. Comparing the two expressions,

$\text C = \text C_1 + \text C_2 + \text C_3$

Thus in a parallel combination the equivalent capacitance is the sum of the individual capacitances.

Given,

• C₁ = 3 µF, C₂ = 6 µF, C₃ = 10 µF
• EMF of the battery, V = 10 V

(i) From the figure, C₁ and C₂ are in series and this branch is in parallel with C₃.

$$\begin{aligned}\text C' = \frac{\text C_1 \text C_2}{\text C_1 + \text C_2} \\[1em] = \frac{3 \times 6}{3 + 6} \\[1em] \text C' = 2\ \mu\text F \\[1em] \text C_\text{eq} = \text C' + \text C_3 \\[1em] = 2 + 10 \\[1em] \text C_\text{eq} = 12\ \mu\text F\end{aligned}$$

(ii) The energy stored is

$$\begin{aligned}\text U = \frac{1}{2}\text C_\text{eq}\text V^2 \\[1em] \text U = \frac{1}{2}(12 \times 10^{-6})(10)^2 \\[1em] \text U = 6 \times 10^{-4}\ \text J\end{aligned}$$

Hence, the equivalent capacitance is 12 µF and the energy stored is 6 × 10^-4 J.

The battery has been disconnected, so the charge Q on the plates remains unchanged.

(i) The capacitance becomes K times its original value,

$\text C' = \text{KC}$

(ii) Since Q is constant and V = Q/C,

$$\begin{aligned}\text V' = \frac{\text Q}{\text{KC}} \\[1em] \text V' = \frac{\text V}{\text K}\end{aligned}$$

(iii) The energy stored is U = Q²/2C, so

$$\begin{aligned}\text U' = \frac{\text Q^2}{2(\text{KC})} \\[1em] \text U' = \frac{\text U}{\text K}\end{aligned}$$

Hence, C' = KC, V' = V/K and U' = U/K. As K > 1, the capacitance increases while the potential difference and the stored energy both decrease.

In a series combination the capacitors are joined end to end. On giving a charge +Q to the first plate, equal and opposite charges appear on the plates of each capacitor by induction, so every capacitor carries the same charge Q, while the potential differences across them are different.

The potential differences are

$\text V_1 = \frac{\text Q}{\text C_1}, \qquad \text V_2 = \frac{\text Q}{\text C_2}, \qquad \text V_3 = \frac{\text Q}{\text C_3}$

If V is the total potential difference between the end points A and B, then

$$\begin{aligned}\text V = \text V_1 + \text V_2 + \text V_3 \\[1em] \text V = \text Q\left(\frac{1}{\text C_1} + \frac{1}{\text C_2} + \frac{1}{\text C_3}\right)\end{aligned}$$

For a single equivalent capacitor C carrying the same charge Q at the same potential difference V, we have V = Q/C. Comparing,

$\frac{1}{\text C} = \frac{1}{\text C_1} + \frac{1}{\text C_2} + \frac{1}{\text C_3}$

Thus in a series combination the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances.

Given,

• C₁ = 6 µF, C₂ = 3 µF, C₃ = 8 µF
• EMF of the battery, V = 50 V

(a) From the figure, C₁ and C₂ are in series between X and Y, and C₃ is in parallel with this branch.

$$\begin{aligned}\text C' = \frac{\text C_1 \text C_2}{\text C_1 + \text C_2} \\[1em] = \frac{6 \times 3}{6 + 3} \\[1em] \text C' = 2\ \mu\text F \\[1em] \text C_\text{XY} = \text C' + \text C_3 \\[1em] = 2 + 8 \\[1em] \text C_\text{XY} = 10\ \mu\text F\end{aligned}$$

(b) The full 50 V acts across the series branch, and both capacitors of that branch carry the same charge,

$$\begin{aligned}\text q = \text C'\text V \\[1em] = (2)(50) \\[1em] \text q = 100\ \mu\text C\end{aligned}$$

Hence, the equivalent capacitance between X and Y is 10 µF and the charge on C₁ is 100 µC.

1 : 100

Explanation:
The minimum capacitance is obtained when all ten are joined in series,

$$\begin{aligned}\text C_\text{min} = \frac{5}{10} \\[1em] \text C_\text{min} = 0.5\ \mu\text F\end{aligned}$$

and the maximum when all ten are joined in parallel,

$$\begin{aligned}\text C_\text{max} = 10 \times 5 \\[1em] \text C_\text{max} = 50\ \mu\text F\end{aligned}$$

Hence

$$\begin{aligned}\frac{\text C_\text{min}}{\text C_\text{max}} = \frac{0.5}{50} \\[1em] \frac{\text C_\text{min}}{\text C_\text{max}} = \frac{1}{100}\end{aligned}$$

potential difference between the two plates

Explanation:
The capacitance of a parallel-plate capacitor is

$\text C = \frac{\text K \epsilon_0 \text A}{\text d}$

which depends only on the plate area A, the separation d and the dielectric constant K of the medium. On changing the potential difference the charge changes in the same proportion, so the ratio Q/V, that is C, is unaltered.

Given,

• C₁ = 50 µF, C₂ = 50 µF, C₃ = 25 µF

From the figure, C₁ and C₂ are joined in parallel, and this combination is in series with C₃.

For the parallel part,

$$\begin{aligned}\text C' = \text C_1 + \text C_2 \\[1em] = 50 + 50 \\[1em] \text C' = 100\ \mu\text F\end{aligned}$$

For the series combination with C₃,

$$\begin{aligned}\text C_\text{AB} = \frac{\text C'\text C_3}{\text C' + \text C_3} \\[1em] = \frac{100 \times 25}{100 + 25} \\[1em] \text C_\text{AB} = \frac{2500}{125} \\[1em] \text C_\text{AB} = 20\ \mu\text F\end{aligned}$$

Hence, the equivalent capacitance of the circuit is 20 µF.

becomes 120 µF

Explanation:
When the dielectric completely fills the space between the plates, the capacitance becomes K times its value with air,

$$\begin{aligned}\text C = \text{KC}_0 \\[1em] = 6 \times 20 \\[1em] \text C = 120\ \mu\text F\end{aligned}$$

Given,

• C₁ = 8 µF, C₂ = 12 µF, C₃ = 12 µF

From the figure, C₂ and C₃ are joined in parallel, and this combination is in series with C₁.

For the parallel part,

$$\begin{aligned}\text C' = \text C_2 + \text C_3 \\[1em] = 12 + 12 \\[1em] \text C' = 24\ \mu\text F\end{aligned}$$

For the series combination with C₁,

$$\begin{aligned}\text C_\text{AB} = \frac{\text C_1 \text C'}{\text C_1 + \text C'} \\[1em] = \frac{8 \times 24}{8 + 24} \\[1em] \text C_\text{AB} = \frac{192}{32} \\[1em] \text C_\text{AB} = 6\ \mu\text F\end{aligned}$$

Hence, the equivalent capacitance between A and B is 6 µF.

32

Reason —

Given,

• Required capacitance = 2 µF, across V = 1.0 kV = 1000 V
• Each available capacitor = 1 µF, withstanding at most 300 V

To withstand 1000 V, the capacitors must be joined in series. The number in each row must be

$$\begin{aligned}\text n = \frac{1000}{300} \\[1em] \text n = 3.33\end{aligned}$$

so at least n = 4 capacitors are needed in a row. The capacitance of one row is

$\text C_\text{row} = \frac{1}{4}\ \mu\text F$

If m such rows are joined in parallel to give 2 µF,

$$\begin{aligned}\text m \times \frac{1}{4} = 2 \\[1em] \text m = 8\end{aligned}$$

Hence the minimum number of capacitors is

$$\begin{aligned}\text n \times \text m = 4 \times 8 \\[1em] \text n \times \text m = 32\end{aligned}$$

independent of the distance between the plates

Reason —

Each plate lies in the field produced by the other plate alone, which is

$$\begin{aligned}\text E = \frac{\sigma}{2\epsilon_0} \\[1em] \text E = \frac{\text Q}{2\epsilon_0 \text A}\end{aligned}$$

Hence the force on a plate carrying charge Q is

$$\begin{aligned}\text F = \text{QE} \\[1em] \text F = \frac{\text Q^2}{2\epsilon_0 \text A}\end{aligned}$$

The capacitor is isolated, so Q is fixed. This expression contains no distance term, hence the electrostatic force is independent of the separation between the plates.

1.2 nC

Reason —

Given,

• C = 90 pF, V = 20 V, K = 5/3
• The battery remains connected, so V is constant

With the dielectric inserted, the charge on the plates becomes

$$\begin{aligned}\text Q = (\text{KC})\text V \\[1em] = \left(\frac{5}{3} \times 90 \times 10^{-12}\right)(20) \\[1em] = 3 \times 10^{-9}\ \text C \\[1em] \text Q = 3\ \text{nC}\end{aligned}$$

The charge induced on the faces of the dielectric is

$\text Q_\text{ind} = \text Q\left(1 - \frac{1}{\text K}\right)$

Substituting the values,

$$\begin{aligned}\text Q_\text{ind} = 3\left(1 - \frac{3}{5}\right) \\[1em] = 3 \times \frac{2}{5} \\[1em] \text Q_\text{ind} = 1.2\ \text{nC}\end{aligned}$$

60 µA, 60 µA

Reason —

Given,

• C = 20 µF = 20 × 10^-6 F
• $\frac{\text{dV}}{\text{dt}} = 3$ V s^-1

The charge on the capacitor is q = CV. Differentiating with respect to time,

$$\begin{aligned}\frac{\text{dq}}{\text{dt}} = \text C\frac{\text{dV}}{\text{dt}} \\[1em] \text I_c = (20 \times 10^{-6})(3) \\[1em] = 60 \times 10^{-6}\ \text A \\[1em] \text I_c = 60\ \mu\text A\end{aligned}$$

The displacement current through the gap between the plates is always equal to the conduction current in the wires, so

$$\begin{aligned}\text I_d = \text I_c \\[1em] \text I_d = 60\ \mu\text A\end{aligned}$$

40 µF and 10 µF

Reason —

From the graph, the slope q/V gives the capacitance. The line A of greater slope is the parallel combination and the line B the series combination,

$$\begin{aligned}\text C_p = \frac{500}{10} \\[1em] \text C_p = 50\ \mu\text F \\[1em] \text C_s = \frac{80}{10} \\[1em] \text C_s = 8\ \mu\text F\end{aligned}$$

If the two capacitances are C₁ and C₂,

$\text C_1 + \text C_2 = 50 \qquad \text{and} \qquad \frac{\text C_1 \text C_2}{\text C_1 + \text C_2} = 8$

so that

$$\begin{aligned}\text C_1 \text C_2 = 8 \times 50 \\[1em] \text C_1 \text C_2 = 400\end{aligned}$$

Hence C₁ and C₂ are the roots of

$$\begin{aligned}\text x^2 - 50\text x + 400 = 0 \\[1em] \text x = \frac{50 \pm 30}{2} \\[1em] \text x = 40\ \text{or}\ 10\end{aligned}$$

8.5

Reason —

Given,

• V = 500 V, E_max = 10⁶ V m^-1
• A = 10^-4 m², C = 15 pF = 15 × 10^-12 F

The least permissible separation follows from E = V/d,

$$\begin{aligned}\text d = \frac{\text V}{\text E_\text{max}} \\[1em] = \frac{500}{10^6} \\[1em] \text d = 5 \times 10^{-4}\ \text m\end{aligned}$$

For a capacitor filled with a dielectric,

$$\begin{aligned}\text C = \frac{\text K \epsilon_0 \text A}{\text d} \\[1em] \text K = \frac{\text{Cd}}{\epsilon_0 \text A}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text K = \frac{(15 \times 10^{-12})(5 \times 10^{-4})}{(8.86 \times 10^{-12})(10^{-4})} \\[1em] \text K = \frac{7.5 \times 10^{-15}}{8.86 \times 10^{-16}} \\[1em] \text K \approx 8.5\end{aligned}$$

200 µC

Reason —

Given, emf = 72 V, resistances 6 Ω, 4 Ω, 2 Ω and 10 Ω, and C = 10 µF.

In the steady state no current flows through the branch containing the capacitor, so the capacitor is simply connected across the 10 Ω resistor.

The 4 Ω resistor is in parallel with the series pair (2 Ω + 10 Ω = 12 Ω),

$$\begin{aligned}\text R' = \frac{4 \times 12}{4 + 12} \\[1em] \text R' = 3\ \Omega\end{aligned}$$

The total resistance of the circuit is

$$\begin{aligned}\text R = 6 + 3 \\[1em] \text R = 9\ \Omega \\[1em] \text I = \frac{72}{9} \\[1em] \text I = 8\ \text A\end{aligned}$$

The potential difference across the parallel section is

$$\begin{aligned}\text V' = \text{IR}' \\[1em] = 8 \times 3 \\[1em] \text V' = 24\ \text V\end{aligned}$$

The current in the 12 Ω branch is

$$\begin{aligned}\text I' = \frac{24}{12} \\[1em] \text I' = 2\ \text A\end{aligned}$$

so the potential difference across the 10 Ω resistor, and hence across the capacitor, is

$$\begin{aligned}\text V_C = 2 \times 10 \\[1em] \text V_C = 20\ \text V\end{aligned}$$

Therefore

$$\begin{aligned}\text Q = \text{CV}_C \\[1em] = (10)(20) \\[1em] \text Q = 200\ \mu\text C\end{aligned}$$

3.75 × 10^-6 J

Reason —

Given,

• C₁ = 5 µF, C₂ = 2 µF, charge q = 5 µC = 5 × 10^-6 C

The charge remains constant while the plates are pulled apart, so the work done equals the increase in the stored energy,

$$\begin{aligned}\text W = \frac{\text q^2}{2\text C_2} - \frac{\text q^2}{2\text C_1} \\[1em] \text W = \frac{\text q^2}{2}\left(\frac{1}{\text C_2} - \frac{1}{\text C_1}\right)\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text W = \frac{(5 \times 10^{-6})^2}{2}\left(\frac{1}{2 \times 10^{-6}} - \frac{1}{5 \times 10^{-6}}\right) \\[1em] = \frac{25 \times 10^{-12}}{2}\left(5 \times 10^5 - 2 \times 10^5\right) \\[1em] = (12.5 \times 10^{-12})(3 \times 10^5) \\[1em] \text W = 3.75 \times 10^{-6}\ \text J\end{aligned}$$

$\frac{(\text n + 1)\text V}{(\text K + \text n)}$

Reason —

The two capacitors in parallel have a total capacitance C + nC = (n + 1)C, so the total charge stored is

$\text Q = (\text n + 1)\text{CV}$

The battery is then removed, so this charge remains constant. On introducing the dielectric in the first capacitor, its capacitance becomes KC, and the total capacitance is

$$\begin{aligned}\text C' = \text{KC} + \text{nC} \\[1em] \text C' = (\text K + \text n)\text C\end{aligned}$$

Hence the new common potential difference is

$$\begin{aligned}\text V' = \frac{\text Q}{\text C'} \\[1em] = \frac{(\text n + 1)\text{CV}}{(\text K + \text n)\text C} \\[1em] \text V' = \frac{(\text n + 1)\text V}{(\text K + \text n)}\end{aligned}$$

$\text {2π}\sqrt{\frac{\text L}{\sqrt{\text g^2 + \left(\frac{\text {qE}}{\text m}\right)^2}}}$

Reason —

Given,

• Length of the pendulum = L
• Mass of the bob = m, charge on the bob = q
• Electric field between the plates = E, directed horizontally

Forces acting on the bob : the plates of the capacitor are vertical, so the field E between them is horizontal. The bob therefore experiences

(i) its weight mg, acting vertically downwards, and

(ii) the electric force qE, acting horizontally, along the direction of the field.

Resultant acceleration : the two forces are mutually perpendicular, so their resultant is obtained by the theorem of parallelogram of vectors,

$\text F = \sqrt{(\text{mg})^2 + (\text{qE})^2}$

The effective acceleration of the bob is this resultant force divided by the mass,

$$\begin{aligned}\text g_\text{eff} = \frac{\text F}{\text m} \\[1em] \text g_\text{eff} = \frac{\sqrt{(\text{mg})^2 + (\text{qE})^2}}{\text m}\end{aligned}$$

Taking m inside the radical,

$$\begin{aligned}\text g_\text{eff} = \sqrt{\frac{(\text{mg})^2 + (\text{qE})^2}{\text m^2}} \\[1em] \text g_\text{eff} = \sqrt{\text g^2 + \left(\frac{\text{qE}}{\text m}\right)^2}\end{aligned}$$

Time period : the bob now oscillates about the new equilibrium position, which is the line of the resultant force, and the time period of a simple pendulum in an effective gravitational field g_eff is

$\text T = 2\pi\sqrt{\frac{\text L}{\text g_\text{eff}}}$

Substituting the value of g_eff,

$\text T = 2\pi\sqrt{\frac{\text L}{\sqrt{\text g^2 + \left(\frac{\text{qE}}{\text m}\right)^2}}}$

The options in which g and qE/m are simply added or subtracted would apply only if the electric force were vertical, that is, parallel or antiparallel to gravity. Here the field is horizontal, so the two accelerations must be combined vectorially.

1 V

Reason —

Given, C = 1 µF, q₁ = + 2 µC and q₂ = + 4 µC.

When two plates carry unequal charges, only the difference of the charges is responsible for the field between them; half of this difference resides as the effective charge of the capacitor. Hence

$\text V = \frac{\text q_1 - \text q_2}{2\text C}$

Substituting the values,

$$\begin{aligned}|\text V| = \frac{|2 - 4|}{2 \times 1} \\[1em] = \frac{2}{2} \\[1em] |\text V| = 1\ \text V\end{aligned}$$

$\frac{7}{11}$ µF

Reason —

Let P be the junction just after C, and let the lower rail be B.

Left branch from P : the 2 µF vertical capacitor is in series with the parallel pair of 2 µF and 2 µF at the bottom,

$$\begin{aligned}2 + 2 = 4\ \mu\text F \\[1em] \frac{2 \times 4}{2 + 4} = \frac{4}{3}\ \mu\text F\end{aligned}$$

Right branch from P : the two 2 µF capacitors of the right-hand group are in series, giving 1 µF, and this is in parallel with the 1 µF capacitor, giving 2 µF. This is reached from P through the top 2 µF capacitor,

$\frac{2 \times 2}{2 + 2} = 1\ \mu\text F$

Between P and B : the two branches are in parallel,

$$\begin{aligned}\text C_\text{PB} = \frac{4}{3} + 1 \\[1em] \text C_\text{PB} = \frac{7}{3}\ \mu\text F\end{aligned}$$

Finally C is in series with this,

$$\begin{aligned}\frac{1}{0.5} = \frac{1}{\text C} + \frac{3}{7} \\[1em] 2 - \frac{3}{7} = \frac{1}{\text C} \\[1em] \frac{11}{7} = \frac{1}{\text C} \\[1em] \text C = \frac{7}{11}\ \mu\text F\end{aligned}$$

Zero

Reason —

The current is the rate of change of charge,

$\text I = \frac{\text{dq}}{\text{dt}}$

which is given by the slope of the q–t graph. Between t = 2 s and t = 6 s the graph is a horizontal straight line, that is, the charge stays constant at 3 µC. Hence at t = 4 s the slope is zero and so the current is zero.

508 pJ

Reason —

Given, C = 12 pF, V = 10 V and K = 6.5.

The initial energy stored is

$$\begin{aligned}\text U_i = \frac{1}{2}\text{CV}^2 \\[1em] = \frac{1}{2}(12)(10)^2 \\[1em] \text U_i = 600\ \text{pJ}\end{aligned}$$

The battery is disconnected, so the charge remains constant while the capacitance becomes KC. Hence

$$\begin{aligned}\text U_f = \frac{\text U_i}{\text K} \\[1em] = \frac{600}{6.5} \\[1em] \text U_f = 92.3\ \text{pJ}\end{aligned}$$

The work done by the capacitor in drawing the slab in is the decrease in its stored energy,

$$\begin{aligned}\text W = \text U_i - \text U_f \\[1em] \text W = 600 - 92.3 \\[1em] \text W \approx 508\ \text{pJ}\end{aligned}$$

12

Reason —

From the figure the three dielectrics occupy three equal parts of the plate area, each of area A/3, with the full separation d. They therefore behave as three capacitors joined in parallel,

$$\begin{aligned}\text C = \frac{\epsilon_0 (\text A/3)}{\text d}(\text K_1 + \text K_2 + \text K_3) \\[1em] = \frac{\epsilon_0 \text A}{3\text d}(10 + 12 + 14) \\[1em] \text C = \frac{36}{3}\frac{\epsilon_0 \text A}{\text d} \\[1em] \text C = \frac{12\thinspace\epsilon_0 \text A}{\text d}\end{aligned}$$

If a single material of dielectric constant K fully filled the capacitor, then C = Kε₀A/d. Comparing the two,

$\text K = 12$

$\frac{3}{8}\cdot\frac{\text q^2}{\text C}$

Reason —

With the switch at A, the capacitor C is charged to a potential V = q/C and carries the total charge q. On turning the switch to B, this capacitor is joined to the uncharged capacitor 3C.

The loss of energy on sharing charge between two capacitors is

$\Delta \text U = \frac{\text C_1 \text C_2}{2(\text C_1 + \text C_2)}(\text V_1 - \text V_2)^2$

Here C₁ = C, C₂ = 3C, V₁ = q/C and V₂ = 0. Substituting,

$$\begin{aligned}\Delta \text U = \frac{\text C(3\text C)}{2(4\text C)}\left(\frac{\text q}{\text C}\right)^2 \\[1em] \Delta \text U = \frac{3\text C}{8}\cdot\frac{\text q^2}{\text C^2} \\[1em] \Delta \text U = \frac{3}{8}\cdot\frac{\text q^2}{\text C}\end{aligned}$$

6 µC and 3 µC

Reason —

The total charge is conserved,

$$\begin{aligned}\text Q = 12 + (-3) \\[1em] \text Q = 9\ \mu\text C\end{aligned}$$

After joining by a wire both spheres attain a common potential, so the charges are shared in the ratio of their capacitances, that is, in the ratio of their radii,

$$\begin{aligned}\frac{\text Q_1}{\text Q_2} = \frac{2\text R/3}{\text R/3} \\[1em] \frac{\text Q_1}{\text Q_2} = \frac{2}{1}\end{aligned}$$

Hence

$$\begin{aligned}\text Q_1 = \frac{2}{3} \times 9 \\[1em] \text Q_1 = 6\ \mu\text C \\[1em] \text Q_2 = \frac{1}{3} \times 9 \\[1em] \text Q_2 = 3\ \mu\text C\end{aligned}$$

$\frac{3}{2}\text {CV}^2$

Reason —

The initial charges are

$$\begin{aligned}\text Q_1 = \text{CV} \\[1em] \text Q_2 = (2\text C)(2\text V) \\[1em] \text Q_2 = 4\text{CV}\end{aligned}$$

Since the positive plate of one is joined to the negative plate of the other, the charges partly neutralise, so the net charge is

$$\begin{aligned}\text Q = 4\text{CV} - \text{CV} \\[1em] \text Q = 3\text{CV}\end{aligned}$$

The equivalent capacitance of the parallel combination is

$$\begin{aligned}\text C_\text{eq} = \text C + 2\text C \\[1em] \text C_\text{eq} = 3\text C\end{aligned}$$

Hence the final energy is

$$\begin{aligned}\text U = \frac{\text Q^2}{2\text C_\text{eq}} \\[1em] = \frac{(3\text{CV})^2}{2(3\text C)} \\[1em] = \frac{9\text C^2\text V^2}{6\text C} \\[1em] \text U = \frac{3}{2}\text{CV}^2\end{aligned}$$

8 : 5

Reason —

The original capacitance is C₀ = ε₀A/d. With a slab of thickness t = d/2 and dielectric constant K = 4 inserted,

$$\begin{aligned}\text C = \frac{\epsilon_0 \text A}{\text d - \text t + \frac{\text t}{\text K}} \\[1em] \text C = \frac{\epsilon_0 \text A}{\text d - \frac{\text d}{2} + \frac{\text d}{8}}\end{aligned}$$

Simplifying the denominator,

$$\begin{aligned}\text d - \frac{\text d}{2} + \frac{\text d}{8} = \frac{8\text d - 4\text d + \text d}{8} \\[1em] \text d - \frac{\text d}{2} + \frac{\text d}{8} = \frac{5\text d}{8} \\[1em] \text C = \frac{\epsilon_0 \text A}{\frac{5\text d}{8}} \\[1em] \text C = \frac{8}{5}\frac{\epsilon_0 \text A}{\text d}\end{aligned}$$

Hence

$\frac{\text C}{\text C_0} = \frac{8}{5}$

$\frac{1}{6}\text {CV}_0^2$

Reason —

The loss of energy when two capacitors at potentials V₁ and V₂ are joined in parallel is

$\Delta \text U = \frac{\text C_1 \text C_2}{2(\text C_1 + \text C_2)}(\text V_1 - \text V_2)^2$

Here C₁ = C, C₂ = C/2, V₁ = V₀ and V₂ = 0. Substituting,

$$\begin{aligned}\Delta \text U = \frac{\text C\left(\frac{\text C}{2}\right)}{2\left(\text C + \frac{\text C}{2}\right)}\text V_0^2 \\[1em] \Delta \text U = \frac{\frac{\text C^2}{2}}{2\left(\frac{3\text C}{2}\right)}\text V_0^2 \\[1em] \Delta \text U = \frac{\text C^2/2}{3\text C}\text V_0^2 \\[1em] \Delta \text U = \frac{1}{6}\text{CV}_0^2\end{aligned}$$

$\frac{l}{3}$

Reason —

The battery remains connected, so V is constant and U = ½CV² is proportional to the capacitance. For the energy to be doubled, the capacitance must be doubled.

Let a length x of the slab be inside the plates. The filled and the empty portions form two capacitors in parallel,

$$\begin{aligned}\text C = \frac{\epsilon_0 \text w(l - \text x)}{\text d} + \frac{\text k\thinspace\epsilon_0 \text w\thinspace\text x}{\text d} \\[1em] = \frac{\epsilon_0 \text w}{\text d}\left[(l - \text x) + 4\text x\right] \\[1em] \text C = \frac{\epsilon_0 \text w(l + 3\text x)}{\text d}\end{aligned}$$

The initial capacitance is C₀ = ε₀wl/d. Setting C = 2C₀,

$$\begin{aligned}l + 3\text x = 2l \\[1em] \text x = \frac{l}{3}\end{aligned}$$

16.36 µC

Reason —

Let the potential of the common lower point O be zero. Both the 6 V cells raise the potentials of the two outer top corners to + 6 V each.

Let V be the potential of the junction to which the 2 µF, the 4 µF and the 5 µF capacitors are joined. Since this junction is isolated, the total charge on the plates meeting there is zero,

$$\begin{aligned}2(\text V - 6) + 4(\text V - 6) + 5(\text V - 0) = 0 \\[1em] 2\text V - 12 + 4\text V - 24 + 5\text V = 0 \\[1em] 11\text V = 36 \\[1em] \text V = \frac{36}{11}\ \text V\end{aligned}$$

Hence the charge on the 5 µF capacitor is

$$\begin{aligned}\text q = 5 \times \frac{36}{11} \\[1em] = \frac{180}{11} \\[1em] \text q = 16.36\ \mu\text C\end{aligned}$$

having a permanent electric dipole moment

Reason —

In a polar molecule the centre of gravity of the positive charges does not coincide with that of the negative charges, so the molecule behaves as an electric dipole even in the absence of any external field, that is, it has a permanent electric dipole moment. Water and HCl are examples. A non-polar molecule, on the other hand, acquires a dipole moment only when an external field displaces its charges.

$\frac{\text R_2}{\text R_1}$

Reason —

On connecting by a wire, both spheres come to the same potential V,

$$\begin{aligned}\frac{\text Q_1}{4\pi\epsilon_0 \text R_1} = \frac{\text Q_2}{4\pi\epsilon_0 \text R_2} \\[1em] \frac{\text Q_1}{\text Q_2} = \frac{\text R_1}{\text R_2}\end{aligned}$$

In terms of the surface charge densities, Q = 4πR²σ, so

$\frac{\text Q_1}{\text Q_2} = \frac{\text R_1^2 \sigma_1}{\text R_2^2 \sigma_2}$

Equating the two expressions,

$$\begin{aligned}\frac{\text R_1^2 \sigma_1}{\text R_2^2 \sigma_2} = \frac{\text R_1}{\text R_2} \\[1em] \frac{\sigma_1}{\sigma_2} = \frac{\text R_2}{\text R_1}\end{aligned}$$

$\frac{1}{2}\epsilon_0\text E^2\text {Ad}$

Reason —

The energy stored per unit volume of the field, that is, the energy density, is

$\text u = \frac{1}{2}\epsilon_0 \text E^2$

The field exists only in the space between the plates, whose volume is Ad. Hence the total energy stored is

$$\begin{aligned}\text U = \text u \times (\text{Ad}) \\[1em] \text U = \frac{1}{2}\epsilon_0 \text E^2 \text{Ad}\end{aligned}$$

2C

Reason —

The two capacitors marked C in the upper and the lower arms are joined between the same pair of points, so they are in parallel, giving 2C.

The third capacitor C is connected across the two ends of the same right-hand junction, that is, both of its plates are joined to the same point by the connecting wires. It is therefore short-circuited and carries no charge (both sides of the vertical capacitor are at the same potential), so it does not contribute to the equivalent capacitance.

$$\begin{aligned}\text C_\text{eq} = \text C + \text C \\[1em] \text C_\text{eq} = 2\text C\end{aligned}$$

0.9 mA

Reason —

Given, ρ = 200 Ω m, C = 2 pF = 2 × 10^-12 F, V = 40 V and K = 50.

For the capacitor,

$$\begin{aligned}\text C = \frac{\text K \epsilon_0 \text A}{\text d} \\[1em] \frac{\text A}{\text d} = \frac{\text C}{\text K \epsilon_0}\end{aligned}$$

The resistance offered by the material between the plates is

$$\begin{aligned}\text R = \frac{\rho \text d}{\text A} \\[1em] \text R = \frac{\rho\thinspace\text K \epsilon_0}{\text C}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text R = \frac{(200)(50)(8.85 \times 10^{-12})}{2 \times 10^{-12}} \\[1em] = \frac{8.85 \times 10^{-8}}{2 \times 10^{-12}} \\[1em] \text R = 4.425 \times 10^4\ \Omega\end{aligned}$$

Hence the leakage current is

$$\begin{aligned}\text I = \frac{\text V}{\text R} \\[1em] = \frac{40}{4.425 \times 10^4} \\[1em] \text I = 9.04 \times 10^{-4}\ \text A \\[1em] \text I \approx 0.9\ \text{mA}\end{aligned}$$

$\frac{\epsilon_0\text A}{\text d}\left(\frac{1}{2} + \frac{\text K_1\text K_2}{\text K_1 + \text K_2}\right)$

Reason —

The arrangement is equivalent to two capacitors in parallel, each of plate area A/2.

Left half (air throughout the gap d) :

$$\begin{aligned}\text C_1 = \frac{\epsilon_0 (\text A/2)}{\text d} \\[1em] \text C_1 = \frac{\epsilon_0 \text A}{2\text d}\end{aligned}$$

Right half : here the two slabs, each of thickness d/2, lie one above the other, so they form two capacitors in series,

$$\begin{aligned}\text C' = \frac{\text K_1 \epsilon_0 (\text A/2)}{\text d/2} \\[1em] \text C' = \frac{\text K_1 \epsilon_0 \text A}{\text d} \\[1em] \text C'' = \frac{\text K_2 \epsilon_0 \text A}{\text d} \\[1em] \text C_2 = \frac{\text C'\text C''}{\text C' + \text C''} \\[1em] \text C_2 = \frac{\epsilon_0 \text A}{\text d}\left(\frac{\text K_1 \text K_2}{\text K_1 + \text K_2}\right)\end{aligned}$$

Adding the two parallel parts,

$$\begin{aligned}\text C = \text C_1 + \text C_2 \\[1em] \text C = \frac{\epsilon_0 \text A}{\text d}\left(\frac{1}{2} + \frac{\text K_1 \text K_2}{\text K_1 + \text K_2}\right)\end{aligned}$$

1 : 2 : 2

Reason —

From the figure, C₂ and C₃ are joined in series between A and B, and C₁ is connected directly between A and B. The battery of emf V is applied across A and B.

Charge on C₁ : it has the full potential difference V,

$$\begin{aligned}\text q_1 = \text C_1 \text V \\[1em] \text q_1 = 2\text V\end{aligned}$$

Charges on C₂ and C₃ : their series capacitance is

$$\begin{aligned}\text C' = \frac{6 \times 12}{6 + 12} \\[1em] \text C' = 4\ \mu\text F\end{aligned}$$

and being in series they carry equal charges,

$$\begin{aligned}\text q_2 = \text q_3 = \text C'\text V \\[1em] \text q_2 = \text q_3 = 4\text V\end{aligned}$$

Hence

$$\begin{aligned}\text q_1 : \text q_2 : \text q_3 = 2\text V : 4\text V : 4\text V \\[1em] \text q_1 : \text q_2 : \text q_3 = 1 : 2 : 2\end{aligned}$$

0.95

Reason —

Given, V₀ = 20 V, R = 10 Ω, V = 2 V at t = 1 µs = 10^-6 s.

During charging through a resistance, the potential difference across the capacitor grows as

$\text V = \text V_0\left(1 - \text e^{-\text t/\text{RC}}\right)$

Substituting the values,

$$\begin{aligned}2 = 20\left(1 - \text e^{-\text t/\text{RC}}\right) \\[1em] \text e^{-\text t/\text{RC}} = 1 - 0.1 \\[1em] \text e^{-\text t/\text{RC}} = 0.9\end{aligned}$$

Taking the natural logarithm of both sides,

$$\begin{aligned}\frac{\text t}{\text{RC}} = \ln\left(\frac{1}{0.9}\right) \\[1em] = \ln\left(\frac{10}{9}\right) \\[1em] \frac{\text t}{\text{RC}} = 0.105 \\[1em] \text{RC} = \frac{10^{-6}}{0.105} \\[1em] \text{RC} = 9.52 \times 10^{-6}\ \text s\end{aligned}$$

Hence

$$\begin{aligned}\text C = \frac{9.52 \times 10^{-6}}{10} \\[1em] \text C = 0.952 \times 10^{-6}\ \text F \\[1em] \text C \approx 0.95\ \mu\text F\end{aligned}$$

$\frac{\text {kA}}{2\ln\left(\frac{2\epsilon_0 + \text {kd}}{2\epsilon_0}\right)}$

Reason —

Consider a thin layer of thickness dx at a distance x from one plate. It behaves as an elementary capacitor of capacitance

$\text{dC} = \frac{\epsilon(\text x)\text A}{\text{dx}}$

All such layers are in series, so the reciprocals add,

$\frac{1}{\text C_1} = \int_0^{\text d/2}\frac{\text{dx}}{\text A(\epsilon_0 + \text{kx})}$

Putting u = ε₀ + kx, so that du = k dx,

$$\begin{aligned}\frac{1}{\text C_1} = \frac{1}{\text{kA}}\left[\ln(\epsilon_0 + \text{kx})\right]_0^{\text d/2} \\[1em] \frac{1}{\text C_1} = \frac{1}{\text{kA}}\ln\left(\frac{\epsilon_0 + \frac{\text{kd}}{2}}{\epsilon_0}\right) \\[1em] \frac{1}{\text C_1} = \frac{1}{\text{kA}}\ln\left(\frac{2\epsilon_0 + \text{kd}}{2\epsilon_0}\right)\end{aligned}$$

By the symmetry of the given permittivity about x = d/2, the second half gives an equal contribution, so that

$\frac{1}{\text C_2} = \frac{1}{\text C_1}$

The two halves are themselves in series, hence

$$\begin{aligned}\frac{1}{\text C} = \frac{1}{\text C_1} + \frac{1}{\text C_2} \\[1em] \frac{1}{\text C} = \frac{2}{\text{kA}}\ln\left(\frac{2\epsilon_0 + \text{kd}}{2\epsilon_0}\right) \\[1em] \text C = \frac{\text{kA}}{2\ln\left(\frac{2\epsilon_0 + \text{kd}}{2\epsilon_0}\right)}\end{aligned}$$

$\frac{3\text V}{\text K + 2}$

Reason —

In parallel the total capacitance is 2C + C = 3C, so the total charge stored is

$\text Q = 3\text{CV}$

The battery is then removed, so this charge is conserved. On filling the capacitor of capacity C with the dielectric, its capacitance becomes KC, and the total capacitance becomes

$$\begin{aligned}\text C' = 2\text C + \text{KC} \\[1em] \text C' = (\text K + 2)\text C\end{aligned}$$

Hence the new common potential difference is

$$\begin{aligned}\text V' = \frac{\text Q}{\text C'} \\[1em] = \frac{3\text{CV}}{(\text K + 2)\text C} \\[1em] \text V' = \frac{3\text V}{\text K + 2}\end{aligned}$$

$\tan^{-1}\left[\frac{\text q}{\text {mg}} \times \frac{\text C_2(\text V_1 + \text V_2)}{(\text C_1 + \text C_2)(\text d - \text t)}\right]$

Reason —

The plates are at potentials + V₂ and − V₁, so the total potential difference between them is V₁ + V₂.

The gap of width d contains a dielectric slab of thickness t and an air region of width (d − t). These two portions form two capacitors in series; let C₁ be the capacitance of the air portion, in which the bob hangs, and C₂ that of the dielectric portion. In a series combination the potential difference divides in the inverse ratio of the capacitances, so the part appearing across the air region is

$\text V_\text{air} = (\text V_1 + \text V_2)\frac{\text C_2}{\text C_1 + \text C_2}$

The field acting on the bob is therefore

$$\begin{aligned}\text E = \frac{\text V_\text{air}}{\text d - \text t} \\[1em] \text E = \frac{\text C_2(\text V_1 + \text V_2)}{(\text C_1 + \text C_2)(\text d - \text t)}\end{aligned}$$

In the equilibrium position the string makes an angle θ with the vertical such that the horizontal electric force balances the horizontal component of the tension,

$$\begin{aligned}\tan\theta = \frac{\text{qE}}{\text{mg}} \\[1em] \tan\theta = \frac{\text q}{\text{mg}} \times \frac{\text C_2(\text V_1 + \text V_2)}{(\text C_1 + \text C_2)(\text d - \text t)}\end{aligned}$$

Note: The figure does not label which of C₁ and C₂ belongs to which region. The printed answer corresponds to taking C₁ as the capacitance of the air portion and C₂ as that of the dielectric portion.

$\text C' = \frac{4\text K}{\text K + 3}\text C_0$

Reason —

With a slab of thickness t and dielectric constant K inserted,

$\text C' = \frac{\epsilon_0 \text A}{\text d - \text t + \frac{\text t}{\text K}}$

Putting t = 3d/4,

$$\begin{aligned}\text C' = \frac{\epsilon_0 \text A}{\text d - \frac{3\text d}{4} + \frac{3\text d}{4\text K}} \\[1em] = \frac{\epsilon_0 \text A}{\text d\left(\frac{1}{4} + \frac{3}{4\text K}\right)} \\[1em] = \frac{\epsilon_0 \text A}{\text d}\cdot\frac{4\text K}{\text K + 3} \\[1em] \text C' = \frac{4\text K}{\text K + 3}\text C_0\end{aligned}$$

$\frac{\cos^2\beta}{\cos^2\alpha}$

Reason —

The electric field between the plates is perpendicular to the plates, so the electron experiences a force only in that direction. The component of its velocity parallel to the plates therefore remains unchanged.

If v₁ and v₂ are the speeds at entry and exit, then

$$\begin{aligned}\text v_1\cos\alpha = \text v_2\cos\beta \\[1em] \frac{\text v_1}{\text v_2} = \frac{\cos\beta}{\cos\alpha}\end{aligned}$$

Since the kinetic energy is proportional to the square of the speed,

$$\begin{aligned}\frac{\text K_1}{\text K_2} = \left(\frac{\text v_1}{\text v_2}\right)^2 \\[1em] \frac{\text K_1}{\text K_2} = \frac{\cos^2\beta}{\cos^2\alpha}\end{aligned}$$

2.33

Reason —

Given,

• A = 30π cm² = 30π × 10^-4 m², d = 1 mm = 10^-3 m
• Dielectric strength, E_max = 3.6 × 10⁷ V m^-1
• Q_max = 7 × 10^-6 C

The maximum permissible potential difference is

$$\begin{aligned}\text V_\text{max} = \text E_\text{max}\thinspace\text d \\[1em] = (3.6 \times 10^7)(10^{-3}) \\[1em] \text V_\text{max} = 3.6 \times 10^4\ \text V\end{aligned}$$

Hence the capacitance is

$$\begin{aligned}\text C = \frac{\text Q_\text{max}}{\text V_\text{max}} \\[1em] = \frac{7 \times 10^{-6}}{3.6 \times 10^4} \\[1em] \text C = 1.944 \times 10^{-10}\ \text F\end{aligned}$$

Now C = Kε₀A/d, so

$\text K = \frac{\text{Cd}}{\epsilon_0 \text A}$

With ε₀ = 8.85 × 10^-12 C² N^-1 m^-2,

$$\begin{aligned}\epsilon_0 \text A = (8.85 \times 10^{-12})(30\pi \times 10^{-4}) \\[1em] \epsilon_0 \text A = 8.34 \times 10^{-14} \\[1em] \text K = \frac{(1.944 \times 10^{-10})(10^{-3})}{8.34 \times 10^{-14}} \\[1em] \text K = 2.33\end{aligned}$$

10

Reason —

The energy stored is U = q²/2C, so for a given capacitor U ∝ q².

Let the original charge be q. When it becomes q + 2, the energy becomes 1.44 times its original value,

$\frac{(\text q + 2)^2}{\text q^2} = 1.44$

Taking the square root of both sides,

$$\begin{aligned}\frac{\text q + 2}{\text q} = 1.2 \\[1em] \text q + 2 = 1.2\thinspace\text q \\[1em] 0.2\thinspace\text q = 2 \\[1em] \text q = 10\ \text C\end{aligned}$$

2 : 1

Reason —

For a slab of thickness t and dielectric constant K,

$\text C = \frac{\epsilon_0 \text A}{\text d - \text t + \frac{\text t}{\text K}}$

For a metal sheet K = ∞, so t/K = 0. With t = d/2,

$$\begin{aligned}\text C = \frac{\epsilon_0 \text A}{\text d - \frac{\text d}{2}} \\[1em] = \frac{2\epsilon_0 \text A}{\text d} \\[1em] \text C = 2\text C_0\end{aligned}$$

Hence the new capacitance is twice the original one, that is, 2 : 1.

$\frac{\text C_1\text C_2}{(\text C_1 + \text C_2)}\text V$

Reason —

The initial charge on C₁ is Q = C₁V. The battery is removed, so on closing the switch this charge is merely shared between the two capacitors, which are then in parallel. The common potential difference is

$$\begin{aligned}\text V' = \frac{\text Q}{\text C_1 + \text C_2} \\[1em] \text V' = \frac{\text C_1 \text V}{\text C_1 + \text C_2}\end{aligned}$$

Hence the charge on C₂ is

$$\begin{aligned}\text Q_2 = \text C_2 \text V' \\[1em] \text Q_2 = \frac{\text C_1 \text C_2}{(\text C_1 + \text C_2)}\text V\end{aligned}$$

Graph (a)

Reason —

For a capacitor of fixed capacitance,

$\text V = \frac{\text Q}{\text C}$

so V is directly proportional to Q, and the graph must be a straight line passing through the origin. When Q reaches its final value of 5 C,

$$\begin{aligned}\text V = \frac{5}{2 \times 10^{-6}} \\[1em] \text V = 2.5 \times 10^6\ \text V\end{aligned}$$

The only graph which is a straight line through the origin ending at 2.5 × 10⁶ V for Q = 5 C is graph (a).

Note: The answer printed in the textbook is (d), which shows a curved graph saturating at 5 × 10⁶ V. This contradicts the relation V = Q/C, which requires a straight line reaching 2.5 × 10⁶ V. The correct option is (a).

$\frac{\text n}{\text n - 1}$

Reason —

The capacitance of the isolated sphere is

$\text C_0 = 4\pi\epsilon_0 \text R_1$

When it is enclosed by a concentric earthed sphere of radius R₂, the arrangement becomes a spherical capacitor of capacitance

$\text C = 4\pi\epsilon_0 \frac{\text R_1 \text R_2}{\text R_2 - \text R_1}$

It is given that C = nC₀, so

$$\begin{aligned}4\pi\epsilon_0 \frac{\text R_1 \text R_2}{\text R_2 - \text R_1} = \text n(4\pi\epsilon_0 \text R_1) \\[1em] \frac{\text R_2}{\text R_2 - \text R_1} = \text n \\[1em] \text R_2 = \text n\text R_2 - \text n\text R_1 \\[1em] \text n\text R_1 = (\text n - 1)\text R_2 \\[1em] \frac{\text R_2}{\text R_1} = \frac{\text n}{\text n - 1}\end{aligned}$$

$\frac{5}{13}$

Reason —

With the key K closed, both identical capacitors of capacitance C are in parallel across the source, so

$$\begin{aligned}\text E_1 = 2\left(\frac{1}{2}\text{CV}^2\right) \\[1em] \text E_1 = \text{CV}^2\end{aligned}$$

and the charge on each is CV.

When K is opened, one capacitor remains connected to the source while the other is isolated with its charge CV. The dielectric of K = 5 is then introduced in both, so each capacitance becomes 5C.

Capacitor still connected to the source : the potential difference stays V,

$$\begin{aligned}\text U_1 = \frac{1}{2}(5\text C)\text V^2 \\[1em] \text U_1 = \frac{5}{2}\text{CV}^2\end{aligned}$$

Isolated capacitor : the charge stays CV,

$$\begin{aligned}\text U_2 = \frac{(\text{CV})^2}{2(5\text C)} \\[1em] \text U_2 = \frac{\text{CV}^2}{10}\end{aligned}$$

Hence

$$\begin{aligned}\text E_2 = \frac{5}{2}\text{CV}^2 + \frac{1}{10}\text{CV}^2 \\[1em] = \frac{25 + 1}{10}\text{CV}^2 \\[1em] \text E_2 = \frac{26}{10}\text{CV}^2 \\[1em] \frac{\text E_1}{\text E_2} = \frac{\text{CV}^2}{\frac{26}{10}\text{CV}^2} \\[1em] = \frac{10}{26} \\[1em] \frac{\text E_1}{\text E_2} = \frac{5}{13}\end{aligned}$$

1.5

Reason —

Given, C₁ = 40 µF and C₂ = 40K µF after filling one of them with the dielectric, and C_eq = 24 µF.

For a series combination,

$$\begin{aligned}\frac{1}{\text C_\text{eq}} = \frac{1}{\text C_1} + \frac{1}{\text C_2} \\[1em] \frac{1}{24} = \frac{1}{40} + \frac{1}{40\text K} \\[1em] \frac{1}{40\text K} = \frac{1}{24} - \frac{1}{40} \\[1em] = \frac{5 - 3}{120} \\[1em] = \frac{2}{120} \\[1em] \frac{1}{40\text K} = \frac{1}{60} \\[1em] 40\text K = 60 \\[1em] \text K = 1.5\end{aligned}$$

$\frac{(\text q_1 - \text q_2)}{2\text C}$

Reason —

When two plates carrying unequal charges are brought close together, the charge distributes itself so that the facing surfaces carry $\pm\frac{\text q_1 - \text q_2}{2}$ and the outer surfaces carry $\frac{\text q_1 + \text q_2}{2}$ each.

Only the charge on the facing surfaces produces the field in the gap, so the effective charge of the capacitor is

$\text q = \frac{\text q_1 - \text q_2}{2}$

Hence the potential difference is

$$\begin{aligned}\text V = \frac{\text q}{\text C} \\[1em] \text V = \frac{\text q_1 - \text q_2}{2\text C}\end{aligned}$$

The capacitance is decreased by a factor of $\frac{1}{\text K + 1}$ .

Reason —

Initially : the dielectric fills the whole gap d, so

$\text C_i = \frac{\text K\epsilon_0 \text A}{\text d}$

Finally : the separation becomes 2d, of which a thickness d is the dielectric and the remaining d is air. These form two capacitors in series,

$$\begin{aligned}\text C_f = \frac{\left(\frac{\text K\epsilon_0 \text A}{\text d}\right)\left(\frac{\epsilon_0 \text A}{\text d}\right)}{\frac{\text K\epsilon_0 \text A}{\text d} + \frac{\epsilon_0 \text A}{\text d}} \\[1em] \text C_f = \frac{\text K}{\text K + 1}\frac{\epsilon_0 \text A}{\text d}\end{aligned}$$

Therefore

$$\begin{aligned}\frac{\text C_f}{\text C_i} = \frac{\frac{\text K}{\text K + 1}\frac{\epsilon_0 \text A}{\text d}}{\frac{\text K\epsilon_0 \text A}{\text d}} \\[1em] \frac{\text C_f}{\text C_i} = \frac{1}{\text K + 1}\end{aligned}$$

so the capacitance is decreased by the factor 1/(K + 1), which makes the second statement correct. The battery remains connected, so the voltage between the plates stays V and the third statement is wrong. Since the field inside the dielectric is not reduced by a factor 2K, and the work done clearly involves K through the change in capacitance, the first and the fourth statements are also wrong.

2.25 × 10^-6 J

Reason —

The charge on C₁ before connecting is

$$\begin{aligned}\text Q = \text C_1 \text V \\[1em] = (900 \times 10^{-12})(100) \\[1em] \text Q = 9 \times 10^{-8}\ \text C\end{aligned}$$

The battery is disconnected, so this charge is shared between the two equal capacitors joined in parallel. The common potential difference is

$$\begin{aligned}\text V' = \frac{\text Q}{\text C_1 + \text C_2} \\[1em] = \frac{9 \times 10^{-8}}{1800 \times 10^{-12}} \\[1em] \text V' = 50\ \text V\end{aligned}$$

The energy stored by the system is

$$\begin{aligned}\text U = \frac{1}{2}(\text C_1 + \text C_2)\text V'^2 \\[1em] \text U = \frac{1}{2}(1800 \times 10^{-12})(50)^2 \\[1em] \text U = \frac{1}{2}(1800 \times 10^{-12})(2500) \\[1em] \text U = 2.25 \times 10^{-6}\ \text J\end{aligned}$$

$\frac{\text {CV}}{\epsilon_0}$

Reason —

By Gauss's theorem, the flux of the electric field through a closed surface is

$\phi = \frac{\text q_\text{enclosed}}{\epsilon_0}$

The closed surface encloses only the positive plate, whose charge is

$\text q = \text{CV}$

Hence

$\phi = \frac{\text{CV}}{\epsilon_0}$

(A) is false but (R) is true.

Reason —

The charge given to a conductor resides entirely on its outer surface, so it makes no difference whether the sphere is solid or hollow. The capacitance of either is

$\text C = 4\pi\epsilon_0 \text a$

which depends only on the radius. Since the two spheres have equal radii, they have equal capacitances, and being at the same potential they carry equal charges. Hence the assertion is false.

The reason, however, is a correct statement, since the capacitance of a metallic sphere does depend on its radius.

2 µF

Reason —

The two 3 µF capacitors on the right are joined between the same pair of points, so they are in parallel,

$$\begin{aligned}\text C' = 3 + 3 \\[1em] \text C' = 6\ \mu\text F\end{aligned}$$

This combination is in series with the 3 µF capacitor on the left,

$$\begin{aligned}\text C_\text{AB} = \frac{3 \times 6}{3 + 6} \\[1em] = \frac{18}{9} \\[1em] \text C_\text{AB} = 2\ \mu\text F\end{aligned}$$

63 pF

Reason —

Given, the conducting walls each of area 50 cm × 50 cm are 5 cm apart, so d = 5 cm = 0.05 m.

Volume of liquid poured in 10 s,

$$\begin{aligned}\text V = 250 \times 10 \\[1em] \text V = 2500\ \text{cm}^3\end{aligned}$$

The base area is 50 × 5 = 250 cm², so the height of the liquid is

$$\begin{aligned}\text h = \frac{2500}{250} \\[1em] \text h = 10\ \text{cm}\end{aligned}$$

The filled and the empty portions form two capacitors in parallel, since the liquid rises across the full separation between the two walls.

Filled portion : area 50 cm × 10 cm = 500 cm² = 0.05 m²,

$$\begin{aligned}\text C_1 = \frac{\text K\epsilon_0 \text A_1}{\text d} \\[1em] = \frac{3(9 \times 10^{-12})(0.05)}{0.05} \\[1em] \text C_1 = 27 \times 10^{-12}\ \text F\end{aligned}$$

Empty portion : area 50 cm × 40 cm = 2000 cm² = 0.2 m²,

$$\begin{aligned}\text C_2 = \frac{\epsilon_0 \text A_2}{\text d} \\[1em] = \frac{(9 \times 10^{-12})(0.2)}{0.05} \\[1em] \text C_2 = 36 \times 10^{-12}\ \text F\end{aligned}$$

Hence

$$\begin{aligned}\text C = \text C_1 + \text C_2 \\[1em] = 27 + 36 \\[1em] \text C = 63\ \text{pF}\end{aligned}$$

1.50

Reason —

Initially the capacitance is

$$\begin{aligned}\text C_0 = \frac{\epsilon_0 \text A}{\text d} \\[1em] \text C_0 = \frac{\epsilon_0 \text A}{0.6}\end{aligned}$$

The slab of thickness t = 0.6 cm fills the whole original gap, and the plate is then moved out by 0.2 cm, so the new separation is d' = 0.8 cm. With the slab inside,

$$\begin{aligned}\text C = \frac{\epsilon_0 \text A}{\text d' - \text t + \frac{\text t}{\text K}} \\[1em] \text C = \frac{\epsilon_0 \text A}{0.8 - 0.6 + \frac{0.6}{\text K}}\end{aligned}$$

For the capacitance to remain the same,

$$\begin{aligned}0.2 + \frac{0.6}{\text K} = 0.6 \\[1em] \frac{0.6}{\text K} = 0.4 \\[1em] \text K = \frac{0.6}{0.4} \\[1em] \text K = 1.50\end{aligned}$$

$\frac{11\epsilon_0\text A}{18\thinspace\text d}$

Reason —

The three stairs lie at distances d, 2d and 3d from the flat plate, and each has area A/3. They behave as three capacitors joined in parallel,

$$\begin{aligned}\text C = \frac{\epsilon_0 (\text A/3)}{\text d} + \frac{\epsilon_0 (\text A/3)}{2\text d} + \frac{\epsilon_0 (\text A/3)}{3\text d} \\[1em] = \frac{\epsilon_0 \text A}{3\text d}\left(1 + \frac{1}{2} + \frac{1}{3}\right) \\[1em] = \frac{\epsilon_0 \text A}{3\text d}\left(\frac{6 + 3 + 2}{6}\right) \\[1em] \text C = \frac{\epsilon_0 \text A}{3\text d}\cdot\frac{11}{6} \\[1em] \text C = \frac{11\epsilon_0 \text A}{18\thinspace\text d}\end{aligned}$$

Note: In the textbook, options (c) and (d) both print $\frac{11\epsilon_0\text A}{20\thinspace\text d}$, which is a printing error. The correct answer, option (a), is unaffected.

P → 3; Q → 2; R → 4; S → 1

Reason —

Each adjacent pair of sheets forms a capacitor of capacitance C₀ = ε₀a²/d.

(P) With S₂ and S₃ unconnected, the three gaps S₁–S₂, S₂–S₃ and S₃–S₄ are in series,

$\text C = \frac{\text C_0}{3}$

(Q) Shorting S₂ to S₃ removes the middle gap, leaving only the gaps S₁–S₂ and S₃–S₄ in series,

$\text C = \frac{\text C_0}{2}$

(R) With S₂ shorted to S₄, the gap S₃–S₄ comes in parallel with the gap S₂–S₃, giving 2C₀, and this is in series with the gap S₁–S₂,

$$\begin{aligned}\text C = \frac{\text C_0 \times 2\text C_0}{\text C_0 + 2\text C_0} \\[1em] \text C = \frac{2\text C_0}{3}\end{aligned}$$

(S) With S₃ shorted to S₁ and S₂ shorted to S₄, all the three gaps lie between the same pair of nodes and are in parallel,

$\text C = 3\text C_0$

A, C and E only

Reason —

The battery remains connected, so the potential difference V is constant. On bringing the plates closer, d decreases and hence

$\text C = \frac{\epsilon_0 \text A}{\text d}\ \text{increases}$

so statement C is correct, and statement D is wrong since Q/V = C has changed.

Since Q = CV with V fixed, the charge increases, so statement A is correct.

The energy stored is U = ½CV², which increases with C, so statement B is wrong. The product QV = CV² also increases, so statement E is correct.

Hence only A, C and E are correct.

Given,

• C₁ = 5 µF = 5 × 10^-6 F, charged to V₁ = 220 V
• C₂ = 2.5 µF = 2.5 × 10^-6 F, uncharged, so V₂ = 0

The supply is disconnected, so the charge on the first capacitor is merely redistributed between the two. The loss of energy in such a redistribution is

$\Delta \text U = \frac{\text C_1 \text C_2}{2(\text C_1 + \text C_2)}(\text V_1 - \text V_2)^2$

Substituting the values,

$$\begin{aligned}\Delta \text U = \frac{(5 \times 10^{-6})(2.5 \times 10^{-6})}{2(7.5 \times 10^{-6})}(220)^2 \\[1em] = \frac{12.5 \times 10^{-12}}{15 \times 10^{-6}}(48400) \\[1em] = (0.8333 \times 10^{-6})(48400) \\[1em] \Delta \text U = 0.0403\ \text J \\[1em] \Delta \text U \approx \frac{4}{100}\ \text J\end{aligned}$$

Hence, X = 4.

Note: The question states that the second capacitor is connected "in series". A capacitor joined in series with a charged one cannot receive any redistributed charge, so no energy change would occur. The printed answer follows from the two capacitors being joined in parallel, and the word "series" is a misprint.

Given,

• Plate area, A = 100 m²
• Plate separation, d = 10 m
• Thickness of the dielectric, t = 5 m, with K = 10
• ε₀ = 8.85 × 10^-12 F m^-1

When a slab of thickness t and dielectric constant K partly fills the gap,

$\text C = \frac{\epsilon_0 \text A}{\text d - \text t + \frac{\text t}{\text K}}$

Substituting the values,

$$\begin{aligned}\text C = \frac{(8.85 \times 10^{-12})(100)}{10 - 5 + \frac{5}{10}} \\[1em] \text C = \frac{8.85 \times 10^{-10}}{5.5} \\[1em] \text C = 1.609 \times 10^{-10}\ \text F \\[1em] \text C = 160.9\ \text{pF}\end{aligned}$$

Hence, x = 161.

Given,

• Capacitance, C = 1 µF
• With S at P, the 1 V cell and the 2 V cell are joined through the 1 Ω and the 2 Ω resistors

After a long time the capacitor is fully charged, so no current flows into its branch. A steady current, however, flows round the loop containing the two cells,

$$\begin{aligned}\text I = \frac{2 - 1}{1 + 2} \\[1em] \text I = \frac{1}{3}\ \text A\end{aligned}$$

Taking the lower rail as the zero of potential, the potential of the junction to which the capacitor is joined is obtained by starting from the 2 V cell and subtracting the drop across the 2 Ω resistor,

$$\begin{aligned}\text V = 2 - \text I(2) \\[1em] = 2 - \frac{2}{3} \\[1em] \text V = \frac{4}{3}\ \text V\end{aligned}$$

Hence the charge on the capacitor is

$$\begin{aligned}\text q_1 = \text{CV} \\[1em] = (1)\left(\frac{4}{3}\right) \\[1em] \text q_1 = 1.33\ \mu\text C\end{aligned}$$

Hence, the magnitude of q₁ is 1.33 µC.

Given,

• Capacitance, C = 1 µF
• With S at Q, the 1 V cell is cut out and the 1 Ω resistor is joined directly to the lower rail

Again no current flows into the capacitor branch in the steady state, but a current flows in the loop containing the 2 V cell, the 2 Ω resistor and the 1 Ω resistor,

$$\begin{aligned}\text I = \frac{2}{1 + 2} \\[1em] \text I = \frac{2}{3}\ \text A\end{aligned}$$

The potential of the junction to which the capacitor is joined is the drop across the 1 Ω resistor,

$$\begin{aligned}\text V = \text I(1) \\[1em] \text V = \frac{2}{3}\ \text V\end{aligned}$$

Hence

$$\begin{aligned}\text q_2 = \text{CV} \\[1em] = (1)\left(\frac{2}{3}\right) \\[1em] \text q_2 = 0.67\ \mu\text C\end{aligned}$$

Hence, the magnitude of q₂ is 0.67 µC.

Given,

• C₁ = 12 µF, C₂ = C₃ = 4 µF, C₄ = C₅ = 2 µF
• The cells in the circuit are of 6 V and 2 V

Let the lower rail be taken as the zero of potential.

The upper rail is continuous from the junction of C₁, C₂, C₃ and C₄ right up to the positive terminal of the 2 V cell. Hence this junction is held at

$\text V = 2\ \text V$

The capacitor C₃ is joined between this junction and the lower rail, so the potential difference across it is 2 V. Therefore

$$\begin{aligned}\text q_3 = \text C_3 \text V \\[1em] \text q_3 = (4)(2) \\[1em] \text q_3 = 8\ \mu\text C\end{aligned}$$

Hence, the charge stored in C₃ is 8.00 µC.

Given,

• Capacitance with air, C = 10 µF = 10 × 10^-6 F
• Separation between the plates, d = 10 mm = 1 × 10^-2 m
• Area of each plate, A = 4 cm² = 4 × 10^-4 m²
• Dielectric constants, K₁ = 2 and K₂ = 3, each occupying half the plate area
• New force between the plates, F = 8 N

The two dielectrics occupy the two halves of the plate area, each of area A/2 with the full separation d, so they behave as two capacitors joined in parallel. Their capacitances are

$$\begin{aligned}\text C_1 = \frac{\text K_1 \text C}{2} \\[1em] = \frac{2 \times 10}{2} \\[1em] \text C_1 = 10\ \mu\text F \\[1em] \text C_2 = \frac{\text K_2 \text C}{2} \\[1em] = \frac{3 \times 10}{2} \\[1em] \text C_2 = 15\ \mu\text F\end{aligned}$$

If V is the supply voltage, the charges on the two halves are

$$\begin{aligned}\text Q_1 = \text C_1 \text V \\[1em] \text Q_1 = 10 \times 10^{-6}\thinspace\text V \\[1em] \text Q_2 = \text C_2 \text V \\[1em] \text Q_2 = 15 \times 10^{-6}\thinspace\text V\end{aligned}$$

The force between the plates of a charged parallel-plate capacitor is half the product of the charge on a plate and the electric field between the plates,

$\text F = \frac{1}{2}\text{QE}$

Both halves have the same potential difference V across the same separation d, so the field in each half is

$\text E = \frac{\text V}{\text d}$

The total force on a plate is the sum of the forces on its two halves,

$$\begin{aligned}\text F = \frac{1}{2}\text Q_1 \text E + \frac{1}{2}\text Q_2 \text E \\[1em] \text F = \frac{1}{2}(\text Q_1 + \text Q_2)\frac{\text V}{\text d}\end{aligned}$$

Substituting the charges, d = 1 × 10^-2 m and F = 8 N,

$$\begin{aligned}8 = \frac{1}{2}(10 \times 10^{-6}\thinspace\text V + 15 \times 10^{-6}\thinspace\text V)\frac{\text V}{1 \times 10^{-2}} \\[1em] 8 = \frac{1}{2}(25 \times 10^{-6}\thinspace\text V)\frac{\text V}{1 \times 10^{-2}} \\[1em] 8 = 1.25 \times 10^{-3}\thinspace\text V^2 \\[1em] \text V^2 = \frac{8}{1.25 \times 10^{-3}} \\[1em] \text V^2 = 6400 \\[1em] \text V = 80\ \text V\end{aligned}$$

Hence, the supply voltage is 80 V.

Note: The textbook gives 0.93 × 10^-2 V because it applies the air-capacitor relation F = Q²/2Aε₀ to the dielectric-filled halves. The stated 10 µF also conflicts with A = 4 cm² and d = 10 mm, which give ε₀A/d = 3.54 × 10^-13 F; 80 V follows when the stated capacitance is taken as intended.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: Electric potential is defined as work done per unit charge and has magnitude only, so it is a scalar. Electric intensity is force per unit charge and has both magnitude and direction, so it is a vector.

Reason (R) is incorrect: At the middle point of the line joining an electron and a proton, the potential is the algebraic sum of the two potentials,

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\sum\frac{\text q}{\text r} \\[1em] \text V = 0 \\[1em] \text{since} \quad \sum \text q = -\text q + \text q \\[1em] \sum \text q = 0\end{aligned}$$

The electric intensity, however, is a vector sum. A unit positive charge placed at the middle point is repelled by the proton and attracted by the electron with equal forces acting in the same direction, so the resultant field is not zero.

Therefore, assertion is true but reason is false.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: Electric lines of force originate from a positive charge and terminate on a negative charge, that is, they run from a higher potential towards a lower potential. Hence a positive charge left free in an electric field moves from a point of higher potential to a point of lower potential.

Reason (R) is incorrect: Electric potential is a scalar quantity, since it is defined as the work done per unit charge and possesses magnitude only.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A surface on which the potential has the same value at every point is called an equipotential surface, and the surface of a conductor is such a surface.

Reason (R) is also correct and explains the assertion: A conductor allows charge to flow freely through it. If two points on a conductor were at different potentials, charge would flow from the point of higher potential to that of lower potential until the two potentials became equal. Hence the whole surface settles down at one and the same potential.

Therefore, both are true and the reason correctly explains the assertion.

If both assertion and reason are false.

Explanation

Assertion (A) is incorrect: The potential of a charged conductor depends on its shape as well, and not merely on the charge and the volume. Two conductors of equal charge and equal volume but of different shapes may be at different potentials.

Reason (R) is also incorrect: For the same reason, the potential does not depend only on the charge and the volume of the conductor. For a spherical conductor, for instance,

$\text V = \frac{\text Q}{4\pi\epsilon_0 \text R}$

so it is the linear dimension R, fixed by the shape, that enters the expression.

Therefore, both assertion and reason are false.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: The earth is a very good conductor of an extremely large size, so its capacitance is enormous. Giving or taking away a small charge therefore produces practically no change in its potential, and the potential of the earth is taken as zero. The potentials of all other bodies are measured with reference to it.

Reason (R) is incorrect: An electric field does exist at the surface of the earth; in fact the atmosphere possesses a significant electric field. The zero of potential is a matter of choosing a reference level, not of the absence of a field.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Inside a charged metallic sphere every point is at the same potential, so in moving a charge from one point to another inside it,

$$\begin{aligned}\text W = \text q\thinspace\Delta \text V \\[1em] \text W = 0\end{aligned}$$

Outside the sphere the potential varies from point to point, so a potential gradient exists and W ≠ 0.

Reason (R) is also correct and explains the assertion: It states precisely this property — the potential is the same at each point inside the sphere but different for different points outside it — which is the reason why the work is zero in the one case and non-zero in the other.

Therefore, both are true and the reason correctly explains the assertion.

If both assertion and reason are false.

Explanation

Assertion (A) is incorrect: The potential falls off at different rates in the two cases. For a point charge

$\text V \propto \frac{1}{\text r}$

whereas for a small electric dipole

$\text V \propto \frac{1}{\text r^2}$

so the potential of a dipole decreases much more rapidly.

Reason (R) is also incorrect: The inverse square relation does not hold in all cases; it applies to a small dipole, while for a point charge the potential is inversely proportional to the first power of the distance.

Therefore, both assertion and reason are false.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: The electric field is directed towards the north, so the potential changes only along the north–south direction. Along the east and west directions there is no component of the field, so

$\frac{\text{dV}}{\text{dr}} = 0$

that is, the rate of change of potential is zero along east and west.

Reason (R) is also correct: The electric field due to a point charge does surround the charge in the space around it.

However, this general statement about the field of a point charge does not account for the vanishing of the potential gradient along the east–west direction, which follows from the direction of the field at that point.

Therefore, both are true but the reason is not the correct explanation of the assertion.