Electric Charges & Fields

K : 1

Reason — In vacuum (or air) the force between two point-charges is $\text F_0 = \frac{1}{4\pi\varepsilon_0}\frac{\text q_1\text q_2}{\text r^2}$, while in a medium of dielectric constant K it is $\text F = \frac{1}{4\pi\varepsilon_0\text K}\frac{\text q_1\text q_2}{\text r^2}$. Dividing one by the other,

$\frac{\text F_0}{\text F} = \text K$

Hence the two forces are in the ratio K : 1, that is, the force decreases by a factor K when a dielectric is introduced.

5

Reason — The dielectric constant is defined as $\text K = \frac{\varepsilon}{\varepsilon_0}$, the ratio of the absolute permittivity of the material to the permittivity of free space. For all dielectrics the value of K is greater than 1, its least value being K = 1 for vacuum.

N^-1-m^-2 C²

Reason — From Coulomb's law,

$$\begin{aligned}\text F = \frac{1}{4\pi\varepsilon_0}\frac{\text q_1\text q_2}{\text r^2} \\[1em] \varepsilon_0 = \frac{1}{4\pi}\frac{\text q_1\text q_2}{\text F\text r^2}\end{aligned}$$

Putting the units of charge, force and distance,

$$\begin{aligned}\varepsilon_0 = \frac{\text C \times \text C}{\text N \times \text m^2} \\[1em] \varepsilon_0 = \text C^2\ \text N^{-1}\ \text m^{-2}\end{aligned}$$

which is the same as N^-1-m^-2 C². Its value is 8.85 × 10^-12 C² N^-1-m^-2.

10^-43

Reason — Electrical forces are much stronger than gravitational forces. For two electrons the electrical force is 10⁴³ times the gravitational force between them, that is,

$\frac{\text F_{\text{electrostatic}}}{\text F_{\text{gravitational}}} = 10^{43}$

Hence the ratio of the gravitational force to the electrostatic force is 10^-43.

J C^-1

Reason — The intensity of electric field is the force per unit positive charge, $\text E = \frac{\text F}{\text q_0}$, so its unit is newton/coulomb (N C^-1). The same unit may also be written as volt/metre (V m^-1) and as J C^-1 m^-1. The unit J C^-1 is the volt, which is the unit of electric potential and not of electric field.

E_A > E_B

Reason — The relative closeness of the electric lines of force in different regions of space indicates the relative strength of the electric field there. In the figure the lines of force are crowded together near A and are farther apart near B. Hence the field is stronger at A than at B, that is, E_A > E_B.

forces equal in magnitude

Reason — The force on a charge placed in an electric field is F = qE. The proton and the electron carry charges of equal magnitude e, so the magnitudes of the forces on them are the same, qE. Since their charges are of opposite sign, the two forces are oppositely directed. Their accelerations, $\text a = \frac{\text{qE}}{\text m}$, are unequal because the masses of the proton and the electron are very different.

$\frac{1}{\text x^3}$

Reason — At a large distance from a short electric dipole, the intensity of the electric field is

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{2\text p}{\text x^3} \quad \text{(axial)}, \qquad \text E = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\text x^3} \quad \text{(equatorial)}$

In either position $\text E \propto \frac{1}{\text x^3}$. On comparing the fields of a point-charge and of a dipole, the $\frac{1}{\text x^2}$ dependence is replaced by $\frac{1}{\text x^3}$ dependence.

$\frac{\text F}{8}$

Reason — At a point in the end-on position, at a large distance r from the dipole,

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{2\text p}{\text r^3}$

The force on a charge q placed there is F = qE, so that $\text F \propto \frac{1}{\text r^3}$. If the distance is doubled,

$$\begin{aligned}\frac{\text F'}{\text F} = \left(\frac{\text r}{2\text r}\right)^3 \\[1em] \frac{\text F'}{\text F} = \frac{1}{8} \\[1em] \text F' = \frac{\text F}{8}\end{aligned}$$

only a torque

Reason — In a uniform electric field the charge + q of the dipole experiences a force qE along the field and the charge − q an equal force qE opposite to the field. These two forces are equal and opposite, so the net translatory force on the dipole is zero. However, they act at different points and therefore form a couple whose moment is the torque

$\tau = \text{pE}\sin\theta$

Hence the dipole experiences only a torque, which tends to align it in the direction of the field.

$\vec{\text p} \times \vec{\text E}$

Reason — The magnitude of the torque acting on a dipole of moment p placed at an angle θ to a uniform field E is τ = pE sin θ. Written in vector form this becomes

$\vec{\tau} = \vec{\text p} \times \vec{\text E}$

The cross product correctly gives both the magnitude pE sin θ and the direction of the torque, which is perpendicular to the plane containing $\vec{\text p}$ and $\vec{\text E} \space$.

charge is not affected but mass changes

Reason — By the invariance property of charge, the value of a charge does not change with velocity. If a static charge q₀ is moved with a velocity v, its value in the moving state remains q₀. The mass, however, is a variant quantity and changes with speed according to

$\text m = \frac{\text m_0}{\sqrt{1 - \frac{\text v^2}{\text c^2}}}$

Hence the charge of the moving body is unaffected while its mass changes.

First glass rod will acquire negative charge while the second will acquire positive charge

Reason — Fur precedes glass, and glass precedes ebonite in the series. On rubbing two materials, the earlier one becomes positive and the later one negative.

Fur comes before glass, so on rubbing with fur the first glass rod gains electrons and becomes negatively charged. Glass comes before ebonite, so on rubbing with ebonite the second glass rod loses electrons and becomes positively charged.

1.6 × 10^-20 C

Reason — By the quantisation of electric charge, the charge on a body must be an integral multiple of the elementary charge e = 1.6 × 10^-19 C, that is, q = ± ne where n = 1, 2, 3, ... No charge is found in a fraction of e.

For 1.6 × 10^-20 C,

$$\begin{aligned}\text n = \frac{1.6 \times 10^{-20}}{1.6 \times 10^{-19}} \\[1em] \text n = 0.1\end{aligned}$$

which is not an integer. Hence a charge of 1.6 × 10^-20 C is not possible.

neutral

Reason — Repulsion is the real test of a body being charged, while attraction may also occur between a charged body and an uncharged body because of induction.

The pairs (2, 3) and (4, 5) repel, so balls 2, 3 and balls 4, 5 are charged. Balls 2 and 4 attract each other, so they carry charges of opposite sign. Ball 1 attracts both 2 and 4, which would be impossible if ball 1 were charged, since it would then repel one of them. Hence ball 1 must be neutral, the attraction being due to electrostatic induction.

$\frac{\text F}{8}$

Reason — Initially the force of attraction between the spheres is

$$\begin{aligned}\text F = \frac{1}{4\pi\varepsilon_0}\frac{(4\text q)(2\text q)}{\text r^2} \\[1em] \text F = \frac{1}{4\pi\varepsilon_0}\frac{8\text q^2}{\text r^2}\end{aligned}$$

When two identical spheres are brought in contact, the charge is shared equally between them, so each acquires

$$\begin{aligned}\text q' = \frac{4\text q + (-2\text q)}{2} \\[1em] \text q' = + \text q\end{aligned}$$

Placing them again at the same distance r,

$$\begin{aligned}\text F' = \frac{1}{4\pi\varepsilon_0}\frac{\text q^2}{\text r^2} \\[1em] \text F' = \frac{\text F}{8}\end{aligned}$$

$-\frac{\text Q}{4}$

Reason — Let the two charges Q be at A and B, a distance 2a apart, with q at the mid-point. The charge q is automatically in equilibrium by symmetry, so the condition is that each charge Q must be in equilibrium.

For the charge Q at A, the repulsion due to the charge Q at B and the force due to q at a distance a must cancel,

$$\begin{aligned}\frac{1}{4\pi\varepsilon_0}\frac{\text Q^2}{(2\text a)^2} + \frac{1}{4\pi\varepsilon_0}\frac{\text{Qq}}{\text a^2} = 0 \\[1em] \frac{\text Q}{4} + \text q = 0 \\[1em] \text q = -\frac{\text Q}{4}\end{aligned}$$

−q

Reason — The charge q at position l experiences a force due to the charge 4q at position 0, which is at a distance l, and due to the charge Q at position $\frac{\text l}{2}$, which is at a distance $\frac{\text l}{2}$. By the principle of superposition, for the resultant force on q to be zero,

$$\begin{aligned}\frac{1}{4\pi\varepsilon_0}\frac{(4\text q)\text q}{\text l^2} + \frac{1}{4\pi\varepsilon_0}\frac{\text{Qq}}{\left(\frac{\text l}{2}\right)^2} = 0 \\[1em] \frac{4\text q}{\text l^2} + \frac{4\text Q}{\text l^2} = 0 \\[1em] \text Q = -\text q\end{aligned}$$

increase

Reason — The charge given to the soap bubble is a conductor's charge and spreads over its entire surface. Every element of this charge repels every other element, since all the charge is of the same sign. This outward electrostatic repulsion acts against the surface tension of the film, so the bubble expands and its size increases.

$\frac{9\text F}{16}$

Reason — Initially the magnitude of the Coulomb force is

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

When 25% of the charge of A is transferred to B,

$$\begin{aligned}\text q_\text A = \text q - \frac{\text q}{4} \\[1em] \text q_\text A = \frac{3\text q}{4} \\[1em] \text q_\text B = -\text q + \frac{\text q}{4} \\[1em] \text q_\text B = -\frac{3\text q}{4}\end{aligned}$$

The new force is

$$\begin{aligned}\text F' = \frac{1}{4\pi\varepsilon_0}\frac{\left(\frac{3\text q}{4}\right)\left(\frac{3\text q}{4}\right)}{\text r^2} \\[1em] = \frac{9}{16}\times\frac{1}{4\pi\varepsilon_0}\frac{\text q^2}{\text r^2} \\[1em] \text F' = \frac{9\text F}{16}\end{aligned}$$

shall increase along the positive x-axis

Reason — The charge q₁ lies on the negative x-axis and the net force on it due to the positive charges q₂ and q₃ is along the + x direction. Two positive charges could attract q₁ towards the origin only if q₁ itself is negative.

The positive charge Q is now placed at (x, 0) on the positive x-axis, that is, on the far side of the origin from q₁. It therefore attracts the negative charge q₁ along the + x direction as well. By the principle of superposition this force adds to the existing one, so the force on q₁ shall increase along the positive x-axis.

$\text K \frac{\text Q^2}{\text L^2}$

Reason — For a regular hexagon of side L, the distance of every vertex from the centre is also L.

If charges + Q were placed at all the six vertices, the forces exerted by them on the charge − Q at the centre would cancel in pairs, since the charges at diametrically opposite vertices are equidistant and produce equal and opposite forces. The resultant would then be zero.

Here one vertex is vacant. Hence the resultant force is simply the force that would have been exerted by the missing charge + Q, with its direction reversed, that is, directed away from the vacant vertex, towards the diametrically opposite occupied vertex,

$$\begin{aligned}\text F = \frac{1}{4\pi\varepsilon_0}\frac{\text Q \times \text Q}{\text L^2} \\[1em] \text F = \text K\frac{\text Q^2}{\text L^2}\end{aligned}$$

The dipole will experience a force towards left

Reason — The field shown is non-uniform, the lines of force being crowded on the left and farther apart on the right, so that the field is stronger on the left. The dipole is placed with $\vec{\text p}$ pointing to the right, that is, parallel to the field.

The charge − q lies in the stronger part of the field and the charge + q in the weaker part. The force on − q, directed to the left, is therefore greater in magnitude than the force on + q, directed to the right. Hence the net force on the dipole is towards the left.

$\frac{2\text A}{\text B}$

Reason — The particle starts from rest at x = 0 and comes to rest again after travelling a distance d. Since the kinetic energy is zero at both ends, the total work done by the electric force over this distance is zero.

$$\begin{aligned}\text W = \int_0^\text d \text{qE}\ \text{dx} \\[1em] = \text q\int_0^\text d (\text A - \text{Bx})\ \text{dx} \\[1em] \text W = 0\end{aligned}$$

Carrying out the integration,

$$\begin{aligned}\text q\left[\text{Ax} - \frac{\text{Bx}^2}{2}\right]_0^\text d = 0 \\[1em] \text{Ad} - \frac{\text{Bd}^2}{2} = 0 \\[1em] \text d\left(\text A - \frac{\text{Bd}}{2}\right) = 0 \\[1em] \text d = \frac{2\text A}{\text B}\end{aligned}$$

v ∝ x^-1/2

Reason — For two identical charged balls hanging from a common point, with the separation x much smaller than the length l, the equilibrium condition gives

$$\begin{aligned}\tan\theta \approx \sin\theta = \frac{\text x}{2\text l} \\[1em] = \frac{\text F}{\text{mg}} \\[1em] \frac{\text x}{2\text l} = \frac{1}{4\pi\varepsilon_0}\frac{\text q^2}{\text x^2 \text{mg}} \\[1em] \text x^3 \propto \text q^2\end{aligned}$$

The charge leaks at a constant rate, so $\frac{\text{dq}}{\text{dt}}$ is a constant. Differentiating x³ ∝ q² with respect to time,

$3\text x^2 \frac{\text{dx}}{\text{dt}} \propto 2\text q\frac{\text{dq}}{\text{dt}} \quad \Rightarrow \quad \text v \propto \frac{\text q}{\text x^2}$

Since $\text q \propto \text x^{3/2}$,

$\text v \propto \frac{\text x^{3/2}}{\text x^2} = \text x^{-1/2}$

$\left(\frac{\text r}{\sqrt[3]{2}}\right)$

Reason — Let y be the vertical distance of the balls below the point of suspension. For a small angle θ,

$$\begin{aligned}\tan\theta \approx \frac{\text r}{2\text y} = \frac{\text F}{\text{mg}} \\[1em] \tan\theta = \frac{1}{4\pi\varepsilon_0}\frac{\text q^2}{\text r^2\text{mg}} \\[1em] \text r^3 = \frac{1}{4\pi\varepsilon_0}\frac{2\text q^2\text y}{\text{mg}} \quad \text{that is,} \quad \text r^3 \propto \text y\end{aligned}$$

When the strings are clamped at half the height, y is replaced by $\frac{\text y}{2}$ while q and m are unchanged,

$$\begin{aligned}\frac{\text r'^3}{\text r^3} =\frac{\frac{\text{y}}{2}}{\text{y}}\\[1em] \frac{\text r'^3}{\text r^3} = \frac{1}{2} \\[1em] \text r' = \frac{\text r}{\sqrt[3]{2}}\end{aligned}$$

$\frac{3\sqrt{3}\text Q}{8\pi^2\varepsilon_0\text R^2}\ \hat{\text i}$

Reason — The rod is bent into an arc of radius R subtending an angle of 120° at the centre of curvature O. Its length is

$\text L = \text R \times \frac{2\pi}{3}$

so that the magnitude of the linear charge density is

$$\begin{aligned}|\lambda| = \frac{\text Q}{\text L} \\[1em] |\lambda| = \frac{3\text Q}{2\pi\text R}\end{aligned}$$

For an arc subtending an angle 2α at the centre, the components perpendicular to the bisector cancel in pairs and the resultant field at the centre is

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{2|\lambda|\sin\alpha}{\text R}$

Here 2α = 120°, so α = 60° and $\sin\alpha = \frac{\sqrt{3}}{2}$,

$$\begin{aligned}\text E = \frac{1}{4\pi\varepsilon_0}\frac{2}{\text R}\times\frac{3\text Q}{2\pi\text R}\times\frac{\sqrt{3}}{2} \\[1em] \text E = \frac{3\sqrt{3}\text Q}{8\pi^2\varepsilon_0\text R^2}\end{aligned}$$

The field is directed towards the negatively charged arc, along the positive x-axis; hence it is directed along $\hat{\text i}$.

$\sqrt{2}$ aq along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)

Reason — The algebraic sum of the charges is q + (−2q) + q = 0, so the system is equivalent to an electric dipole.

The charge − 2q at the origin may be treated as two charges of − q each placed at the origin. One of them together with + q at (0, a, 0) forms a dipole of moment qa directed along + y, and the other together with + q at (a, 0, 0) forms a dipole of moment qa directed along + x.

The two component dipole moments are equal in magnitude and mutually perpendicular, so their resultant is

$$\begin{aligned}\text p = \sqrt{(\text{qa})^2 + (\text{qa})^2} \\[1em] \text p = \sqrt{2}\ \text{qa}\end{aligned}$$

and it is inclined at 45° to both axes, that is, it is directed along the line joining (0, 0, 0) and (a, a, 0).

Additivity, conservation and quantisation of charge.

Explanation:
(i) Additivity of charge — charges add up algebraically, the positive and the negative charges cancelling each other's effect.

(ii) Conservation of charge — the total charge of an isolated system remains constant.

(iii) Quantisation of charge — the charge on a body is always an integral multiple of the elementary charge e.

For two point-charges q₁ and q₂ placed a distance r apart in vacuum (or air),

$\text F = \frac{1}{4\pi\varepsilon_0}\frac{\text q_1\text q_2}{\text r^2}$

and in a medium of dielectric constant K,

$\text F = \frac{1}{4\pi\varepsilon_0\text K}\frac{\text q_1\text q_2}{\text r^2}$

where $\frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9$ N-m² C^-2.

Yes.

Explanation:
The Coulomb forces exerted by two charges on each other are equal in magnitude and opposite in direction,

$\vec{\text F}_{21} = -\vec{\text F}_{12}$

so they form an action-reaction pair and obey Newton's third law of motion.

Yes.

Explanation:
Coulomb's law is found to hold for point-charges separated by distances ranging from very large distances down to atomic distances (~ 10^-11 m) and nuclear distances (~ 10^-15 m).

One coulomb is that charge which, when placed at a distance of 1 metre from an equal and similar charge in vacuum (or air), repels it with a force of 9.0 × 10⁹ newton.

Since the charge on one electron is 1.6 × 10^-19 C, the number of electrons carrying a total charge of 1 coulomb is

$$\begin{aligned}\text n = \frac{1}{1.6 \times 10^{-19}} \\[1em] \text n = 6.25 \times 10^{18}\end{aligned}$$

Hence, 6.25 × 10¹⁸ electrons carry a total charge of 1 coulomb.

The elementary charge is the smallest amount of free charge found in nature, of which all existing charges are integral multiples. It is denoted by e and is the magnitude of the charge on an electron.

Its value is e = 1.6 × 10^-19 coulomb.

Fe; Cl^-.

Explanation:
Fe⁺⁺⁺ is formed by the removal of three electrons from a neutral iron atom, so Fe has more electrons. Cl^- is formed by the addition of one electron to a neutral chlorine atom, so Cl^- has more electrons than Cl.

C² N^-1 m^-2.

Explanation:
The value of the permittivity of free space is 8.85 × 10^-12 C² N^-1-m^-2.

$\varepsilon_\text r = \frac{\varepsilon}{\varepsilon_0}$

Explanation:
The relative permittivity is the ratio of the absolute permittivity of the medium to the permittivity of free space. It is also known as the dielectric constant K of the medium, so that we may also write $\text K = \frac{\varepsilon}{\varepsilon_0}$. It is a dimensionless quantity.

The dielectric constant of a medium is the ratio of the force between two given charges placed a certain distance apart in vacuum to the force between the same charges placed the same distance apart in that medium,

$\text K = \frac{\text F_{\text{vacuum}}}{\text F_{\text{medium}}}$

[ML³T^-4A^-2].

Explanation:
From Coulomb's law, $\frac{1}{4\pi\varepsilon_0} = \frac{\text F\text r^2}{\text q_1\text q_2}$.

Taking dimensions, with [q] = [AT],

$$\begin{aligned}\left[\frac{1}{4\pi\varepsilon_0}\right] = \frac{[\text{MLT}^{-2}][\text L^2]}{[\text{AT}][\text{AT}]} \\[1em] \left[\frac{1}{4\pi\varepsilon_0}\right] = [\text{ML}^3\text T^{-4}\text A^{-2}]\end{aligned}$$

Attractive.

Explanation:
Since A and B repel each other, and repulsion is the real test of a body being charged, A and B carry like charges. Since A attracts C, the charge on C is opposite in sign to that on A, and hence opposite in sign to that on B as well. Therefore the force between B and C is one of attraction.

No.

Explanation:
Coulomb's law in this form holds only for point-charges. The spheres are large and are placed close together, so each induces a charge on the other. Because of this induction the charge distributions on the spheres no longer remain uniform, and the charges cannot be treated as though they were concentrated at the centres. Hence the force is not exactly $\frac{\text q_1\text q_2}{4\pi\varepsilon_0\text r^2}$.

10¹⁹ excess electrons.

Explanation:
The conductor carries a negative charge, so it has an excess of electrons. For excess electrons, the quantisation relation is q = −ne,

$$\begin{aligned}\text n = \frac{|\text q|}{\text e} \\[1em] = \frac{1.6}{1.6 \times 10^{-19}} \\[1em] \text n = 10^{19}\end{aligned}$$

A shortage of 15 electrons.

Explanation:
The conductor carries a positive charge, which means a deficiency (shortage) of electrons. The number of electrons corresponding to this charge is

$$\begin{aligned}\text n = \frac{\text q}{\text e} \\[1em] = \frac{2.4 \times 10^{-18}}{1.6 \times 10^{-19}} \\[1em] \text n = 15\end{aligned}$$

The force becomes 16 times its initial value.

Explanation:
By Coulomb's law $\text F = \frac{1}{4\pi\varepsilon_0}\frac{\text q^2}{\text r^2}$.

On doubling each charge the numerator becomes (2q)(2q) = 4q², and on halving the distance the denominator becomes $\left(\frac{\text r}{2}\right)^2 = \frac{\text r^2}{4}$. Hence

$$\begin{aligned}\frac{\text F'}{\text F} = 4 \times 4 \\[1em] \frac{\text F'}{\text F} = 16\end{aligned}$$

For $\frac{\text F}{3}$: $\sqrt{3}\ \text r$; for $3\text F$: $\frac{\text r}{\sqrt{3}}$.

Explanation:
Since $\text F \propto \frac{1}{\text r^2}$, we have $\text F\text r^2 = \text F'\text r'^2$.

For the force to become $\frac{\text F}{3}$:

$$\begin{aligned}\text F\text r^2 = \frac{\text F}{3}\text r'^2 \\[1em] \text r' = \sqrt{3}\ \text r\end{aligned}$$

For the force to become 3F :

$$\begin{aligned}\text F\text r^2 = 3\text F\text r'^2 \\[1em] \text r' = \frac{\text r}{\sqrt{3}}\end{aligned}$$

Zero.

Explanation:
The two charges are + 2 μC and − 2 μC, which are equal and opposite. When they are mutually touched, the charge is shared equally and each acquires

$$\begin{aligned}\text q' = \frac{(+2\ \mu\text C) + (-2\ \mu\text C)}{2} \\[1em] \text q' = 0\end{aligned}$$

Since both the bodies are left uncharged, the new force between them is zero.

The force decreases.

Explanation:
In a medium of dielectric constant K,

$\text F = \frac{1}{4\pi\varepsilon_0\text K}\frac{\text q_1\text q_2}{\text r^2} \quad \text{that is,} \quad \text F \propto \frac{1}{\text K}$

N C^-1; [MLT^-3A^-1].

Explanation:
The intensity of electric field is the force per unit positive charge, $\text E = \frac{\text F}{\text q_0}$.

SI unit : newton/coulomb (N C^-1), which is also written as volt/metre (V m^-1).

Dimensions :

$$\begin{aligned}[\text E] = \frac{[\text{MLT}^{-2}]}{[\text{AT}]} \\[1em] [\text E] = [\text{MLT}^{-3}\text A^{-1}]\end{aligned}$$

A charged circular loop behaves as a point-charge when the point of observation lies on the axis of the loop at a distance much larger than the radius of the loop.

Explanation:
At such a distance the whole charge of the loop may be treated as concentrated at its centre, and the field falls off as $\frac{1}{\text r^2}$, exactly as for a point-charge.

The electric lines of force always begin on positive charges and end on negative charges, and the tangent at any point on a line of force gives the direction of the electric field there.

If a line of force had a sudden break, the electric field would have to be zero at that place. But the field due to the combined effect of the two opposite charges cannot vanish at points lying on a line of force. Hence an electrostatic line of force must be a continuous curve without any break.

9.0 × 10⁹ N C^-1.

Explanation:
The intensity of the electric field due to a point-charge in air is

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

Substituting q = 1 C and r = 1 m,

$$\begin{aligned}\text E = \frac{9.0 \times 10^9 \times 1}{(1)^2} \\[1em] \text E = 9.0 \times 10^9\ \text N\ \text C^{-1}\end{aligned}$$

The electric field is directed vertically upwards.

Explanation:
The weight of the particle acts vertically downwards, so the electric force qE on it must act vertically upwards to balance the weight.

As the charge on the particle is positive, the force on it is in the direction of the field.

Zero.

Explanation:
The four equal charges of + 100 μC lie at the corners of a square, and the centre of the square is equidistant from all of them. The fields due to the charges at the two ends of either diagonal are equal in magnitude and opposite in direction, and hence cancel in pairs. Therefore the resultant electric field at the centre of the square is zero.

An electric dipole is a pair of equal and opposite point-charges placed at a short distance apart.

Examples : The HCl molecule and the H₂O molecule are electric dipoles, since in each of them the centre of the positive charges and the centre of the negative charges do not coincide, although the net charge on the molecule is zero.

The dimensions of electric dipole moment are [LTA].

Explanation:
The electric dipole moment is p = q × 2l, that is, charge multiplied by distance.

$$\begin{aligned}[\text p] = [\text{AT}][\text L] \\[1em] [\text p] = [\text{LTA}]\end{aligned}$$

Its SI unit is coulomb-metre (C-m).

(i) End-on position :

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{2\text p}{\text r^3}$

directed along the dipole axis, pointing from − q to + q.

(ii) Broad-side-on position :

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\text r^3}$

directed antiparallel to the dipole moment $\vec{\text p}$.

Note: The printed answer gives $\frac{1}{4\pi\varepsilon_0}\frac{2\text p}{\text r^3}$ for the broad-side-on position also. This is a misprint: the chapter's own derivation of the equatorial field gives $\frac{1}{4\pi\varepsilon_0}\frac{\text p}{\text r^3}$, half the end-on value.

No, the electric field does not decrease at the same rate in the two cases.

Explanation:
For a point-charge the field falls off as the inverse square of the distance, $\text E \propto \frac{1}{\text r^2}$, whereas for a short electric dipole it falls off as the inverse cube, $\text E \propto \frac{1}{\text r^3}$.

Hence the field of a dipole decreases more rapidly with distance than the field of a point-charge. This happens because the fields of the two equal and opposite charges of the dipole tend to cancel each other at large distances.

During landing, the tyres of an aircraft rub against the airstrip and become highly charged by friction.

If the tyres were made of ordinary rubber, which is an insulator, this charge would stay at the place where it is produced and would go on accumulating, which could produce a spark. Since the special rubber used is slightly conducting, the charge does not accumulate on the tyres but leaks away through them to the earth.

While the vehicle is in motion, its body becomes charged because of friction with the air.

If a large amount of charge accumulates on the body, it can produce a spark and the inflammable material being carried may catch fire. To prevent this mis-happening, metallic chains are hung from the vehicle so as to touch the ground, so that any charge accumulated on the vehicle continuously leaks to the earth.

An electron carries a charge of − 1.6 × 10^-19 C. The removal of one electron from the metal leaves a charge of + 1.6 × 10^-19 C on it.

Hence, by the quantisation of charge, the number of electrons to be removed to give the metal a charge of + 1.0 × 10^-7 C is

$$\begin{aligned}\text n = \frac{\text q}{\text e} \\[1em] = \frac{1.0 \times 10^{-7}}{1.6 \times 10^{-19}} \\[1em] \text n = 6.25 \times 10^{11}\end{aligned}$$

Hence, 6.25 × 10¹¹ electrons must be removed.

Let the two parts be q' and (q − q'), placed a constant distance r apart. The force between them is

$\text F = \frac{1}{4\pi\varepsilon_0}\frac{\text q'(\text q - \text q')}{\text r^2}$

For F to be maximum, $\frac{\text{dF}}{\text{dq}'} = 0$.

Differentiating with respect to q', keeping q and r constant,

$$\begin{aligned}\frac{\text{dF}}{\text{dq}'} = \frac{1}{4\pi\varepsilon_0\text r^2}\frac{\text d}{\text{dq}'}\left(\text q\text q' - \text q'^2\right) \\[1em] \frac{\text{dF}}{\text{dq}'} = \frac{1}{4\pi\varepsilon_0\text r^2}\left(\text q - 2\text q'\right)\end{aligned}$$

Putting this equal to zero,

$$\begin{aligned}\text q - 2\text q' = 0 \\[1em] \text q' = \frac{\text q}{2}\end{aligned}$$

Hence, the charge q should be divided into two equal parts.

Yes.

Explanation:
By the quantisation of charge, the smallest possible charge that can exist in the free state is the elementary charge e = 1.6 × 10^-19 C. Hence the smallest possible electric force between two charges 1 m apart in vacuum is

$$\begin{aligned}\text F_{\text{min}} = \frac{1}{4\pi\varepsilon_0}\frac{\text e^2}{\text r^2} \\[1em] \text F_{\text{min}} = \left(9 \times 10^9\right)\frac{\left(1.6 \times 10^{-19}\right)^2}{(1)^2} \\[1em] \text F_{\text{min}} = 9 \times 10^9 \times 2.56 \times 10^{-38} \\[1em] \text F_{\text{min}} = 2.3 \times 10^{-28}\ \text N\end{aligned}$$

Hence, the lower limit of the electric force is 2.3 × 10^-28 N.

Electric line of force : An electric line of force is that imaginary smooth curve drawn in an electric field along which a free, isolated positive charge moves. The tangent drawn at any point on the line of force gives the direction of the electric field at that point.

Two important properties :

(i) Two lines of force can never intersect each other, for if they did there would be two different directions of the electric field at the point of intersection, which is impossible.

(ii) The relative closeness of the lines of force indicates the relative strength of the field — where the lines are closer the field is stronger, and where they are farther apart the field is weaker.

Lines of force for isolated test-charges : For an isolated positive charge the lines of force are straight and directed radially outwards, while for an isolated negative charge they are straight and directed radially inwards.

If the two positive charges are not equal, the neutral point will not lie at the middle of the line joining them.

At the neutral point the fields due to the two charges must be equal and opposite. If q₁ and q₂ are the magnitudes of the charges and r₁ and r₂ their respective distances from the neutral point, then

$$\begin{aligned}\frac{1}{4\pi\varepsilon_0}\frac{\text q_1}{\text r_1^{\ 2}} = \frac{1}{4\pi\varepsilon_0}\frac{\text q_2}{\text r_2^{\ 2}} \\[1em] \frac{\text q_1}{\text r_1^{\ 2}} = \frac{\text q_2}{\text r_2^{\ 2}}\end{aligned}$$

Hence the neutral point lies nearer to the smaller of the two charges.

The tangent drawn at any point on a line of force gives the direction of the electric field at that point.

If two lines of force intersected at a point, then two tangents could be drawn at the point of intersection, which would mean two different directions of the electric field at the same point. This is impossible, since the electric field at a given point can have only one direction. Hence electric lines of force never intersect each other.

The resultant electric field is parallel to the line joining the two charges at :

(i) all points lying on the right bisector (the equatorial line) of the line joining the charges, where the resultant field is antiparallel to the dipole axis, and

(ii) all points lying along the axis of the line joining the two charges, where the resultant field is along the dipole axis.

The lines of force do not enter the plate at all, which means that the electric field inside the metallic plate is zero.

Further, the lines of force meet the surface of the plate normally. This means that just outside the plate the electric field is normal to its surface, that is, the electric potential is uniform over the whole surface of the plate.

These conclusions are not true for a plate of non-metal, because an insulator has practically no free electrons and hence the field is not reduced to zero inside it.

Not always.

Explanation:
If the electric lines of force are straight and the positive charge is released from rest, the force on it is always along the same straight line, so the charge does move along the line of force.

But if the lines of force are curved, the charge is accelerated at every point along the tangent to the line of force at that point. Because it already possesses a velocity acquired earlier, its resultant path deviates from the line of force. Hence in a curved field the charged particle does not follow the path of the line of force.

Forces : The magnitudes of the charges on the electron and the proton are equal, each being e. Since F = qE, the electric forces on them, eE, are equal in magnitude but opposite in direction, because their charges are of opposite sign.

Accelerations : The accelerations are not equal, since

$$\begin{aligned}\text a = \frac{\text F}{\text m} \\[1em] \text a = \frac{\text{eE}}{\text m}\end{aligned}$$

The electron, being much lighter than the proton, has a much greater acceleration.

Since the electric field vanishes at a point lying between the two charges, the fields there due to the two charges must be oppositely directed. This is possible only if the two charges are of the same sign.

If C is the point of zero field between A and B, then

$$\begin{aligned}\frac{1}{4\pi\varepsilon_0}\frac{\text q_1}{(\text{AC})^2} = \frac{1}{4\pi\varepsilon_0}\frac{\text q_2}{(\text{BC})^2} \\[1em] \frac{\text q_1}{\text q_2} = \left(\frac{\text{AC}}{\text{BC}}\right)^2\end{aligned}$$

Hence the magnitudes of the charges are in the ratio of the squares of their distances from the point of zero field.

The two essential conditions are :

(i) The charges q₁ and q₂ must be of opposite signs, for only then can their fields be oppositely directed at a point lying outside the two charges.

(ii) The magnitude of the charge nearer to that point must be smaller than that of the other charge, so that the two fields can become equal there.

If the charges are a distance r apart and the point of zero field is at a distance x from the smaller charge q₁, then their magnitudes are related by

$\frac{|\text q_1|}{|\text q_2|} = \left(\frac{\text x}{\text r + \text x}\right)^2$

Zero.

Explanation:
The three charges + q are equal and are placed at the corners of an equilateral triangle inscribed in the circle, so the centre of the circle is equidistant from all of them. Hence the three fields at the centre are equal in magnitude, and the angle between each pair of them is 120°.

Three equal vectors inclined at 120° to one another add up to zero. Hence the resultant intensity of the electric field at the centre of the circle is zero.

Electric dipole moment : The dipole moment of an electric dipole consisting of charges − q and + q separated by a small distance 2l is a vector $\vec{\text p}$ of magnitude

$\text p = \text q \times 2\text l$

Its direction is along the line joining the two charges, pointing from the negative charge towards the positive charge, that is, along the dipole axis.

SI unit : coulomb-metre (C-m).

The uniform field $\vec{\text E} \space$exerts a force qE on the charge + q along the field and an equal force qE on the charge − q opposite to the field.

(i) Net translatory force : These two forces are equal and opposite, so the net translatory force on the dipole is zero.

(ii) Torque : The two forces do not act along the same line, so they form a couple. The torque acting on the dipole is

$\tau = \text{pE}\sin\theta$

On being released, this restoring couple turns the dipole so as to align it parallel to the field $\vec{\text E} \space$.

(i) Dipole placed parallel to the field : Here θ = 0°, so the torque τ = pE sin θ is zero, and the net translatory force in a uniform field is also zero. Hence the dipole will not move; it is in the position of stable equilibrium.

(ii) Dipole placed perpendicular to the field : Here θ = 90°, so the torque is maximum, τ = pE, while the net translatory force is still zero. Hence the dipole rotates about its centre until it becomes parallel to the electric field.

In a non-uniform field the forces on the two charges of the dipole are unequal, so a net force also acts on it in addition to the torque.

(i) Dipole parallel to the field : The torque is zero, but the forces on + q and − q are unequal. Hence a resultant force acts on the dipole and it undergoes translational motion in the direction of the net force.

(ii) Dipole perpendicular to the field : Here a torque as well as a net force acts on the dipole. Hence it has both translational and rotational motion.

(i) For q < 0 : The lines of force are straight and are directed radially inwards, since a free positive test-charge placed in the field would move radially towards the negative charge.

(ii) For q > 0 : The lines of force are straight and are directed radially outwards, since a free positive test-charge would be repelled radially away from the charge.

q₁ is negative; q₂ is positive.

Explanation:
The lines of force originate on positive charges and terminate on negative charges. In the figure the lines of force start from the charge q₂ at B and end on the charge q₁ at A. Hence q₂ is positive and q₁ is negative.

q₁ is negative; q₂ is positive.

Explanation:
Since the electric lines of force begin on positive charges and end on negative charges, the charge at B from which the lines of force emerge is positive, while the charge at A on which the lines of force terminate is negative.

Quantisation of electric charge : Millikan's oil-drop experiment and many other experiments have shown that in nature electric charges are found to be made up of integral multiples of a smallest amount of charge. This smallest amount is 1.6 × 10^-19 coulomb, denoted by e, and is the magnitude of the charge of an electron.

Hence the charge q on any body must be given by

$\text q = \pm\ \text{ne}, \qquad \text{where n} = 1,\ 2,\ 3,\ ...$

No charge is ever found to be a fraction of e, such as 0.7e or 2.5e. In other words, electric charge exists in the form of discrete packets and not in continuous amounts, and cannot be divided indefinitely. This property is called the quantisation, or atomicity, of charge. Since e is the smallest unit of charge, it is called the elementary charge.

The charges of some natural elementary particles are

charge of electron = -e,
charge of proton = +e,
charge of $\alpha$-particle = +2e

Why quantisation may be ignored for large charges : The value of the elementary charge is extremely small compared with the charges met with in ordinary experiments. A charge of even one microcoulomb contains about 10¹³ elementary charges.

For example, nearly 3 × 10¹⁸ elementary charges flow per second at one end of a 100-watt bulb filament, and an equal number flow out at the other end. When the number of elementary charges is so enormous, the addition or removal of a few of them makes no perceptible difference. The charge then appears to vary continuously. Hence, while dealing with large charges, the quantisation of charge can be ignored and the charge treated as a continuous quantity.

(i) Law of conservation of charge : The total amount of charge in an isolated system remains constant. Charge may be transferred from one part of the system to another, but the net charge always remains the same. It means that charge can neither be created nor be destroyed.

Examples :

(a) When a glass rod is rubbed with silk, a positive charge appears on the rod and at the same time an equal negative charge appears on the silk. The net charge on the rod-silk system is zero both before and after rubbing.

(b) When an energetic γ-ray photon (charge = zero) interacts with the nuclear electric field inside an atom, it is converted into an electron (charge = − e) and a positron (charge = + e),

$\gamma \longrightarrow \text e^- + \text e^+ \qquad \text{(pair production)}$

The net charge is zero both before and after the event.

(c) In radioactive decay the uranium nucleus (charge = + 92e) is converted into a thorium nucleus (charge = + 90e) with the emission of an α-particle (charge = + 2e),

$$\begin{aligned}_{92}\text U^{238} \longrightarrow \\ _{90}\text{Th}^{234} + \\ _{2}\text{He}^{4}\end{aligned}$$

The total charge is + 92e both before and after the decay.

(ii) Principle of superposition : Coulomb's law gives the electric force between two point-charges. If a system contains a number of interacting charges, then the net force on any one charge due to the other charges is the vector sum of all the forces exerted on it by all the other charges. The individual forces are unaffected by the presence of the other charges.

For a system of n point-charges q₁, q₂, ..., q_n, the force on q₁ due to all the other charges is

$$\begin{aligned}\vec{\text F}_1 = \vec{\text F}_{12} + \vec{\text F}_{13} + \dots + \vec{\text F}_{1\text n} \\[1em] \vec{\text F}_1 = \frac{\text q_1}{4\pi\varepsilon_0}\left[\frac{\text q_2}{\text r_{12}^{\ 2}}\hat{\text r}_{12} + \frac{\text q_3}{\text r_{13}^{\ 2}}\hat{\text r}_{13} + \dots + \frac{\text q_\text n}{\text r_{1\text n}^{\ 2}}\hat{\text r}_{1\text n}\right]\end{aligned}$$

Here $\text r_{1\text j}$ is the distance between $\text q_1$ and $\text q_\text j$, and $\hat{\text r}_{1\text j}$ is the unit vector directed from $\text q_\text j$ towards $\text q_1$, for $\text j = 2, 3, \ldots, \text n$.

Thus the force which two charges exert on each other is not changed by the presence of the other charges.

Coulomb's law : Two stationary point-charges repel or attract each other with a force which is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This force acts along the line joining the two charges.

Thus, if two point-charges q₁ and q₂ are situated a distance r apart,

$\text F \propto \frac{\text q_1\text q_2}{\text r^2} \quad \text{or} \quad \text F = \text k\frac{\text q_1\text q_2}{\text r^2}$

In SI units, with force in newton, distance in metre and charge in coulomb, and the charges placed in vacuum (or air),

$$\begin{aligned}\text k = \frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9\ \text{N-m}^2\ \text C^{-2} \\[1em] \text F = \frac{1}{4\pi\varepsilon_0}\frac{\text q_1\text q_2}{\text r^2}\ \text{newton}\end{aligned}$$

where ε₀ = 8.85 × 10^-12 C² N^-1-m^-2 is the permittivity of free space.

Similarities between electrostatic and gravitational interactions :

(i) Both are inverse-square forces, $\text F \propto \frac{1}{\text r^2}$, the charge q playing the same role in Coulomb's law as the mass m does in the law of gravitation.

(ii) Both are central forces, acting along the line joining the two bodies, and both obey Newton's third law of motion.

Dissimilarities between electrostatic and gravitational interactions :

Property	Electrostatic interaction	Gravitational interaction
Nature	It may be attractive or repulsive because charge is of two types, positive and negative.	It is always attractive because mass is of one type.
Effect of medium	It depends on the medium and decreases K times in a medium of dielectric constant K.	It is independent of the medium; G is a universal constant.

Coulomb's law in vector form : Let two charges q₁ and q₂ be placed at the points A and B with position vectors $\vec{\text r}_1$ and $\vec{\text r}_2$ respectively.

Let $\hat{\text r}_{12}$ be the unit vector pointing from q₂ towards q₁ and $\hat{\text r}_{21}$ the unit vector pointing from q₁ towards q₂, and let r be the distance between the charges,

$\text r = \left|\vec{\text r}_1 - \vec{\text r}_2\right|$

Then the force on the charge q₁ due to the charge q₂ is

$\vec{\text F}_{12} = \frac{1}{4\pi\varepsilon_0}\frac{\text q_1\text q_2}{\text r^2}\ \hat{\text r}_{12} \qquad \dots(\text i)$

and the force on the charge q₂ due to the charge q₁ is

$\vec{\text F}_{21} = \frac{1}{4\pi\varepsilon_0}\frac{\text q_1\text q_2}{\text r^2}\ \hat{\text r}_{21} \qquad \dots(\text{ii})$

These two equations hold for all conditions of q₁ and q₂. If the charges are of like sign (q₁q₂ > 0), the forces are repulsive, and if they are of opposite sign (q₁q₂ < 0), the forces are attractive.

Note: The lower panel of printed Fig. 7 reverses the displacement arrows. Geometrically, $\hat{\text r}_{12}$ must point from B to A and $\hat{\text r}_{21}$ from A to B.

Importance of the vector form : It gives the following additional information.

(i) Since $\hat{\text r}_{12}$ and $\hat{\text r}_{21}$ are unit vectors in opposite directions, $\hat{\text r}_{21} = -\hat{\text r}_{12}$. It then follows from equations (i) and (ii) that

$\vec{\text F}_{21} = -\vec{\text F}_{12}$

that is, the forces exerted by the two charges on each other are equal and opposite, so that Newton's third law of motion is obeyed.

(ii) The equations also express that the forces act along the line joining the two charges. That is, the electrostatic force is a central force.

Intensity of electric field due to a point-charge :

Suppose an isolated point-charge of + q coulomb is placed at a point O in a medium whose dielectric constant is K. Let P be a point at a distance r metre from O, at which the intensity of the electric field is to be determined.

Let a test-charge q₀ be placed at the point P. According to Coulomb's law, the electric force acting on q₀ is

$\text F = \frac{1}{4\pi\varepsilon_0\text K}\frac{\text q\text q_0}{\text r^2}\ \text{newton}$

The intensity of the electric field at a point is the force experienced by a unit positive point-charge placed at that point,

$\text E = \frac{\text F}{\text q_0}$

Substituting the value of F,

$\text E = \frac{1}{4\pi\varepsilon_0\text K}\frac{\text q}{\text r^2}\ \text{newton/coulomb}$

For vacuum (or air), K = 1, so that

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\text r^2}\ \text{newton/coulomb}$

where $\frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9$ N-m² C^-2.

In vector notation,

$\vec{\text E} = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\text r^2}\ \hat{\text r}$

where $\hat{\text r}$ is a unit vector pointing from the source-charge towards the point of observation.

The force on the test-charge due to the source-charge + q is a force of repulsion directed along OP. Hence the direction of $\vec{\text E} \space$at P is also along OP produced, that is, away from the charge + q. If the source-charge at O were − q, the direction of the field at P would have been along PO, that is, towards the charge.

The charge − q of mass m moves on a circular path of radius R around the fixed charge + Q. The necessary centripetal force is provided by the electrostatic force of attraction between the two charges.

Electrostatic force of attraction :

$\text F = \frac{1}{4\pi\varepsilon_0}\frac{\text Q\text q}{\text R^2}$

Centripetal force required :

$\text F_\text c = \frac{\text m\text v^2}{\text R}$

Equating the two,

$$\begin{aligned}\frac{\text m\text v^2}{\text R} = \frac{1}{4\pi\varepsilon_0}\frac{\text Q\text q}{\text R^2} \\[1em] \text v^2 = \frac{1}{4\pi\varepsilon_0}\frac{\text Q\text q}{\text m\text R} \qquad \dots(\text i)\end{aligned}$$

Period of revolution : The particle covers the circumference 2πR in one revolution, so

$$\begin{aligned}\text T = \frac{2\pi\text R}{\text v} \\[1em] \text T^2 = \frac{4\pi^2\text R^2}{\text v^2}\end{aligned}$$

Substituting the value of v² from equation (i),

$$\begin{aligned}\text T^2 = 4\pi^2\text R^2 \times \frac{4\pi\varepsilon_0\text m\text R}{\text Q\text q} \\[1em] \text T^2 = \frac{16\pi^3\varepsilon_0\text m\text R^3}{\text Q\text q}\end{aligned}$$

Taking the square root,

$$\begin{aligned}\text T = \sqrt{\frac{16\pi^3\varepsilon_0\text m\text R^3}{\text Q\text q}} \\[1em] \text T = 4\left[\frac{\pi^3\varepsilon_0\text m\text R^3}{\text Q\text q}\right]^{1/2}\end{aligned}$$

Hence, the period of revolution is $\text T = 4\left[\frac{\pi^3\varepsilon_0\text m\text R^3}{\text Q\text q}\right]^{1/2}$, which is the required result.

(i) The response of the student is correct.

Given,

• Mass of the oil drop, m = 1.08 × 10^-14 kg
• Electric field, E = 1.68 × 10⁵ N C^-1
• g = 10 m s^-2

For a stationary charged oil drop the upward electric force must balance the weight of the drop,

$\text{qE} = \text{mg}$

that is, the net force on the drop must be zero. Hence

$$\begin{aligned}\text q = \frac{\text{mg}}{\text E} \\[1em] = \frac{1.08 \times 10^{-14} \times 10}{1.68 \times 10^{5}} \\[1em] \text q = 6.4 \times 10^{-19}\ \text C\end{aligned}$$

(ii) The response of the student is incorrect.

The electric field between the two parallel metal plates is

$\text E = \frac{\text V}{\text d}$

where V is the potential difference across the plates and d the separation between them. The student increases the distance d while keeping the source potential V unchanged, which decreases E instead of increasing it. To increase the field strength the plates should be brought closer together, or the source potential increased.

(iii) The response of the student is correct.

By the quantisation of charge, q = ± ne, where n is an integer. Here

$$\begin{aligned}\text n = \frac{|\text q|}{\text e} \\[1em] = \frac{7.2 \times 10^{-19}}{1.6 \times 10^{-19}} \\[1em] \text n = 4.5\end{aligned}$$

Since n comes out to be 4.5, which is not an integer, the charge cannot be in a decimal or fraction. Hence such a charge cannot be given to an oil drop.

Electric dipole : An electric dipole is a pair of equal and opposite point-charges placed at a short distance apart.

Electric dipole moment : The product of one charge and the distance between the charges is called the magnitude of the electric dipole moment,

$\vec{\text p} = \text q \times 2\vec{\text l}$

It is a vector quantity directed along the line joining the two charges, pointing from the negative charge to the positive charge. Its SI unit is coulomb-metre (C-m).

Electric field on the perpendicular bisector (broad-side-on position) :

Let the dipole AB consist of charges − q and + q separated by a distance 2l, and let P be a point on the right bisector of the dipole at a distance r from its mid-point O. The distance of P from each charge is $\sqrt{\text r^2 + \text l^2}$.

The magnitudes of the intensities at P due to the two charges are

$$\begin{aligned}\text E_1 = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\left(\text r^2 + \text l^2\right)} \quad \text{(along BP)} \\[1em] \text E_2 = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\left(\text r^2 + \text l^2\right)} \quad \text{(along PA)}\end{aligned}$$

The magnitudes of E₁ and E₂ are equal, but their directions are different. Resolving each into components parallel and perpendicular to AB, the components perpendicular to AB, E₁ sin θ and E₂ sin θ, are equal and opposite and hence cancel each other, while the components parallel to AB, E₁ cos θ and E₂ cos θ, are in the same direction and add up.

Hence the resultant intensity at P is

$$\begin{aligned}\text E = \text E_1\cos\theta + \text E_2\cos\theta \\[1em] \text E = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\left(\text r^2 + \text l^2\right)}\ 2\cos\theta\end{aligned}$$

From the figure,

$\cos\theta = \frac{\text l}{\sqrt{\text r^2 + \text l^2}}$

Substituting,

$$\begin{aligned}\text E = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\left(\text r^2 + \text l^2\right)} \times \frac{2\text l}{\left(\text r^2 + \text l^2\right)^{1/2}} \\[1em] \text E = \frac{1}{4\pi\varepsilon_0}\frac{2\text q\text l}{\left(\text r^2 + \text l^2\right)^{3/2}}\end{aligned}$$

But 2ql = p, the electric dipole moment,

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\left(\text r^2 + \text l^2\right)^{3/2}}$

The direction of $\vec{\text E} \space$at every point of the equatorial line is antiparallel to the dipole axis.

For a short dipole, or when r >> l, the term l² may be neglected in comparison with r²,

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\text r^3}, \qquad \text{in vector form} \quad \vec{\text E} = -\frac{1}{4\pi\varepsilon_0}\frac{\vec{\text p}}{\text r^3}$

Electric dipole moment is a vector quantity.

Its magnitude is p = q × 2l, the product of one charge and the distance between the charges. Its direction is along the line joining the two charges, pointing from the negative charge − q to the positive charge + q. This direction is known as the direction of the dipole axis.

Electric field at a point on the axis of the dipole (end-on position) :

Let the dipole AB be made up of charges − q and + q separated by a distance 2l, and let P be an exterior point on the axis at a distance r from the mid-point O of the dipole, where r > l. The distance of P from + q is (r − l) and from − q is (r + l).

The magnitudes of the intensities at P due to the two charges are

$$\begin{aligned}\text E_1 = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{(\text r - \text l)^2} \quad \text{(along the dipole axis)} \\[1em] \text E_2 = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{(\text r + \text l)^2} \quad \text{(opposite to the dipole axis)}\end{aligned}$$

E₁ and E₂ are along the same line in opposite directions, and E₁ > E₂. Hence the resultant intensity is their difference, directed along the dipole axis,

$$\begin{aligned}\text E = \text E_1 - \text E_2 \\[1em] = \frac{\text q}{4\pi\varepsilon_0}\left[\frac{1}{(\text r - \text l)^2} - \frac{1}{(\text r + \text l)^2}\right] \\[1em] = \frac{\text q}{4\pi\varepsilon_0}\left[\frac{(\text r + \text l)^2 - (\text r - \text l)^2}{\left(\text r^2 - \text l^2\right)^2}\right] \\[1em] = \frac{\text q}{4\pi\varepsilon_0}\frac{4\text l\text r}{\left(\text r^2 - \text l^2\right)^2} \\[1em] \text E = \frac{1}{4\pi\varepsilon_0}\frac{2(2\text q\text l)\text r}{\left(\text r^2 - \text l^2\right)^2}\end{aligned}$$

But 2ql = p, the electric dipole moment,

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{2\text p\text r}{\left(\text r^2 - \text l^2\right)^2}$

The direction of $\vec{\text E} \space$at every axial point outside the interval between the charges is along the dipole axis, from the negative charge towards the positive charge.

For a short dipole, or when r >> l, l² may be neglected in comparison with r²,

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{2\text p}{\text r^3}, \qquad \text{in vector form} \quad \vec{\text E} = \frac{1}{4\pi\varepsilon_0}\frac{2\vec{\text p}}{\text r^3}$

Dipole moment of an electric dipole : The dipole moment of an electric dipole is a vector whose magnitude is the product of one charge and the distance between the two charges,

$\text p = \text q \times 2\text l$

and whose direction is along the dipole axis, from the negative charge to the positive charge.

Field of a short dipole along its axis : For a point at a distance r on the axis, the intensities due to + q and − q are oppositely directed, so that

$$\begin{aligned}\text E_{\text{axial}} = \frac{\text q}{4\pi\varepsilon_0}\left[\frac{1}{(\text r - \text l)^2} - \frac{1}{(\text r + \text l)^2}\right] \\[1em] \text E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\frac{2\text p\text r}{\left(\text r^2 - \text l^2\right)^2}\end{aligned}$$

For a short dipole, l² is negligible compared with r²,

$\text E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\frac{2\text p}{\text r^3} \qquad \dots(\text i)$

Field of a short dipole along its equatorial axis : For a point at the same distance r on the right bisector, the components perpendicular to the dipole axis cancel and the parallel components add up,

$$\begin{aligned}\text E_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\left(\text r^2 + \text l^2\right)}\ 2\cos\theta \\[1em] \text E_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\left(\text r^2 + \text l^2\right)^{3/2}}\end{aligned}$$

For a short dipole,

$\text E_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\text r^3} \qquad \dots(\text{ii})$

Comparison : Dividing equation (i) by equation (ii),

$$\begin{aligned}\frac{\text E_{\text{axial}}}{\text E_{\text{equatorial}}} = \frac{\frac{2\text p}{4\pi\varepsilon_0\text r^3}}{\frac{\text p}{4\pi\varepsilon_0\text r^3}} \\[1em] \frac{\text E_{\text{axial}}}{\text E_{\text{equatorial}}} = 2\end{aligned}$$

Hence, at the same distance r from a short dipole, the electric field intensity along the axis is twice the intensity along the equatorial axis, the two being oppositely directed.

Given,

• Mass of each ball = M, charges + Q and − Q
• Length of each string = L
• Separation of the suspension points, S₁S₂ = 3x
• Separation of the balls, AB = x

Since the balls carry opposite charges they attract each other, so each string is drawn inwards through a horizontal distance

$\frac{3\text x - \text x}{2} = \text x$

If θ is the angle each string makes with the vertical, then

$\sin\theta = \frac{\text x}{\text L} \qquad \dots(\text i)$

Forces acting on the ball A :

• Weight Mg, vertically downwards
• Tension T along the string
• Electrostatic force of attraction F, horizontally towards B, where

$\text F = \frac{1}{4\pi\varepsilon_0}\frac{\text Q^2}{\text x^2} \qquad \dots(\text{ii})$

For the equilibrium of the ball, resolving the tension into horizontal and vertical components,

$$\begin{aligned}\text T\sin\theta = \text F \qquad \dots(\text{iii}) \\[1em] \text T\cos\theta = \text{Mg} \qquad \dots(\text{iv})\end{aligned}$$

(a) For the tension, substituting sin θ from equation (i) and F from equation (ii) into equation (iii),

$$\begin{aligned}\text T \times \frac{\text x}{\text L} = \frac{1}{4\pi\varepsilon_0}\frac{\text Q^2}{\text x^2} \\[1em] \text T = \frac{\text Q^2\text L}{4\pi\varepsilon_0\text x^3}\end{aligned}$$

(b) For the angle, dividing equation (iii) by equation (iv),

$$\begin{aligned}\tan\theta = \frac{\text F}{\text{Mg}} \\[1em] \tan\theta = \frac{1}{4\pi\varepsilon_0}\frac{\text Q^2}{\text x^2\text{Mg}} \\[1em] \tan\theta = \frac{\text Q^2}{4\pi\varepsilon_0\text x^2\text{Mg}}\end{aligned}$$

Hence both the required results are proved.

Given,

• Length of the rod = a, pivoted at its centre
• Charges at the left and right ends = 2q and q respectively
• A fixed sphere of charge + Q lies a distance d directly below each end
• The block of weight W is at a distance x from the right end

Upward electrostatic forces on the two ends : Each end-charge is repelled vertically upwards by the fixed charge + Q below it, at a distance d,

$\text F_\text L = \frac{1}{4\pi\varepsilon_0}\frac{\text Q(2\text q)}{\text d^2}, \qquad \text F_\text R = \frac{1}{4\pi\varepsilon_0}\frac{\text Q\text q}{\text d^2}$

(i) Position of the block :

Since F_L > F_R, the charges alone tend to raise the left end, so the block must lie on the left of the pivot. Its distance from the pivot is $\left(\text x - \frac{\text a}{2}\right)$, since it is at a distance x from the right end.

Taking moments about the pivot at the centre, each end being at a distance $\frac{\text a}{2}$ from it,

$$\begin{aligned}\text F_\text L \times \frac{\text a}{2} = \text F_\text R \times \frac{\text a}{2} + \text W\left(\text x - \frac{\text a}{2}\right) \\[1em] \frac{\text a}{2}\left(\text F_\text L - \text F_\text R\right) = \text W\left(\text x - \frac{\text a}{2}\right)\end{aligned}$$

Substituting the forces,

$$\begin{aligned}\frac{\text a}{2}\times\frac{1}{4\pi\varepsilon_0}\frac{\text Q\text q}{\text d^2} = \text W\left(\text x - \frac{\text a}{2}\right) \\[1em] \text x - \frac{\text a}{2} = \frac{\text a}{2}\times\frac{1}{4\pi\varepsilon_0}\frac{\text q\text Q}{\text W\text d^2} \\[1em] \text x = \frac{\text a}{2}\left[1 + \frac{1}{4\pi\varepsilon_0}\frac{\text q\text Q}{\text W\text d^2}\right]\end{aligned}$$

(ii) Value of d for no vertical force on the bearing :

The rod is massless, so the vertical forces on it are the two upward electrostatic forces and the downward weight W of the block. If the bearing is to exert no vertical force,

$$\begin{aligned}\text F_\text L + \text F_\text R = \text W \\[1em] \frac{1}{4\pi\varepsilon_0}\frac{2\text q\text Q}{\text d^2} + \frac{1}{4\pi\varepsilon_0}\frac{\text q\text Q}{\text d^2} = \text W \\[1em] \frac{3\text q\text Q}{4\pi\varepsilon_0\text d^2} = \text W \\[1em] \text d^2 = \frac{3\text q\text Q}{4\pi\varepsilon_0\text W} \\[1em] \text d = \sqrt{\frac{3\text q\text Q}{4\pi\varepsilon_0\text W}}\end{aligned}$$

(i) Ratio of the magnitudes of the charges :

The number of lines of force starting from or ending on a charge is proportional to the magnitude of that charge. In the figure, twelve lines of force originate from q₁ while only six of them terminate on q₂, the remaining six going off to infinity. Hence

$$\begin{aligned}\frac{|\text q_1|}{|\text q_2|} = \frac{12}{6} \\[1em] \frac{|\text q_1|}{|\text q_2|} = 2\end{aligned}$$

(ii) Signs of the charges :

The lines of force originate on positive charges and terminate on negative charges. The arrows in the figure point away from q₁ and towards q₂. Hence q₁ is positive and q₂ is negative.

(iii) Position of the neutral point :

Given, the separation between the charges is $\text d = 5\left(\sqrt{2} - 1\right)$ cm.

Since the charges are of opposite signs and |q₁| > |q₂|, the neutral point must lie outside the two charges, on the side of the smaller charge q₂. Let it be at a distance y from q₂, so that its distance from q₁ is (d + y).

At the neutral point the two field intensities are equal in magnitude,

$$\begin{aligned}\frac{1}{4\pi\varepsilon_0}\frac{|\text q_1|}{(\text d + \text y)^2} = \frac{1}{4\pi\varepsilon_0}\frac{|\text q_2|}{\text y^2} \\[1em] \frac{(\text d + \text y)^2}{\text y^2} = \frac{|\text q_1|}{|\text q_2|} \\[1em] \frac{(\text d + \text y)^2}{\text y^2} = 2 \\[1em] \frac{\text d + \text y}{\text y} = \sqrt{2} \\[1em] \text d = \text y\left(\sqrt{2} - 1\right) \\[1em] \text y = \frac{\text d}{\sqrt{2} - 1} \\[1em] = \frac{5\left(\sqrt{2} - 1\right)}{\sqrt{2} - 1} \\[1em] \text y = 5\ \text{cm}\end{aligned}$$

Hence, the neutral point lies at a distance of 5 cm from q₂, on the side away from q₁.

(iv) Will the positive charge follow the line of force?

No. The lines of force here are curved. A free positive charge is accelerated at every point along the tangent to the line of force at that point, but it already possesses a velocity acquired earlier. The resultant path therefore deviates from the line of force. A charged particle follows a line of force only when the lines of force are straight and the particle starts from rest.

(i) Value of Q in terms of q :

From the figure the system consists of the charge Q at the origin O and two charges + q, one at A on the X-axis and the other at B on the Y-axis, each at a distance a from O.

A system of charges is equivalent to an electric dipole if the algebraic sum of its charges is zero. Hence

$$\begin{aligned}\text Q + \text q + \text q = 0 \\[1em] \text Q = -2\text q\end{aligned}$$

(ii) Electric dipole moment of the system :

The charge − 2q at O may be regarded as two charges of − q each placed at O. One of them together with the charge + q at A forms a dipole of moment qa directed along OA, and the other together with the charge + q at B forms a dipole of moment qa directed along OB.

The two component dipole moments are equal in magnitude and mutually perpendicular, so their resultant is

$$\begin{aligned}\text p = \sqrt{(\text q\text a)^2 + (\text q\text a)^2} \\[1em] \text p = \sqrt{2}\ \text q\text a\end{aligned}$$

Its direction makes an angle of 45° with OA, that is, it is directed along the line joining O to the mid-point of AB.

(iii) SI unit of electric dipole moment :

The SI unit of electric dipole moment is the coulomb-metre (C-m).

(iv) Net electric force on a dipole in a uniform electric field :

In a uniform field the charge + q experiences a force qE along the field and the charge − q an equal force qE opposite to the field. These two forces are equal and opposite, so the net electric force on the dipole is zero.

(v) Condition for maximum torque :

The torque on a dipole placed in a uniform electric field is

$\tau = \text{pE}\sin\theta$

This is maximum when sin θ = 1, that is, when θ = 90°.

Hence, the dipole experiences maximum torque, τ_max = pE, when $\vec{\text p}$ is perpendicular to $\vec{\text E} \space$.

Given,

• Total negative charge, q = 1 C
• Charge on one electron, e = 1.6 × 10^-19 C

By the quantisation of charge, q = ne, where n is the number of electrons.

$$\begin{aligned}\text n = \frac{\text q}{\text e} \\[1em] \text n = \frac{1}{1.6 \times 10^{-19}} \\[1em] \text n = 6.25 \times 10^{18}\end{aligned}$$

Hence, 6.25 × 10¹⁸ electrons are required.

Given,

• Number of electrons removed, n = 6.25 × 10¹²
• Charge on one electron, e = 1.6 × 10^-19 C
• Mass of one electron, m_e = 9.0 × 10^-31 kg

(i) Charge acquired by the glass rod :

Removal of electrons leaves a deficiency of electrons, so the rod becomes positively charged. By the quantisation of charge,

$$\begin{aligned}\text q = \text{ne} \\[1em] \text q = 6.25 \times 10^{12} \times 1.6 \times 10^{-19} \\[1em] \text q = 1.0 \times 10^{-6}\ \text C \\[1em] \text q = 1.0\ \mu\text C\end{aligned}$$

Hence, the rod acquires a charge of 1.0 μC, which is positive.

(ii) Change in the mass of the rod :

Since the electrons carry away their mass with them, the mass of the rod decreases by

$$\begin{aligned}\Delta\text m = \text n\text m_\text e \\[1em] \Delta\text m = 6.25 \times 10^{12} \times 9.0 \times 10^{-31} \\[1em] \Delta\text m = 5.625 \times 10^{-18}\ \text{kg}\end{aligned}$$

Hence, the mass of the rod decreases by 5.625 × 10^-18 kg.

Yes, the law of conservation of charge is obeyed in the given reaction.

Explanation:
By the law of conservation of charge, the total charge of an isolated system remains constant before and after the reaction.

The charge of a nucleus is Ze, where Z is the atomic number written as the subscript.

Total charge before the reaction :

$$\begin{aligned}\text q_\text i = (20\text e) + (1\text e) \\[1em] \text q_\text i = 21\text e\end{aligned}$$

Total charge after the reaction :

$$\begin{aligned}\text q_\text f = (21\text e) + (0) \\[1em] \text q_\text f = 21\text e\end{aligned}$$

Since q_i = q_f, the net charge is the same before and after the reaction.

Given,

• Charge on a proton, q₁ = + 1.6 × 10^-19 C
• Charge on an electron, q₂ = − 1.6 × 10^-19 C
• Separation, r = 0.8 × 10^-15 m

By Coulomb's law, the magnitude of the force is

$\text F = \frac{1}{4\pi\varepsilon_0}\frac{|\text q_1\text q_2|}{\text r^2}$

Substituting the values,

$$\begin{aligned}\text F = \frac{\left(9.0 \times 10^9\right)\left(1.6 \times 10^{-19}\right)\left(1.6 \times 10^{-19}\right)}{\left(0.8 \times 10^{-15}\right)^2} \\[1em] \text F = \frac{9.0 \times 10^9 \times 2.56 \times 10^{-38}}{6.4 \times 10^{-31}} \\[1em] \text F = \frac{2.304 \times 10^{-28}}{6.4 \times 10^{-31}} \\[1em] \text F = 360\ \text N\end{aligned}$$

Hence, the Coulombian force is 360 N and it is attractive, since the two charges are of opposite signs.

Given,

• Charge on each particle, e = 1.6 × 10^-19 C
• Mass of electron, m_e = 9.0 × 10^-31 kg
• Mass of proton, m_p = 1.7 × 10^-27 kg
• g = 9.8 m s^-2

The Coulomb force between the two particles is to be equal to the weight of one particle,

$$\begin{aligned}\frac{1}{4\pi\varepsilon_0}\frac{\text e^2}{\text r^2} = \text{mg} \\[1em] \text r = \sqrt{\frac{1}{4\pi\varepsilon_0}\frac{\text e^2}{\text{mg}}}\end{aligned}$$

For two electrons :

$$\begin{aligned}\text r^2 = \frac{\left(9.0 \times 10^9\right)\left(1.6 \times 10^{-19}\right)^2}{9.0 \times 10^{-31} \times 9.8} \\[1em] \text r^2 = \frac{2.304 \times 10^{-28}}{8.82 \times 10^{-30}} \\[1em] \text r^2 = 26.12 \\[1em] \text r = 5.1\ \text m\end{aligned}$$

For two protons :

$$\begin{aligned}\text r^2 = \frac{\left(9.0 \times 10^9\right)\left(1.6 \times 10^{-19}\right)^2}{1.7 \times 10^{-27} \times 9.8} \\[1em] \text r^2 = \frac{2.304 \times 10^{-28}}{1.666 \times 10^{-26}} \\[1em] \text r^2 = 1.383 \times 10^{-2} \\[1em] \text r = 0.117\ \text m\end{aligned}$$

Hence, the separation is 5.1 m for two electrons and 0.117 m for two protons.

Given,

• Separation, r = 0.1 m
• Force of repulsion, F = 18 N
• Sum of the charges, q₁ + q₂ = 9 μC

By Coulomb's law,

$$\begin{aligned}\text F = \frac{1}{4\pi\varepsilon_0}\frac{\text q_1\text q_2}{\text r^2} \\[1em] 18 = \frac{\left(9.0 \times 10^9\right)\text q_1\text q_2}{(0.1)^2} \\[1em] \text q_1\text q_2 = \frac{18 \times 0.01}{9.0 \times 10^9} \\[1em] \text q_1\text q_2 = 2 \times 10^{-11}\ \text C^2 \\[1em] \text q_1\text q_2 = 20\ (\mu\text C)^2\end{aligned}$$

So the two charges satisfy

$\text q_1 + \text q_2 = 9\ \mu\text C, \qquad \text q_1\text q_2 = 20\ (\mu\text C)^2$

They are therefore the roots of the quadratic equation

$$\begin{aligned}\text q^2 - (9\ \mu\text C)\text q + 20\ (\mu\text C)^2 = 0 \\[1em] \text q^2 - (5\ \mu\text C)\text q - (4\ \mu\text C)\text q + 20\ (\mu\text C)^2 = 0 \\[1em] (\text q - 5\ \mu\text C)(\text q - 4\ \mu\text C) = 0 \\[1em] \text q = 5\ \mu\text C \quad \text{or} \quad \text q = 4\ \mu\text C\end{aligned}$$

Hence, the two charges are 5 μC and 4 μC.

Given,

• Separation, r = 1 mm = 1 × 10^-3 m
• Force of repulsion, F = 4.0 × 10^-5 N
• If the charge on one particle is q, that on the other is 4q
• Charge on an electron, $\text e = \frac{5}{3} \times 10^{-19}$ C

By Coulomb's law,

$$\begin{aligned}\text F = \frac{1}{4\pi\varepsilon_0}\frac{\text q \times 4\text q}{\text r^2} \\[1em] 4.0 \times 10^{-5} = \frac{\left(9.0 \times 10^9\right)\left(4\text q^2\right)}{\left(1 \times 10^{-3}\right)^2} \\[1em] 4\text q^2 = \frac{4.0 \times 10^{-5} \times 10^{-6}}{9.0 \times 10^9} \\[1em] 4\text q^2 = \frac{4.0 \times 10^{-11}}{9.0 \times 10^9} \\[1em] \text q^2 = 1.111 \times 10^{-21} \\[1em] \text q = 3.333 \times 10^{-11}\ \text C\end{aligned}$$

The other charge is 4q = 1.333 × 10^-10 C.

Number of electrons on the first particle :

$$\begin{aligned}\text n_1 = \frac{\text q}{\text e} \\[1em] \text n_1 = \frac{3.333 \times 10^{-11}}{\frac{5}{3} \times 10^{-19}} \\[1em] \text n_1 = 2 \times 10^{8}\end{aligned}$$

Number of electrons on the second particle :

$$\begin{aligned}\text n_2 = 4\text n_1 \\[1em] \text n_2 = 8 \times 10^{8}\end{aligned}$$

Hence, the two particles carry 2 × 10⁸ and 8 × 10⁸ electrons respectively.

Given,

• Initial charges, q₁ = + 2 μC and q₂ = + 6 μC
• Initial force of repulsion, F = 12 N
• Additional charge given to each = − 4 μC

New charges :

$$\begin{aligned}\text q_1' = 2\ \mu\text C - 4\ \mu\text C \\[1em] \text q_1' = -2\ \mu\text C \\[1em] \text q_2' = 6\ \mu\text C - 4\ \mu\text C \\[1em] \text q_2' = +2\ \mu\text C\end{aligned}$$

Since the separation r remains unchanged, by Coulomb's law the magnitude of the force $|\text F| \propto |\text q_1\text q_2|$,

$$\begin{aligned}\frac{|\text F'|}{|\text F|} = \frac{|\text q_1'\text q_2'|}{|\text q_1\text q_2|} \\[1em] \frac{|\text F'|}{|\text F|} = \frac{2 \times 2}{2 \times 6} \\[1em] \frac{|\text F'|}{|\text F|} = \frac{1}{3} \\[1em] |\text F'| = \frac{12}{3} \\[1em] |\text F'| = 4\ \text N\end{aligned}$$

Hence, the new force is 4 N, and it is a force of attraction since the charges are now of opposite signs.

Given,

• Charge on ball A, q_A = + 40 μC = 40 × 10^-6 C
• Charge on ball B, q_B = − 10 μC = − 10 × 10^-6 C
• Separation, r = 2.0 m

Initial force :

$$\begin{aligned}\text F = \frac{1}{4\pi\varepsilon_0}\frac{|\text q_\text A\text q_\text B|}{\text r^2} \\[1em] \text F = \frac{\left(9.0 \times 10^9\right)\left(40 \times 10^{-6}\right)\left(10 \times 10^{-6}\right)}{(2.0)^2} \\[1em] \text F = \frac{9.0 \times 10^9 \times 4.0 \times 10^{-10}}{4} \\[1em] \text F = \frac{3.6}{4} \\[1em] \text F = 0.90\ \text N\end{aligned}$$

Since the charges are of opposite signs, this is a force of attraction.

After the balls are touched : The balls are identical, so the charge is shared equally,

$$\begin{aligned}\text q' = \frac{\text q_\text A + \text q_\text B}{2} \\[1em] \text q' = \frac{40 - 10}{2} \\[1em] \text q' = +15\ \mu\text C \\[1em] \text F' = \frac{\left(9.0 \times 10^9\right)\left(15 \times 10^{-6}\right)^2}{(2.0)^2} \\[1em] \text F' = \frac{9.0 \times 10^9 \times 2.25 \times 10^{-10}}{4} \\[1em] \text F' = \frac{2.025}{4} \\[1em] \text F' = 0.51\ \text N\end{aligned}$$

Both the balls now carry like charges, so this is a force of repulsion.

If one ball is connected to earth : Its charge flows to the earth and it becomes uncharged. With one of the charges reduced to zero, the force between the balls is zero.

Hence, the forces are 0.90 N (attraction), 0.51 N (repulsion) and zero respectively.

Given,

• Spheres A and B carry equal charges, each q, separated by a distance r
• Force of repulsion between A and B, F = 2.0 × 10^-5 N
• Sphere C is identical and initially uncharged

$$\begin{aligned}\text F = \frac{1}{4\pi\varepsilon_0}\frac{\text q^2}{\text r^2} \\[1em] \text F = 2.0 \times 10^{-5}\ \text N \qquad \dots(\text i)\end{aligned}$$

Charge after C touches A : Since A and C are identical, the charge q is shared equally,

$\text q_\text A = \text q_\text C = \frac{\text q}{2}$

The charge on B remains q. The sphere C is now placed at the mid-point, so it is at a distance $\frac{\text r}{2}$ from both A and B.

Force on C due to A :

$$\begin{aligned}\text F_1 = \frac{1}{4\pi\varepsilon_0}\frac{\left(\frac{\text q}{2}\right)\left(\frac{\text q}{2}\right)}{\left(\frac{\text r}{2}\right)^2} \\[1em] \text F_1 = \frac{1}{4\pi\varepsilon_0}\frac{\text q^2}{\text r^2} \\[1em] \text F_1 = \text F \quad \text{(directed from A to B)}\end{aligned}$$

Force on C due to B :

$$\begin{aligned}\text F_2 = \frac{1}{4\pi\varepsilon_0}\frac{\left(\frac{\text q}{2}\right)(\text q)}{\left(\frac{\text r}{2}\right)^2} \\[1em] \text F_2 = \frac{1}{4\pi\varepsilon_0}\frac{2\text q^2}{\text r^2} \\[1em] \text F_2 = 2\text F \quad \text{(directed from B to A)}\end{aligned}$$

Net force on C : The two forces are oppositely directed,

$$\begin{aligned}\text F_{\text{net}} = 2\text F - \text F \\[1em] \text F_{\text{net}} = \text F \\[1em] \text F_{\text{net}} = 2.0 \times 10^{-5}\ \text N\end{aligned}$$

Hence, the net electric force on C is 2.0 × 10^-5 N, directed towards A.

Given,

• Charge placed below the bob, Q = 2.0 × 10^-4 C
• Separation, r = 10 cm = 0.10 m
• Mass of the bob, m = 100 g = 0.1 kg
• g = 9.8 m s^-2

The string becomes loose when its tension falls to zero. This happens when the upward electrostatic force of repulsion on the bob just balances its weight,

$\frac{1}{4\pi\varepsilon_0}\frac{\text Q\text q}{\text r^2} = \text{mg}$

Substituting the values,

$$\begin{aligned}\frac{\left(9.0 \times 10^9\right)\left(2.0 \times 10^{-4}\right)\text q}{(0.10)^2} = 0.1 \times 9.8 \\[1em] \frac{1.8 \times 10^6 \times \text q}{0.01} = 0.98 \\[1em] 1.8 \times 10^8 \times \text q = 0.98 \\[1em] \text q = \frac{0.98}{1.8 \times 10^8} \\[1em] \text q = 5.4 \times 10^{-9}\ \text C\end{aligned}$$

Hence, the bob should be given a charge of 5.4 × 10^-9 C, of the same sign as Q so that the force is repulsive.

Given,

• The n^th particle is at x_n = 10n cm = 0.1n m, with n = 1, 2, 3, ..., 10
• The charges are 1 × 10^-8 C, 8 × 10^-8 C, 27 × 10^-8 C, ..., 1000 × 10^-8 C, that is, q_n = n³ × 10^-8 C
• Test charge at the origin, q₀ = 1 C

All the charges are positive and lie on the same side of the origin, so by the principle of superposition all the forces on the charge at the origin are in the same direction and simply add up.

Force due to the n^th particle :

$$\begin{aligned}\text F_\text n = \frac{1}{4\pi\varepsilon_0}\frac{\text q_\text n\text q_0}{\text x_\text n^{\ 2}} \\[1em] \text F_\text n = \left(9 \times 10^9\right)\frac{\text n^3 \times 10^{-8}}{(0.1\text n)^2} \\[1em] \text F_\text n = \left(9 \times 10^9\right)\frac{\text n^3 \times 10^{-8}}{0.01\text n^2} \\[1em] \text F_\text n = \left(9 \times 10^9\right)\left(10^{-6}\right)\text n \\[1em] \text F_\text n = 9 \times 10^3\ \text n\ \text N\end{aligned}$$

Total force :

$$\begin{aligned}\text F = \sum_{\text n = 1}^{10}\text F_\text n \\[1em] \text F = 9 \times 10^3 \sum_{\text n=1}^{10}\text n \\[1em] \text F = 9 \times 10^3 \times \frac{10 \times 11}{2} \\[1em] \text F = 9 \times 10^3 \times 55 \\[1em] \text F = 4.95 \times 10^5\ \text N\end{aligned}$$

Hence, the magnitude of the electric force on the 1 C charge at the origin is 4.95 × 10⁵ N.

Given,

• Each charge = + e = 1.6 × 10^-19 C
• Separation, r = 20 cm = 0.20 m
• Thickness of the dielectric slab, t = 8 cm = 0.08 m
• Dielectric constant, K = 4

When the space between the charges is partly filled with a dielectric slab of thickness t and dielectric constant K, the slab is equivalent to an air distance $\text t\sqrt{\text K}$. Hence the effective separation in air is

$\text r_{\text{eff}} = (\text r - \text t) + \text t\sqrt{\text K}$

Substituting the values,

$$\begin{aligned}\text r_{\text{eff}} = (0.20 - 0.08) + 0.08 \times \sqrt{4} \\[1em] \text r_{\text{eff}} = 0.12 + 0.16 \\[1em] \text r_{\text{eff}} = 0.28\ \text m\end{aligned}$$

The force between the charges is therefore

$$\begin{aligned}\text F = \frac{1}{4\pi\varepsilon_0}\frac{\text e^2}{\text r_{\text{eff}}^{\ 2}} \\[1em] \text F = \frac{\left(9.0 \times 10^9\right)\left(1.6 \times 10^{-19}\right)^2}{(0.28)^2} \\[1em] \text F = \frac{9.0 \times 10^9 \times 2.56 \times 10^{-38}}{0.0784} \\[1em] \text F = \frac{2.304 \times 10^{-28}}{0.0784} \\[1em] \text F = 2.94 \times 10^{-27}\ \text N\end{aligned}$$

Hence, the force between the two charges is 2.94 × 10^-27 N.

Given,

• ABC is an equilateral triangle of side 10 m, D is the mid-point of BC
• q_B = + 100 μC, q_C = − 100 μC, q_D = + 75 μC
• Test charge at A, q = + 1 μC

Distances : AB = AC = 10 m, and AD is the altitude of the equilateral triangle,

$$\begin{aligned}\text{AD} = \frac{\sqrt{3}}{2} \times 10 \\[1em] \text{AD} = 5\sqrt{3}\ \text m \\[1em] (\text{AD})^2 = 75\ \text m^2\end{aligned}$$

Force due to the charge at B : It is a repulsion, directed along BA,

$$\begin{aligned}\text F_\text B = \frac{\left(9 \times 10^9\right)\left(100 \times 10^{-6}\right)\left(1 \times 10^{-6}\right)}{(10)^2} \\[1em] \text F_\text B = \frac{9 \times 10^{-1}}{100} \\[1em] \text F_\text B = 9 \times 10^{-3}\ \text N\end{aligned}$$

Force due to the charge at C : It is an attraction of the same magnitude, directed along AC,

$\text F_\text C = 9 \times 10^{-3}\ \text N$

Resultant of F_B and F_C : These two forces are equal in magnitude and are inclined at 120° to each other, so their resultant is

$$\begin{aligned}\text F_1 = 2\text F_\text B\cos 60° \\[1em] \text F_1 = 2 \times 9 \times 10^{-3} \times \frac{1}{2} \\[1em] \text F_1 = 9 \times 10^{-3}\ \text N\end{aligned}$$

directed parallel to BC, from B towards C.

Force due to the charge at D : It is a repulsion, directed along DA, that is, perpendicular to BC,

$$\begin{aligned}\text F_\text D = \frac{\left(9 \times 10^9\right)\left(75 \times 10^{-6}\right)\left(1 \times 10^{-6}\right)}{75} \\[1em] \text F_\text D = \frac{6.75 \times 10^{-1}}{75} \\[1em] \text F_\text D = 9 \times 10^{-3}\ \text N\end{aligned}$$

Resultant force at A : F₁ and F_D are equal in magnitude and mutually perpendicular,

$$\begin{aligned}\text F = \sqrt{\text F_1^{\ 2} + \text F_\text D^{\ 2}} \\[1em] \text F = \sqrt{2}\times 9 \times 10^{-3} \\[1em] \text F = 9\sqrt{2} \times 10^{-3}\ \text N\end{aligned}$$

and it makes an angle

$$\begin{aligned}\tan\theta = \frac{\text F_\text D}{\text F_1} \\[1em] \tan\theta = 1 \\[1em] \theta = 45°\end{aligned}$$

Hence, the force on the + 1 μC charge at A is $9\sqrt{2} \times 10^{-3}$ N, acting at 45° to a line parallel to BC.

Given,

• q₁ = 1.0 μC, q₂ = − 0.25 μC
• Separation, r = 0.40 m

Since the two charges are of opposite signs, the point of zero force cannot lie between them. It must lie outside the two charges, on the side of the smaller charge q₂.

Let the third charge be placed at a distance x from q₂, so that its distance from q₁ is (0.40 + x).

For no force to act on the third charge, the two forces on it must be equal in magnitude,

$$\begin{aligned}\frac{1}{4\pi\varepsilon_0}\frac{|\text q_1|}{(0.40 + \text x)^2} = \frac{1}{4\pi\varepsilon_0}\frac{|\text q_2|}{\text x^2} \\[1em] \frac{(0.40 + \text x)^2}{\text x^2} = \frac{|\text q_1|}{|\text q_2|} \\[1em] \frac{(0.40 + \text x)^2}{\text x^2} = \frac{1.0}{0.25} \\[1em] \frac{(0.40 + \text x)^2}{\text x^2} = 4\end{aligned}$$

Taking the square root,

$$\begin{aligned}\frac{0.40 + \text x}{\text x} = 2 \\[1em] 0.40 + \text x = 2\text x \\[1em] \text x = 0.40\ \text m\end{aligned}$$

Hence, the third charge should be placed outside the second charge, at a distance of 0.40 m from the − 0.25 μC charge (that is, 0.80 m from the 1.0 μC charge).

Given,

• q₁ = + 9e, q₂ = + e
• Separation, r = 16 cm

Both the charges are positive, so the third charge q can remain in equilibrium only at a point between them, where the two forces on it are oppositely directed.

Let q be placed at a distance x from the charge + 9e, so that its distance from + e is (16 − x) cm.

For the equilibrium of q,

$$\begin{aligned}\frac{1}{4\pi\varepsilon_0}\frac{(9\text e)\text q}{\text x^2} = \frac{1}{4\pi\varepsilon_0}\frac{(\text e)\text q}{(16 - \text x)^2} \\[1em] \frac{9}{\text x^2} = \frac{1}{(16 - \text x)^2}\end{aligned}$$

Taking the square root of both sides,

$$\begin{aligned}\frac{3}{\text x} = \frac{1}{16 - \text x} \\[1em] 3(16 - \text x) = \text x \\[1em] 48 - 3\text x = \text x \\[1em] 4\text x = 48 \\[1em] \text x = 12\ \text{cm}\end{aligned}$$

Hence, the third charge should be placed at a distance of 12 cm from the charge + 9e (and 4 cm from the charge + e).

Given,

• q₁, q₂ and q₃ lie in a line at equal distances, say d apart
• q₂ and q₃ are opposite in sign
• Net force on q₃ is zero

The distance of q₃ from q₂ is d and from q₁ is 2d.

For the net force on q₃ to be zero, the forces due to q₁ and q₂ must be equal in magnitude and opposite in direction,

$$\begin{aligned}\frac{1}{4\pi\varepsilon_0}\frac{\text q_1\text q_3}{(2\text d)^2} + \frac{1}{4\pi\varepsilon_0}\frac{\text q_2\text q_3}{\text d^2} = 0 \\[1em] \frac{\text q_1}{4} + \text q_2 = 0 \\[1em] \text q_1 = -4\text q_2\end{aligned}$$

Hence the magnitude of q₁ is four times that of q₂, and its sign is opposite to that of q₂. Since q₂ and q₃ are opposite in sign, q₁ has the same sign as q₃.

Hence, q₁ has a magnitude four times that of q₂ and its sign is the same as that of q₃.

Given,

• Charges + q and + 4q, a distance L apart

(a) Location, magnitude and nature of the third charge :

The third charge Q must lie between the two charges, since only there can the two forces on it be oppositely directed. Let it be at a distance x from + q, so that its distance from + 4q is (L − x).

For the equilibrium of Q,

$$\begin{aligned}\frac{1}{4\pi\varepsilon_0}\frac{\text q\text Q}{\text x^2} = \frac{1}{4\pi\varepsilon_0}\frac{(4\text q)\text Q}{(\text L - \text x)^2} \\[1em] \frac{1}{\text x^2} = \frac{4}{(\text L - \text x)^2} \\[1em] \frac{\text L - \text x}{\text x} = 2 \\[1em] \text L = 3\text x \\[1em] \text x = \frac{\text L}{3}\end{aligned}$$

Thus Q is at a distance $\frac{\text L}{3}$ from + q and $\frac{2\text L}{3}$ from + 4q.

For the entire system to be in equilibrium, the charge + q must also be in equilibrium. The forces on it are the repulsion due to + 4q at a distance L and the force due to Q at a distance $\frac{\text L}{3}$,

$$\begin{aligned}\frac{1}{4\pi\varepsilon_0}\frac{\text q(4\text q)}{\text L^2} + \frac{1}{4\pi\varepsilon_0}\frac{\text q\text Q}{\left(\frac{\text L}{3}\right)^2} = 0 \\[1em] \frac{4\text q}{\text L^2} + \frac{9\text Q}{\text L^2} = 0 \\[1em] \text Q = -\frac{4\text q}{9}\end{aligned}$$

Hence, the third charge is $-\frac{4\text q}{9}$, that is, it is negative, and it is placed at a distance $\frac{\text L}{3}$ from + q and $\frac{2\text L}{3}$ from + 4q.

(b) The equilibrium is unstable :

Suppose the negative charge Q is displaced slightly along the line towards + q. The distance from + q decreases and the distance from + 4q increases, so the attraction due to + q increases while that due to + 4q decreases. The net force on Q is then towards + q, that is, in the direction of the displacement itself, and Q moves still farther from its equilibrium position.

Since the net force does not bring the charge back to its original position, the equilibrium is unstable.

Given,

• Four charges +q at the corners of a square of side a
• A charge Q is to be placed at the centre

Consider the charge +q at the lower-left corner. The distances of the other charges from it are

• a for the two adjacent corners,
• $\text a\sqrt{2}$ for the opposite corner,
• $\frac{\text a}{\sqrt{2}}$ for the centre of the square.

Forces due to the other three corner charges : Let $\text k = \frac{1}{4\pi\varepsilon_0}$.

The two adjacent charges exert repulsions F₁ and F₂, each of magnitude $\frac{\text{kq}^2}{\text a^2}$, and their components along the diagonal are each $\frac{\text{kq}^2}{\text a^2}\cos 45^\circ$. The charge at the opposite corner exerts a repulsion F₃ = $\frac{\text{kq}^2}{2\text a^2}$ directly along the diagonal.

Hence the resultant outward force F along the diagonal is

$$\begin{aligned}\text F = 2 \times \frac{\text{kq}^2}{\text a^2} \times \frac{1}{\sqrt{2}} + \frac{\text{kq}^2}{2\text a^2} \\[1em] \text F = \frac{\text{kq}^2}{\text a^2}\left(\sqrt{2} + \frac{1}{2}\right)\end{aligned}$$

Force due to the charge at the centre : It acts along the diagonal, of magnitude

$$\begin{aligned}\text F_\text Q = \frac{\text{kq}|\text Q|}{\left(\frac{\text a}{\sqrt{2}}\right)^2} \\[1em] \text F_\text Q = \frac{2\text{kq}|\text Q|}{\text a^2}\end{aligned}$$

Condition of equilibrium : F_Q must be directed inwards and must balance F,

$$\begin{aligned}\frac{2\text{kq}|\text Q|}{\text a^2} = \frac{\text{kq}^2}{\text a^2}\left(\sqrt{2} + \frac{1}{2}\right) \\[1em] |\text Q| = \frac{\text q}{2}\left(\sqrt{2} + \frac{1}{2}\right) \\[1em] |\text Q| = \frac{\text q\left(1 + 2\sqrt{2}\right)}{4}\end{aligned}$$

Since the force due to Q has to be an attraction, Q must be negative.

$\text Q = -\frac{\text q\left(1 + 2\sqrt{2}\right)}{4}$

The charge Q itself is in equilibrium for any value, as the four equal charges at the corners pull it equally from all sides.

Hence, a charge of $-\frac{\text q\left(1 + 2\sqrt{2}\right)}{4}$ should be placed at the centre of the square.

Given,

• The helium nucleus has a positive charge equal to that on an α-particle, q = + 2e = 3.2 × 10^-19 C
• Distance, r = 1 Å = 1 × 10^-10 m

The intensity of the electric field due to a point-charge is

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

Substituting the values,

$$\begin{aligned}\text E = \frac{\left(9.0 \times 10^9\right)\left(3.2 \times 10^{-19}\right)}{\left(1 \times 10^{-10}\right)^2} \\[1em] \text E = \frac{2.88 \times 10^{-9}}{1 \times 10^{-20}} \\[1em] \text E = 2.88 \times 10^{11}\ \text N\ \text C^{-1}\end{aligned}$$

Hence, the intensity of the electric field is 2.88 × 10¹¹ N C^-1, directed radially outwards from the nucleus.

Given,

• Charge, q = 2 μC = 2 × 10^-6 C, placed at (1, −2, 1) m
• Point of observation, P (1, 2, 4) m

Position vector of P relative to the charge :

$$\begin{aligned}\vec{\text r} = (1 - 1)\hat{\text i} + \left(2 - (-2)\right)\hat{\text j} + (4 - 1)\hat{\text k} \\[1em] \vec{\text r} = 4\hat{\text j} + 3\hat{\text k} \\[1em] \text r = \sqrt{0^2 + 4^2 + 3^2} \\[1em] \text r = 5\ \text m\end{aligned}$$

Electric field intensity in vector form :

$$\begin{aligned}\vec{\text E} = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\text r^2}\hat{\text r} \\[1em] \vec{\text E} = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\text r^3}\vec{\text r}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\vec{\text E} = \frac{\left(9 \times 10^9\right)\left(2 \times 10^{-6}\right)}{(5)^3}\left(4\hat{\text j} + 3\hat{\text k}\right) \\[1em] \vec{\text E} = \frac{1.8 \times 10^4}{125}\left(4\hat{\text j} + 3\hat{\text k}\right) \\[1em] \vec{\text E} = 144\left(4\hat{\text j} + 3\hat{\text k}\right)\ \text N\ \text C^{-1}\end{aligned}$$

Hence, the electric field intensity at P is $144\left(4\hat{\text j} + 3\hat{\text k}\right)$ N C^-1.

Given,

• q₁ = + 0.2 C, q₂ = + 0.4 C
• Separation = 0.1 m

(i) At the mid-point : The mid-point is 0.05 m from each charge. Both the charges are positive, so their fields there are oppositely directed.

$$\begin{aligned}\text E_1 = \frac{\left(9 \times 10^9\right)(0.2)}{(0.05)^2} \\[1em] \text E_1 = \frac{1.8 \times 10^9}{2.5 \times 10^{-3}} \\[1em] \text E_1 = 7.20 \times 10^{11}\ \text N\ \text C^{-1}\end{aligned}$$

directed away from q₁, and

$$\begin{aligned}\text E_2 = \frac{\left(9 \times 10^9\right)(0.4)}{(0.05)^2} \\[1em] \text E_2 = \frac{3.6 \times 10^9}{2.5 \times 10^{-3}} \\[1em] \text E_2 = 1.44 \times 10^{12}\ \text N\ \text C^{-1}\end{aligned}$$

directed away from q₂, that is, towards q₁.

$$\begin{aligned}\text E = \text E_2 - \text E_1 \\[1em] \text E = 1.44 \times 10^{12} - 0.72 \times 10^{12} \\[1em] \text E = 7.20 \times 10^{11}\ \text N\ \text C^{-1}\end{aligned}$$

Hence, the field at the mid-point is 7.20 × 10¹¹ N C^-1, directed towards q₁.

(ii) At a point 0.15 m from q₁ and 0.05 m from q₂ : This point lies on the line joining the charges, beyond q₂. Here both fields are directed away from the charges, that is, in the same direction.

$$\begin{aligned}\text E_1 = \frac{\left(9 \times 10^9\right)(0.2)}{(0.15)^2} \\[1em] \text E_1 = \frac{1.8 \times 10^9}{2.25 \times 10^{-2}} \\[1em] \text E_1 = 8.0 \times 10^{10}\ \text N\ \text C^{-1} \\[1em] \text E_2 = \frac{\left(9 \times 10^9\right)(0.4)}{(0.05)^2} \\[1em] \text E_2 = 1.44 \times 10^{12}\ \text N\ \text C^{-1} \\[1em] \text E = \text E_1 + \text E_2 \\[1em] \text E = 0.08 \times 10^{12} + 1.44 \times 10^{12} \\[1em] \text E = 1.52 \times 10^{12}\ \text N\ \text C^{-1}\end{aligned}$$

Hence, the field at this point is 1.52 × 10¹² N C^-1, directed away from q₁ and q₂ both.

Given,

• q_A = + 1000 μC = 1 × 10^-3 C at A, q_B = − 1000 μC = − 1 × 10^-3 C at B
• AB = 2 m

(i) At the mid-point of AB : The mid-point is 1 m from each charge. The field due to the positive charge is directed away from A and that due to the negative charge is directed towards B, so both are along AB and they add up.

$$\begin{aligned}\text E_\text A = \text E_\text B = \frac{\left(9 \times 10^9\right)\left(1 \times 10^{-3}\right)}{(1)^2} \\[1em] \text E_\text A = \text E_\text B = 9 \times 10^6\ \text N\ \text C^{-1} \\[1em] \text E = \text E_\text A + \text E_\text B \\[1em] \text E = 1.8 \times 10^7\ \text N\ \text C^{-1}\end{aligned}$$

Hence, the field at the mid-point is 1.8 × 10⁷ N C^-1, directed along AB.

(ii) At a point equidistant, 4 m, from each charge : Such a point lies on the right bisector of AB, that is, in the broad-side-on position of the dipole. Here r = 4 m and half the dipole length is l = 1 m.

$$\begin{aligned}\text E_\text A = \text E_\text B = \frac{\left(9 \times 10^9\right)\left(1 \times 10^{-3}\right)}{(4)^2} \\[1em] \text E_\text A = \text E_\text B = \frac{9 \times 10^6}{16} \\[1em] \text E_\text A = \text E_\text B = 5.625 \times 10^5\ \text N\ \text C^{-1}\end{aligned}$$

The components perpendicular to AB cancel, and the components parallel to AB add up,

$$\begin{aligned}\text E = 2\text E_\text A\cos\theta, \qquad \text{where } \cos\theta = \frac{\text l}{\text r} \\[1em] \cos\theta = \frac{1}{4} \\[1em] \text E = 2 \times 5.625 \times 10^5 \times \frac{1}{4} \\[1em] \text E = 2.8 \times 10^5\ \text N\ \text C^{-1}\end{aligned}$$

Hence, the field at this point is 2.8 × 10⁵ N C^-1, directed parallel to AB.

Given,

• Excess of electrons, n = 10⁶
• Charge on one electron, e = 1.6 × 10^-19 C
• Electric field, E = 1000 N C^-1, directed towards East

Charge on the body : An excess of electrons means the body carries a negative charge,

$$\begin{aligned}\text q = -\text{ne} \\[1em] \text q = -10^6 \times 1.6 \times 10^{-19} \\[1em] \text q = -1.6 \times 10^{-13}\ \text C\end{aligned}$$

Force on the body : Taking the direction of the field, towards East, as positive, E = + 1000 N C^-1,

$$\begin{aligned}\text F = \text{qE} \\[1em] \text F = \left(-1.6 \times 10^{-13}\right)(+1000) \\[1em] \text F = -1.6 \times 10^{-10}\ \text N\end{aligned}$$

The negative sign shows that the force on the body is directed opposite to the electric field.

Hence, the force is 1.6 × 10^-10 N, directed towards West.

Given,

• Charge on the oil drop, q = 3.2 × 10^-19 C
• Weight of the drop, mg = 1.6 × 10^-14 N

The drop is stationary between the plates, so the upward electric force on it must balance its weight,

$$\begin{aligned}\text{qE} = \text{mg} \\[1em] \text E = \frac{\text{mg}}{\text q} \\[1em] \text E = \frac{1.6 \times 10^{-14}}{3.2 \times 10^{-19}} \\[1em] \text E = 5.0 \times 10^{4}\ \text N\ \text C^{-1}\end{aligned}$$

Hence, the electric field acting on the drop is 5.0 × 10⁴ N C^-1.

Given,

• q_A = + 0.25 μC = + 0.25 × 10^-6 C at A (0, 0, −15 cm)
• q_B = − 0.25 μC = − 0.25 × 10^-6 C at B (0, 0, +15 cm)

Total charge :

$$\begin{aligned}\text q_\text A + \text q_\text B = \left(+0.25\ \mu\text C\right) + \left(-0.25\ \mu\text C\right) \\[1em] \text q_\text A + \text q_\text B = 0\end{aligned}$$

Hence, the total charge of the system is zero.

Electric dipole moment : The two charges are equal and opposite, so they form an electric dipole. The separation between them is

$$\begin{aligned}2\text l = 15\ \text{cm} + 15\ \text{cm} \\[1em] 2\text l = 30\ \text{cm} \\[1em] 2\text l = 0.30\ \text m \\[1em] \text p = \text q \times 2\text l \\[1em] \text p = \left(0.25 \times 10^{-6}\right)(0.30) \\[1em] \text p = 7.5 \times 10^{-8}\ \text{C-m}\end{aligned}$$

The dipole moment is directed from the negative charge towards the positive charge, that is, from B (0, 0, + 15 cm) to A (0, 0, − 15 cm).

Hence, the total charge is zero and the electric dipole moment is 7.5 × 10^-8 C-m, directed along the − Z axis.

Given,

• Charges of the dipole, q = 2.4 μC = 2.4 × 10^-6 C
• Separation, 2l = 2.5 × 10^-3 m
• Distance of the point, r = 0.30 m

Electric dipole moment :

$$\begin{aligned}\text p = \text q \times 2\text l \\[1em] \text p = \left(2.4 \times 10^{-6}\right)\left(2.5 \times 10^{-3}\right) \\[1em] \text p = 6.0 \times 10^{-9}\ \text{C-m}\end{aligned}$$

Field in the broad-side-on position : Since r >> l, the dipole is a short dipole,

$$\begin{aligned}\text E_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\text r^3} \\[1em] \text E_{\text{equatorial}} = \frac{\left(9 \times 10^9\right)\left(6.0 \times 10^{-9}\right)}{(0.30)^3} \\[1em] \text E_{\text{equatorial}} = \frac{54}{2.7 \times 10^{-2}} \\[1em] \text E_{\text{equatorial}} = 2.0 \times 10^{3}\ \text N\ \text C^{-1}\end{aligned}$$

When the dipole is rotated through 90° : The given point, which was on the equatorial line, now lies on the axis of the dipole at the same distance r. Hence

$$\begin{aligned}\text E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\frac{2\text p}{\text r^3} \\[1em] \text E_{\text{axial}} = 2 \times 2.0 \times 10^{3} \\[1em] \text E_{\text{axial}} = 4.0 \times 10^{3}\ \text N\ \text C^{-1}\end{aligned}$$

Hence, the intensity is 2.0 × 10³ N C^-1 before rotation and 4.0 × 10³ N C^-1 after rotation.

Given,

• Charges of the dipole, q = 1 μC = 1 × 10^-6 C
• Separation, 2l = 2 cm = 2 × 10^-2 m
• Electric field, E = 1 × 10⁵ N C^-1
• Angle between $\vec{\text p}$ and $\vec{\text E} \space$, θ = 30°

(i) Magnitude of the electric dipole moment :

$$\begin{aligned}\text p = \text q \times 2\text l \\[1em] \text p = \left(1 \times 10^{-6}\right)\left(2 \times 10^{-2}\right) \\[1em] \text p = 2 \times 10^{-8}\ \text{C-m}\end{aligned}$$

(ii) Torque acting on the dipole :

$$\begin{aligned}\tau = \text{pE}\sin\theta \\[1em] \tau = \left(2 \times 10^{-8}\right)\left(1 \times 10^{5}\right)\sin 30° \\[1em] \tau = 2 \times 10^{-3} \times \frac{1}{2} \\[1em] \tau = 1 \times 10^{-3}\ \text{N-m}\end{aligned}$$

(iii) Maximum torque :

The torque τ = pE sin θ is maximum when sin θ = 1, that is, when θ = 90°, so that the dipole is normal to the field.

$$\begin{aligned}\tau_{\text{max}} = \text{pE} \\[1em] \tau_{\text{max}} = \left(2 \times 10^{-8}\right)\left(1 \times 10^{5}\right) \\[1em] \tau_{\text{max}} = 2 \times 10^{-3}\ \text{N-m}\end{aligned}$$

Hence, p = 2 × 10^-8 C-m, τ = 1 × 10^-3 N-m, and the torque is maximum, 2 × 10^-3 N-m, when the dipole is normal to the field (θ = 90°).

Given,

• Angle between the dipole and the field, θ = 30°
• Electric field, E = 10⁴ N C^-1
• Torque, τ = 9 × 10^-26 N-m

The torque acting on an electric dipole placed in a uniform electric field is

$$\begin{aligned}\tau = \text{pE}\sin\theta \\[1em] \text p = \frac{\tau}{\text E\sin\theta}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text p = \frac{9 \times 10^{-26}}{\left(10^{4}\right)\left(\sin 30°\right)} \\[1em] \text p = \frac{9 \times 10^{-26}}{10^{4} \times \frac{1}{2}} \\[1em] \text p = \frac{9 \times 10^{-26}}{5 \times 10^{3}} \\[1em] \text p = 1.8 \times 10^{-29}\ \text{C-m}\end{aligned}$$

Hence, the moment of the dipole is 1.8 × 10^-29 C-m.

Given,

• Two identical dipoles AB and CD, each of dipole moment p
• Angle between them, 2α = 120°
• Uniform electric field $\vec{\text E} \space$directed along + X

Resultant dipole moment : The two dipole moments are equal in magnitude and are inclined at 120° to each other. Their resultant is

$$\begin{aligned}\text p_{\text{res}} = \sqrt{\text p_{\text{AB}}^2 + \text p_{\text{CD}}^2 + 2\text p_{\text{AB}} \text p_{\text{CD}}\cos 120°} \\[1em] \text p_{\text{res}} = \sqrt{\text p^2 + \text p^2 + 2\text p\text p\cos 120°} \\[1em] \text p_{\text{res}} = \sqrt{2\text p^2 + 2\text p^2\left(-\frac{1}{2}\right)} \\[1em] \text p_{\text{res}} = \sqrt{\text p^2} \\[1em] \text p_{\text{res}} = \text p\end{aligned}$$

Since the two dipole moments are equal, the resultant bisects the angle between them. Using the stated angle of 120°, $\vec{\text p}_{\text{AB}}$ (from B to A) lies along + Y and $\vec{\text p}_{\text{CD}}$ (from D to C) lies at 30° below + X, as in the vector diagram above, which is drawn to this stated angle. Hence the bisector makes an angle of 30° with the + X axis, measured in the anticlockwise direction.

Torque acting on the combination : The angle between $\vec{\text p}_{\text{res}}$ and $\vec{\text E} \space$is θ = 30°,

$$\begin{aligned}\tau = \text p_{\text{res}}\text E\sin\theta \\[1em] \tau = \text{pE}\sin 30° \\[1em] \tau = \frac{1}{2}\text{pE}\end{aligned}$$

The torque $\vec{\tau} = \vec{\text p}_{\text{res}} \times \vec{\text E}$is perpendicular to the plane of the figure. Since $\vec{\text p}_{\text{res}}$ lies in the XY plane above the X-axis and $\vec{\text E} \space$is along + X, the torque is directed along the negative Z-direction.

Hence, the resultant dipole moment is p, at an angle of 30° with the + X axis in the anticlockwise direction, and the torque is $\frac{1}{2}\text{pE}$ along the negative Z-direction.

Note: The textbook's sketch is schematic — its line CD is drawn steeper than the marked 120° — so the working uses the stated angle of 120°.

A parabola.

Explanation:
The electron enters the field at right angles to it, so its velocity along the original direction remains constant while it is uniformly accelerated along the field. This is exactly the condition of projectile motion.

Taking the initial velocity along the x-direction and the acceleration a = eE/m along the y-direction,

$$\begin{aligned}\text x = \text{ut}, \qquad \text y = \frac{1}{2}\text a\text t^2 \\[1em] \text y = \left(\frac{\text{eE}}{2\text m\text u^2}\right)\text x^2\end{aligned}$$

Since y ∝ x², the locus of the electron is a parabola.

Let the dipole consist of charges − q and + q separated by a distance 2l, placed in a uniform electric field $\vec{\text E} \space$with its dipole moment vector $\vec{\text p}$ making an angle θ with the field.

Due to the electric field, the charge + q experiences a force qE in the direction of the field and the charge − q experiences an equal and opposite force qE. Since these two forces are equal and opposite, the net translatory force on the dipole is zero.

However, the two forces act at different points and hence form a couple. The moment of this couple is the torque on the dipole. Its magnitude is force multiplied by the perpendicular distance between the lines of action of the two forces,

$$\begin{aligned}\tau = \text{qE}\left(2\text l\sin\theta\right) \\[1em] \tau = \left(2\text q\text l\right)\text E\sin\theta\end{aligned}$$

But 2ql = p, the magnitude of the dipole moment,

$\tau = \text{pE}\sin\theta$

The torque is perpendicular to the plane containing $\vec{\text p}$ and $\vec{\text E} \space$, its sense being given by the right-hand screw rule. Both the magnitude and the direction are therefore correctly given by the vector product,

$\vec{\tau} = \vec{\text p} \times \vec{\text E}$

Hence, the torque acting on the dipole is $\vec{\tau} = \vec{\text p} \times \vec{\text E}$.

No, this charge cannot be given to a conductor.

By the quantisation of charge, the charge on a body must be an integral multiple of the elementary charge, q = ± ne with e = 1.6 × 10^-19 C. Here

$$\begin{aligned}\text n = \frac{8.6 \times 10^{-19}}{1.6 \times 10^{-19}} \\[1em] \text n = 5.375\end{aligned}$$

Since n is not an integer, the given charge is not an integral multiple of e and hence cannot exist on a conductor.

Zero.

Explanation:

In a uniform electric field the charge + q of the dipole experiences a force qE in the direction of the field and the charge − q experiences an equal force qE opposite to the field. These two forces are equal in magnitude and opposite in direction, so their resultant is zero.

Hence there is no net translatory force on the dipole, and it undergoes no translatory motion in a uniform electric field; only a torque pE sin θ acts on it.

$\vec{\text F}_{12} = \frac{1}{4\pi\varepsilon_0}\frac{\text q_1\text q_2}{\text r^2}\ \hat{\text r}_{12}$

Explanation:
Let two charges q₁ and q₂ be placed at points with position vectors $\vec{\text r}_1$ and $\vec{\text r}_2$, a distance r apart. Let $\hat{\text r}_{12}$ be the unit vector pointing from q₂ towards q₁, and $\hat{\text r}_{21}$ that pointing from q₁ towards q₂. Then $\vec{\text F}_{12}$ is the force on q₁ due to q₂, and the force on q₂ due to q₁ is

$\vec{\text F}_{21} = \frac{1}{4\pi\varepsilon_0}\frac{\text q_1\text q_2}{\text r^2}\ \hat{\text r}_{21}$

Since $\hat{\text r}_{21} = -\hat{\text r}_{12}$, it follows that $\vec{\text F}_{21} = -\vec{\text F}_{12}$, so that Newton's third law of motion is obeyed.

(i) Relative permittivity of water is 81 :

The relative permittivity, or dielectric constant, of a medium is the ratio of the force between two given charges placed a certain distance apart in vacuum to the force between the same charges placed the same distance apart in the medium.

The statement therefore means that the absolute permittivity of water is 81 times the permittivity of free space, ε_w = 81ε₀, so that the electrostatic force between two charges placed in water is $\frac{1}{81}$ of the force between the same charges at the same separation in vacuum.

(ii) Electric lines of force due to an electric dipole :

The lines of force originate on the positive charge and terminate on the negative charge. Their concentration is greatest in the region between the charges, where the electric field is strongest.

The intensity of the electric field due to a point-charge q at a distance r from it is

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

that is,

$\text E \propto \frac{1}{\text r^2}$

Hence, the electric field varies inversely as the square of the distance from the origin. The graph of E against r is therefore a curve which falls off rapidly as r increases and approaches the r-axis asymptotically.

Let the dipole AB consist of charges − q and + q separated by a distance 2l, and let P be a point in the broad-side-on position, that is, on the right bisector of the dipole at a distance r from its mid-point O.

The distance of P from each charge is $\sqrt{\text r^2 + \text l^2}$, so the magnitudes of the two intensities are

$$\begin{aligned}\text E_1 = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\left(\text r^2 + \text l^2\right)} \quad \text{(along BP, due to + q)} \\[1em] \text E_2 = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\left(\text r^2 + \text l^2\right)} \quad \text{(along PA, due to } - \text{ q)}\end{aligned}$$

Resolving each into components parallel and perpendicular to AB, the perpendicular components E₁ sin θ and E₂ sin θ are equal and opposite and cancel each other, while the parallel components E₁ cos θ and E₂ cos θ are in the same direction and add up.

$$\begin{aligned}\text E = 2\text E_1\cos\theta \\[1em] \text E = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\left(\text r^2 + \text l^2\right)}\ 2\cos\theta\end{aligned}$$

From the figure, $\cos\theta = \frac{\text l}{\left(\text r^2 + \text l^2\right)^{1/2}}$,

$$\begin{aligned}\text E = \frac{1}{4\pi\varepsilon_0}\frac{2\text q\text l}{\left(\text r^2 + \text l^2\right)^{3/2}} \\[1em] \text E = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\left(\text r^2 + \text l^2\right)^{3/2}}\end{aligned}$$

since p = 2ql. For a short dipole (r >> l),

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\text r^3}$

The direction of $\vec{\text E} \space$at every point of the equatorial line is antiparallel to the dipole axis.

(i) Two equal positive point charges : The lines of force start on both the charges and spread outwards. At the mid-point N of the line joining the charges the field due to one charge is equal and opposite to that due to the other, so the resultant field is zero. This point, shown by the absence of lines of force around it, is called the neutral point. Far from the charges the lines of force become nearly radial, characteristic of a single charge.

(ii) An electric dipole : The lines of force originate on the positive charge and terminate on the negative charge. The concentration of the lines is greatest in the region between the two charges, showing that the electric field is strongest there.

a deficiency of 5 electrons

Reason — The body carries a positive charge, which means a deficiency of electrons. By the quantisation of charge, q = ne,

$$\begin{aligned}\text n = \frac{\text q}{\text e} \\[1em] \text n = \frac{8 \times 10^{-19}}{1.6 \times 10^{-19}} \\[1em] \text n = 5\end{aligned}$$

Hence the body has a deficiency of 5 electrons.

Given,

• Charge, q = 5 × 10^-6 C
• Force, F = 2 × 10^-3 N

The force on a charge placed in an electric field is F = qE, so that

$\text E = \frac{\text F}{\text q}$

Substituting the values,

$$\begin{aligned}\text E = \frac{2 \times 10^{-3}}{5 \times 10^{-6}} \\[1em] \text E = 400\ \text N\ \text C^{-1}\end{aligned}$$

Hence, the intensity of the electric field is 400 N C^-1.

ε_w = 81ε₀

Reason — The relative permittivity, or dielectric constant, of a medium is defined as

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

For water K = 81, so that

$$\begin{aligned}\frac{\varepsilon_\text w}{\varepsilon_0} = 81 \\[1em] \varepsilon_\text w = 81\varepsilon_0\end{aligned}$$

(i) Expressions for the intensity of the electric field :

For a dipole of charges − q and + q separated by a distance 2a, so that the dipole moment is p = q × 2a, and for a point at a distance r from the centre of the dipole :

(1) Axial position (end-on position) :

$\text E_1 = \frac{1}{4\pi\varepsilon_0}\frac{2\text p\text r}{\left(\text r^2 - \text a^2\right)^2}$

directed along the dipole axis, from − q towards + q.

(2) Broad side-on position (equatorial position) :

$\text E_2 = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\left(\text r^2 + \text a^2\right)^{3/2}}$

directed antiparallel to the dipole axis.

(ii) Ratio of the two intensities for a short dipole :

For a short dipole, a² may be neglected in comparison with r², so that

$$\begin{aligned}\text E_1 = \frac{1}{4\pi\varepsilon_0}\frac{2\text p}{\text r^3}, \qquad \text E_2 = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\text r^3} \\[1em] \frac{\text E_1}{\text E_2} = 2\end{aligned}$$

Hence, E₁ : E₂ = 2 : 1.

Torque on the dipole : When a dipole of moment $\vec{\text p}$ is placed in a uniform electric field $\vec{\text E} \space$with its axis inclined at an angle θ to the field, the two forces qE on its charges are equal and opposite and form a couple. The magnitude of the torque is

$\tau = \text{pE}\sin\theta$

and in vector form,

$\vec{\tau} = \vec{\text p} \times \vec{\text E}$

(i) For maximum torque : τ = pE sin θ is maximum when sin θ = 1, that is, θ = 90°. The dipole must therefore be kept perpendicular to the electric field, $\vec{\text p} \perp \vec{\text E}$, and then τ_max = pE.

(ii) For zero torque : τ = 0 when sin θ = 0, that is, θ = 0°. The dipole must therefore be kept parallel to the electric field, $\vec{\text p} \parallel \vec{\text E}$. This is the position of stable equilibrium.

Quantisation of charge means that the electric charge on a body is not continuous but exists in the form of discrete packets. In nature all charges are found to be integral multiples of a smallest amount of charge, the elementary charge e = 1.6 × 10^-19 C, which is the magnitude of the charge on an electron.

Hence the charge q on any body must be

$\text q = \pm\ \text{ne}, \qquad \text{where n} = 1,\ 2,\ 3,\ ...$

No charge is found in a fraction of e, such as 0.7e or 2.5e; that is, electric charge cannot be divided indefinitely.

Given,

• Weight of the oil drop, mg = 1.6 × 10^-15 N
• Electric field, E = 2 × 10³ N C^-1

The drop remains suspended, so the upward electric force on it balances its weight,

$$\begin{aligned}\text{qE} = \text{mg} \\[1em] \text q = \frac{\text{mg}}{\text E}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text q = \frac{1.6 \times 10^{-15}}{2 \times 10^{3}} \\[1em] \text q = 8.0 \times 10^{-19}\ \text C\end{aligned}$$

Hence, the charge on the drop is 8.0 × 10^-19 C.

Given,

• Electric dipole moment, p = 3 × 10^-29 C-m
• Maximum torque, τ_max = 6 × 10^-25 N-m

The torque on a dipole is τ = pE sin θ, which is maximum when θ = 90°, that is, when the dipole is perpendicular to the field,

$$\begin{aligned}\tau_{\text{max}} = \text{pE} \\[1em] \text E = \frac{\tau_{\text{max}}}{\text p}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text E = \frac{6 \times 10^{-25}}{3 \times 10^{-29}} \\[1em] \text E = 2 \times 10^{4}\ \text N\ \text C^{-1}\end{aligned}$$

Hence, the electric field is 2 × 10⁴ N/C.

Let the dipole consist of charges − q and + q separated by a distance 2l, and let P be a point in the broadside on position at a distance r from the mid-point O of the dipole.

The distance of P from each charge is $\sqrt{\text r^2 + \text l^2}$, so the magnitudes of the intensities at P due to the two charges are equal,

$\text E_1 = \text E_2 = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\left(\text r^2 + \text l^2\right)}$

Resolving each intensity into components parallel and perpendicular to the dipole axis, the perpendicular components are equal and opposite and cancel out, while the parallel components add up. Hence

$$\begin{aligned}\text E = \text E_1\cos\theta + \text E_2\cos\theta \\[1em] \text E = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\left(\text r^2 + \text l^2\right)}\ 2\cos\theta\end{aligned}$$

From the figure,

$\cos\theta = \frac{\text l}{\left(\text r^2 + \text l^2\right)^{1/2}}$

Substituting,

$$\begin{aligned}\text E = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\left(\text r^2 + \text l^2\right)} \times \frac{2\text l}{\left(\text r^2 + \text l^2\right)^{1/2}} \\[1em] \text E = \frac{1}{4\pi\varepsilon_0}\frac{2\text q\text l}{\left(\text r^2 + \text l^2\right)^{3/2}}\end{aligned}$$

But 2ql = p, the electric dipole moment,

$\text E = \left(\frac{1}{4\pi\varepsilon_0}\right)\frac{\text p}{\left(\text r^2 + \text l^2\right)^{3/2}}$

Hence the required result is proved.

Let the dipole AB be made up of charges − q and + q separated by a distance 2l, and let P be a point in the end-on position, that is, on the axis of the dipole at a distance r from its mid-point O.

The distance of P from the charge + q is (r − l) and from the charge − q is (r + l). Hence the magnitudes of the two intensities are

$$\begin{aligned}\text E_1 = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{(\text r - \text l)^2} \quad \text{(along the dipole axis)} \\[1em] \text E_2 = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{(\text r + \text l)^2} \quad \text{(opposite to the dipole axis)}\end{aligned}$$

The two intensities are along the same line in opposite directions, and E₁ > E₂. Hence the resultant intensity is their difference, directed along the dipole axis,

$$\begin{aligned}\text E = \text E_1 - \text E_2 \\[1em] \text E = \frac{\text q}{4\pi\varepsilon_0}\left[\frac{1}{(\text r - \text l)^2} - \frac{1}{(\text r + \text l)^2}\right] \\[1em] \text E = \frac{\text q}{4\pi\varepsilon_0}\left[\frac{(\text r + \text l)^2 - (\text r - \text l)^2}{\left(\text r^2 - \text l^2\right)^2}\right] \\[1em] \text E = \frac{\text q}{4\pi\varepsilon_0}\frac{4\text l\text r}{\left(\text r^2 - \text l^2\right)^2}\end{aligned}$$

Putting p = 2ql,

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{2\text p\text r}{\left(\text r^2 - \text l^2\right)^2}$

For a short dipole, l² may be neglected in comparison with r²,

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{2\text p}{\text r^3}$

directed along the dipole axis, from the negative charge towards the positive charge.

Given,

• q₁ = + 80 μC = 80 × 10^-6 C, q₂ = + 20 μC = 20 × 10^-6 C
• Separation = 6 m, so the mid-point is at r = 3 m from each charge

Both the charges are positive, so their fields at the mid-point are oppositely directed.

Field due to the 80 μC charge :

$$\begin{aligned}\text E_1 = \frac{1}{4\pi\varepsilon_0}\frac{\text q_1}{\text r^2} \\[1em] \text E_1 = \frac{\left(9 \times 10^9\right)\left(80 \times 10^{-6}\right)}{(3)^2} \\[1em] \text E_1 = \frac{7.2 \times 10^5}{9} \\[1em] \text E_1 = 8 \times 10^{4}\ \text N\ \text C^{-1}\end{aligned}$$

Field due to the 20 μC charge :

$$\begin{aligned}\text E_2 = \frac{\left(9 \times 10^9\right)\left(20 \times 10^{-6}\right)}{(3)^2} \\[1em] \text E_2 = \frac{1.8 \times 10^5}{9} \\[1em] \text E_2 = 2 \times 10^{4}\ \text N\ \text C^{-1}\end{aligned}$$

Resultant intensity :

$$\begin{aligned}\text E = \text E_1 - \text E_2 \\[1em] \text E = 8 \times 10^{4} - 2 \times 10^{4} \\[1em] \text E = 6 \times 10^{4}\ \text N\ \text C^{-1}\end{aligned}$$

Hence, the resultant electric field intensity at the mid-point is 6 × 10⁴ N/C, directed from the 80 μC charge towards the 20 μC charge.

F

Reason — By Coulomb's law the initial force is

$\text F = \frac{1}{4\pi\varepsilon_0}\frac{\text q_1\text q_2}{\text r^2}$

When each charge is halved and the distance is also halved,

$$\begin{aligned}\text F' = \frac{1}{4\pi\varepsilon_0}\frac{\left(\frac{\text q_1}{2}\right)\left(\frac{\text q_2}{2}\right)}{\left(\frac{\text r}{2}\right)^2} \\[1em] \text F' = \frac{1}{4\pi\varepsilon_0}\frac{\frac{\text q_1\text q_2}{4}}{\frac{\text r^2}{4}} \\[1em] \text F' = \frac{1}{4\pi\varepsilon_0}\frac{\text q_1\text q_2}{\text r^2} \\[1em] \text F' = \text F\end{aligned}$$

Hence the new force is the same as the original force.

They intersect each other.

Reason — The tangent drawn at any point on a line of force gives the direction of the electric field there. If two lines of force intersected, two tangents could be drawn at the point of intersection, which would mean two different directions of the electric field at the same point. This is impossible, so lines of force never intersect each other.

The other three statements are genuine properties of electric lines of force.

4 N-m

Reason — Given, q = 2 mC = 2 × 10^-3 C, separation 2l = 4 cm = 4 × 10^-2 m, E = 1 × 10⁵ N C^-1 and θ = 30°.

Electric dipole moment :

$$\begin{aligned}\text p = \text q \times 2\text l \\[1em] \text p = \left(2 \times 10^{-3}\right)\left(4 \times 10^{-2}\right) \\[1em] \text p = 8 \times 10^{-5}\ \text{C-m}\end{aligned}$$

Torque :

$$\begin{aligned}\tau = \text{pE}\sin\theta \\[1em] \tau = \left(8 \times 10^{-5}\right)\left(1 \times 10^{5}\right)\sin 30° \\[1em] \tau = 8 \times \frac{1}{2} \\[1em] \tau = 4\ \text{N-m}\end{aligned}$$

(i) No, a body cannot be given a charge of 2.2 × 10^-19 C.

By the quantisation of charge, the charge on a body must be an integral multiple of the elementary charge e = 1.6 × 10^-19 C. Here

$$\begin{aligned}\text n = \frac{2.2 \times 10^{-19}}{1.6 \times 10^{-19}} \\[1em] \text n = 1.375\end{aligned}$$

Since n is not an integer, the given charge is not an integral multiple of e and hence is not possible.

(ii) The torque on an electric dipole placed in a uniform electric field is

$\tau = \text{pE}\sin\theta$

This is maximum when sin θ = 1, that is, when θ = 90°.

Hence, the torque is maximum when the dipole is placed perpendicular to the electric field, and then τ_max = pE.

2 : 1.

Explanation:
For a short electric dipole of moment p, at a distance r from its centre :

At the point A in the axial position :

$\text E_\text A = \frac{1}{4\pi\varepsilon_0}\frac{2\text p}{\text r^3}$

At the point B in the broad side position :

$\text E_\text B = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\text r^3}$

Dividing one by the other,

$$\begin{aligned}\frac{\text E_\text A}{\text E_\text B} = \frac{\frac{2\text p}{4\pi\varepsilon_0\text r^3}}{\frac{\text p}{4\pi\varepsilon_0\text r^3}} \\[1em] \frac{\text E_\text A}{\text E_\text B} = \frac{2}{1}\end{aligned}$$

Hence, E_A : E_B = 2 : 1.

smaller

Reason —

Given, both the particles fall from rest through the same vertical distance h, the magnitude of the electric field being the same in the two cases.

The electric force on a particle of charge q is F = qE, so its acceleration is

$\text a = \frac{\text{qE}}{\text m}$

Starting from rest and covering a distance h,

$$\begin{aligned}\text h = \frac{1}{2}\text a\text t^2 \\[1em] \text h = \frac{1}{2}\left(\frac{\text{qE}}{\text m}\right)\text t^2 \\[1em] \text t = \sqrt{\frac{2\text{hm}}{\text{qE}}} \quad \Rightarrow \quad \text t \propto \sqrt{\text m}\end{aligned}$$

since h, q and E are the same for the electron and for the proton. The mass of the electron is much smaller than that of the proton, so the time of fall of the electron is smaller than that of the proton.

$\frac{9\text F}{16}$

Reason —

Given, the charges are + Q and − Q, a distance r apart, and the force between them is

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

When 25% of the charge of A is transferred to B, the charge transferred is $\frac{\text Q}{4}$, so the new charges are

$$\begin{aligned}\text Q_\text A = \text Q - \frac{\text Q}{4} \\[1em] \text Q_\text A = \frac{3\text Q}{4} \\[1em] \text Q_\text B = -\text Q + \frac{\text Q}{4} \\[1em] \text Q_\text B = -\frac{3\text Q}{4}\end{aligned}$$

The separation is unchanged, so the new force is

$$\begin{aligned}\text F' = \frac{1}{4\pi\varepsilon_0}\frac{\left(\frac{3\text Q}{4}\right)^2}{\text r^2} \\[1em] \text F' = \frac{9}{16} \times \frac{1}{4\pi\varepsilon_0}\frac{\text Q^2}{\text r^2} \\[1em] \text F' = \frac{9\text F}{16}\end{aligned}$$

$\frac{-\text Q}{4}$

Reason —

Given, + Q at x = 0, q at $\text x = \frac{\text d}{2}$ and + Q at x = d.

The charge + Q at x = 0 experiences

• a repulsion due to the charge + Q at x = d, at a distance d,

$\text F_1 = \frac{1}{4\pi\varepsilon_0}\frac{\text Q^2}{\text d^2}$

• a force due to the charge q at a distance $\frac{\text d}{2}$,

$$\begin{aligned}\text F_2 = \frac{1}{4\pi\varepsilon_0}\frac{\text{Qq}}{\left(\frac{\text d}{2}\right)^2} \\[1em] \text F_2 = \frac{1}{4\pi\varepsilon_0}\frac{4\text{Qq}}{\text d^2}\end{aligned}$$

For the net force on the charge at x = 0 to be zero, these two must cancel,

$$\begin{aligned}\frac{\text Q^2}{\text d^2} + \frac{4\text{Qq}}{\text d^2} = 0 \\[1em] \text Q + 4\text q = 0 \\[1em] \text q = -\frac{\text Q}{4}\end{aligned}$$

$\left(63\hat{\text i} - 27\hat{\text j}\right) \times 10^2$

Reason —

Given, $\text q_1 = \sqrt{10}\ \mu\text C$ at (1 m, 0), $\text q_2 = -25\ \mu\text C$ at (4 m, 0), and the point of observation is P (0, 3 m).

Field due to q₁ : The vector from q₁ to P is

$$\begin{aligned}\vec{\text r}_1 = (0 - 1)\hat{\text i} + (3 - 0)\hat{\text j} \\[1em] \vec{\text r}_1 = -\hat{\text i} + 3\hat{\text j} \\[1em] \text r_1 = \sqrt{1 + 9} \\[1em] \text r_1 = \sqrt{10}\ \text m \\[1em] \vec{\text E}_1 = \frac{1}{4\pi\varepsilon_0}\frac{\text q_1}{\text r_1^{\ 3}}\vec{\text r}_1 \\[1em] \vec{\text E}_1 = \frac{\left(9 \times 10^9\right)\left(\sqrt{10} \times 10^{-6}\right)}{10\sqrt{10}}\left(-\hat{\text i} + 3\hat{\text j}\right) \\[1em] \vec{\text E}_1 = 900\left(-\hat{\text i} + 3\hat{\text j}\right) \\[1em] \vec{\text E}_1 = \left(-900\hat{\text i} + 2700\hat{\text j}\right)\ \text V/\text m\end{aligned}$$

Field due to q₂ : The vector from q₂ to P is

$$\begin{aligned}\vec{\text r}_2 = (0 - 4)\hat{\text i} + (3 - 0)\hat{\text j} \\[1em] \vec{\text r}_2 = -4\hat{\text i} + 3\hat{\text j} \\[1em] \text r_2 = \sqrt{16 + 9} \\[1em] \text r_2 = 5\ \text m \\[1em] \vec{\text E}_2 = \frac{\left(9 \times 10^9\right)\left(-25 \times 10^{-6}\right)}{125}\left(-4\hat{\text i} + 3\hat{\text j}\right) \\[1em] \vec{\text E}_2 = -1800\left(-4\hat{\text i} + 3\hat{\text j}\right) \\[1em] = \left(7200\hat{\text i} - 5400\hat{\text j}\right)\ \text V/\text m\end{aligned}$$

Resultant field :

$$\begin{aligned}\vec{\text E} = \vec{\text E}_1 + \vec{\text E}_2 \\[1em] \vec{\text E} = \left(-900 + 7200\right)\hat{\text i} + \left(2700 - 5400\right)\hat{\text j} \\[1em] \vec{\text E} = \left(6300\hat{\text i} - 2700\hat{\text j}\right) \\[1em] \vec{\text E} = \left(63\hat{\text i} - 27\hat{\text j}\right) \times 10^2\ \text V/\text m\end{aligned}$$

$\text E \propto \frac{1}{\text D^4}$

Reason —

Given, the charges − q, + q, + q and − q are placed on the Y-axis at y = − 2d, − d, + d and + 2d respectively.

The charges + q at y = + d and − q at y = + 2d form one dipole, and the charges + q at y = − d and − q at y = − 2d form a second dipole of equal moment pointing in the opposite direction. The algebraic sum of all the charges is zero and the net dipole moment of the whole arrangement is also zero.

Such an arrangement of two equal and opposite dipoles is a quadrupole. Just as the field of a dipole falls off one power of distance faster than that of a point-charge, the field of a quadrupole falls off one power faster than that of a dipole,

$\text{point-charge} : \text E \propto \frac{1}{\text D^2},$

$\text{dipole} : \text E \propto \frac{1}{\text D^3},$

$\text{quadrupole} : \text E \propto \frac{1}{\text D^4}$

Hence at a point on the x-axis with D >> d, the field behaves as $\text E \propto \frac{1}{\text D^4}$.

tan^-1(0.5)

Reason —

Given,

• Mass of the bob, m = 2 g = 2 × 10^-3 kg
• Charge, q = 5.0 μC = 5.0 × 10^-6 C
• Electric field, E = 2000 V/m, horizontal
• g = 10 m/s²

At equilibrium the string makes an angle θ with the vertical. Resolving the tension into horizontal and vertical components,

$\text T\sin\theta = \text{qE}, \qquad \text T\cos\theta = \text{mg}$

Dividing,

$\tan\theta = \frac{\text{qE}}{\text{mg}}$

Substituting the values,

$$\begin{aligned}\tan\theta = \frac{\left(5.0 \times 10^{-6}\right)(2000)}{\left(2 \times 10^{-3}\right)(10)} \\[1em] \tan\theta = \frac{1.0 \times 10^{-2}}{2 \times 10^{-2}} \\[1em] \tan\theta = 0.5 \\[1em] \theta = \tan^{-1}(0.5)\end{aligned}$$

$\frac{\text R}{\sqrt 2}$

Reason —

The electric field on the axis of a uniformly charged ring of radius R, at a distance h from its centre, is

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{\text Q\text h}{\left(\text R^2 + \text h^2\right)^{3/2}}$

For E to be maximum, $\frac{\text{dE}}{\text{dh}} = 0$. Differentiating with respect to h,

$$\begin{aligned}\frac{\text{dE}}{\text{dh}} = \frac{\text Q}{4\pi\varepsilon_0}\left[\frac{\left(\text R^2 + \text h^2\right)^{3/2} - \text h \times \frac{3}{2}\left(\text R^2 + \text h^2\right)^{1/2}(2\text h)}{\left(\text R^2 + \text h^2\right)^{3}}\right] \\[1em] = \frac{\text Q}{4\pi\varepsilon_0}\left[\frac{\left(\text R^2 + \text h^2\right)^{1/2}\left[\left(\text R^2 + \text h^2\right) - 3\text h^2\right]}{\left(\text R^2 + \text h^2\right)^{3}}\right]\end{aligned}$$

Putting this equal to zero,

$$\begin{aligned}\text R^2 + \text h^2 - 3\text h^2 = 0 \\[1em] \text R^2 = 2\text h^2 \\[1em] \text h = \frac{\text R}{\sqrt{2}}\end{aligned}$$

$-\sqrt{3}\thinspace\text{ql}\ \hat{\text j}$

Reason —

From the figure the charge + q lies at the origin, the second charge + q at (l, 0) and the charge − 2q at the apex $\left(\frac{\text l}{2},\ \frac{\sqrt{3}\text l}{2}\right)$ of the equilateral triangle of side l.

The dipole moment of a system of charges is $\vec{\text p} = \sum \text q_\text i\vec{\text r}_\text i$. Since the algebraic sum of the charges is q + q − 2q = 0, this is independent of the choice of origin.

x-component :

$$\begin{aligned}\text p_\text x = \text q(0) + \text q(\text l) + (-2\text q)\left(\frac{\text l}{2}\right) \\[1em] \text p_\text x = \text{ql} - \text{ql} \\[1em] \text p_\text x = 0\end{aligned}$$

y-component :

$$\begin{aligned}\text p_\text y = \text q(0) + \text q(0) + (-2\text q)\left(\frac{\sqrt{3}\text l}{2}\right) \\[1em] \text p_\text y = -\sqrt{3}\ \text{ql} \\[1em] \vec{\text p} = -\sqrt{3}\ \text{ql}\ \hat{\text j}\end{aligned}$$

$\sqrt{\frac{2\text{qE}}{\text{md}}}$

Reason —

Given, the two charges q are of mass m each and are separated by a distance d, so the dipole moment is p = qd.

Moment of inertia about the centre : Each charge is at a distance $\frac{\text d}{2}$ from the mid-point,

$$\begin{aligned}\text I = 2\text m\left(\frac{\text d}{2}\right)^2 \\[1em] \text I = \frac{\text m\text d^2}{2}\end{aligned}$$

Restoring torque : When the dipole is rotated through a small angle θ from its equilibrium orientation,

$\tau = -\text{pE}\sin\theta \approx -\text{pE}\theta \qquad (\text{for small }\theta)$

This is of the form τ = − cθ, so the motion is angular simple harmonic with

$\omega = \sqrt{\frac{\text{pE}}{\text I}}$

Substituting p = qd and $\text I = \frac{\text m\text d^2}{2}$,

$$\begin{aligned}\omega = \sqrt{\frac{\text{qdE}}{\frac{\text m\text d^2}{2}}} \\[1em] \omega = \sqrt{\frac{2\text{qE}}{\text{md}}}\end{aligned}$$

$\sqrt{\frac{3}{\text a}}$

Reason —

Given, $\text E = \text E_0\left(1 - \text{ax}^2\right)$ along the x-direction, and the particle of charge q starts from rest at x = 0.

The particle starts from rest and again comes to rest at some distance d, so its kinetic energy is zero at both positions. By the work-energy theorem the total work done by the electric force over this distance must be zero,

$$\begin{aligned}\text W = \int_0^\text d \text{qE}\ \text{dx} \\[1em] \text W = \text q\text E_0\int_0^\text d\left(1 - \text{ax}^2\right)\text{dx} \\[1em] \text W = 0\end{aligned}$$

Carrying out the integration,

$$\begin{aligned}\text q\text E_0\left[\text x - \frac{\text{ax}^3}{3}\right]_0^\text d = 0 \\[1em] \text d - \frac{\text{ad}^3}{3} = 0 \\[1em] \text d\left(1 - \frac{\text{ad}^2}{3}\right) = 0\end{aligned}$$

Since d ≠ 0 (the initial position is excluded),

$$\begin{aligned}\text d^2 = \frac{3}{\text a} \\[1em] \text d = \sqrt{\frac{3}{\text a}}\end{aligned}$$

10²² m/s²

Reason —

Given,

• Separation, r = 1.6 Å = 1.6 × 10^-10 m
• e = 1.6 × 10^-19 C, m_e = 9 × 10^-31 kg

Electrostatic force of attraction :

$$\begin{aligned}\text F = \frac{1}{4\pi\varepsilon_0}\frac{\text e^2}{\text r^2} \\[1em] \text F = \frac{\left(9 \times 10^9\right)\left(1.6 \times 10^{-19}\right)^2}{\left(1.6 \times 10^{-10}\right)^2} \\[1em] \text F = \frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{2.56 \times 10^{-20}} \\[1em] \text F = 9 \times 10^{-9}\ \text N\end{aligned}$$

Acceleration of the electron :

$$\begin{aligned}\text a = \frac{\text F}{\text m_\text e} \\[1em] \text a = \frac{9 \times 10^{-9}}{9 \times 10^{-31}} \\[1em] \text a = 10^{22}\ \text m/\text s^2\end{aligned}$$

$\vec{\text E} = -\frac{\vec{\text p}}{4\pi\varepsilon_0\text r^3}$

Reason —

For a point in the broad-side-on position, that is, on the equatorial plane at a distance r from the centre of the dipole, the components of the two field intensities perpendicular to the dipole axis cancel while the parallel components add up, giving

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\left(\text r^2 + \text l^2\right)^{3/2}}$

Since r >> the separation of the two charges, l² may be neglected in comparison with r²,

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\text r^3}$

The direction of the field at every point of the equatorial line is antiparallel to the dipole axis, so a minus sign must be put in the vector form,

$\vec{\text E} = -\frac{1}{4\pi\varepsilon_0}\frac{\vec{\text p}}{\text r^3}$

$\text F = \frac{1}{4\pi\varepsilon_0}\frac{\text{Qq}}{\text r^2}$ for r > R

Reason —

A uniformly charged spherical shell behaves, for all points outside it, exactly as though its entire charge Q were concentrated at its centre. Hence for r > R the force on a charge q placed at a distance r from the centre is

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

For points inside the shell, that is, for r < R, the contributions of all the elements of the shell cancel exactly, so the field and hence the force is zero,

$\text F = 0 \qquad \text{for r} \lt \text R$

$\frac{\text x_1}{\text x_2}$

Reason —

Given, from the figure Q₁ is at A on the y-axis at a distance x₁ from O, and Q₂ is at B on the x-axis at a distance x₂ from O.

The field at O due to Q₁ is directed along AO, that is, along the y-axis, and the field due to Q₂ is directed along BO, that is, along the x-axis,

$\text E_1 = \frac{1}{4\pi\varepsilon_0}\frac{\text Q_1}{\text x_1^{\ 2}}, \qquad \text E_2 = \frac{1}{4\pi\varepsilon_0}\frac{\text Q_2}{\text x_2^{\ 2}}$

If the resultant field at O makes an angle α with the x-axis, then

$\tan\alpha = \frac{\text E_1}{\text E_2}$

The hypotenuse AB makes an angle β with the x-axis such that

$$\begin{aligned}\tan\beta = \frac{0 - \text x_1}{\text x_2 - 0} \\[1em] \tan\beta = -\frac{\text x_1}{\text x_2}\end{aligned}$$

For the resultant to be perpendicular to AB, the product of the two slopes must be − 1,

$$\begin{aligned}\frac{\text E_1}{\text E_2} \times \frac{\text x_1}{\text x_2} = 1 \\[1em] \frac{\text E_1}{\text E_2} = \frac{\text x_2}{\text x_1}\end{aligned}$$

Substituting the field intensities,

$$\begin{aligned}\frac{\text Q_1}{\text x_1^{\ 2}} \times \frac{\text x_2^{\ 2}}{\text Q_2} = \frac{\text x_2}{\text x_1} \\[1em] \frac{\text Q_1}{\text Q_2} = \frac{\text x_1^{\ 2}}{\text x_2^{\ 2}} \times \frac{\text x_2}{\text x_1} \\[1em] \frac{\text Q_1}{\text Q_2} = \frac{\text x_1}{\text x_2}\end{aligned}$$

The straight line passing through the origin, that is, option 2.

Reason —

Given, the point mass carries a positive charge q and is released from rest at the edge of the table. The electric field E is uniform and horizontal, along the x-direction, while gravity acts vertically downwards along the y-direction.

Acceleration along the x-direction :

$\text a_\text x = \frac{\text{qE}}{\text m} = \text{a constant}$

Acceleration along the y-direction :

$\text a_\text y = \text g = \text{a constant}$

Since the mass starts from rest, after a time t,

$\text x = \frac{1}{2}\text a_\text x\text t^2, \qquad \text y = \frac{1}{2}\text g\text t^2$

Dividing one by the other, t² cancels,

$$\begin{aligned}\frac{\text y}{\text x} = \frac{\text g}{\text a_\text x} \\[1em] \frac{\text y}{\text x} = \frac{\text{mg}}{\text{qE}} \\[1em] \frac{\text y}{\text x} = \text{a constant}\end{aligned}$$

Hence y is directly proportional to x, and the trajectory of the mass is a straight line passing through the origin.

Note: Both accelerations are constant, so the ratio y/x is fixed and the path must be a straight line. The answer key printed with the question gives option 4, which shows a curved path; the correct option is 2.

Towards the right as its potential energy will decrease.

Reason —

From the figure the lines of force diverge towards the right, so they are crowded on the left and farther apart on the right. Hence the field is non-uniform, being stronger on the left and weaker on the right.

The dipole is placed with + q on the left and − q on the right, so the dipole moment $\vec{\text p}$, which points from − q to + q, is directed towards the left, that is, antiparallel to $\vec{\text E} \space$.

Net force : The charge + q lies in the stronger part of the field and experiences a force qE₁ towards the right, while the charge − q lies in the weaker part and experiences a force qE₂ towards the left. Since E₁ > E₂, the net force is towards the right.

Potential energy : For a dipole, U = − pE cos θ. Here θ = 180°, so U = + pE. As the dipole moves to the right, the field E becomes weaker, and hence U decreases.

Therefore the dipole moves towards the right and its potential energy decreases.

$\text d = \left(\frac{\text q^2\text l}{2\pi\varepsilon_0\text{mg}}\right)^{\frac{1}{3}}$

Reason —

Given, two balls each of mass m and charge q hang from a common point by threads of length l, and each thread makes a small angle θ with the vertical. Let the equilibrium separation be d.

Forces on either ball : the weight mg vertically downwards, the tension T along the thread, and the electrostatic repulsion

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

For equilibrium,

$$\begin{aligned}\text T\sin\theta = \text F, \qquad \text T\cos\theta = \text{mg} \\[1em] \tan\theta = \frac{\text F}{\text{mg}}\end{aligned}$$

For a small angle, $\tan\theta \approx \sin\theta = \frac{\text d/2}{\text l} = \frac{\text d}{2\text l}$,

$$\begin{aligned}\frac{\text d}{2\text l} = \frac{1}{4\pi\varepsilon_0}\frac{\text q^2}{\text d^2\text{mg}} \\[1em] \text d^3 = \frac{2\text l\text q^2}{4\pi\varepsilon_0\text{mg}} \\[1em] \text d^3 = \frac{\text q^2\text l}{2\pi\varepsilon_0\text{mg}} \\[1em] \text d = \left(\frac{\text q^2\text l}{2\pi\varepsilon_0\text{mg}}\right)^{\frac{1}{3}}\end{aligned}$$

$\frac{1}{4\pi\varepsilon_0}\frac{\text q}{\left(2\text l^2\right)}\left(2\sqrt 2 - 1\right)$

Reason —

Taking O as the origin, with the x-axis horizontal and the y-axis vertical, the positions and charges read off from the figure are

• B (−l, 0) : + q, C (0, l) : + 2q, G (l, 0) : + 2q, F (0, −l) : + q — each at a distance l from O
• A (−l, l) : − q, D (l, l) : + q, E (−l, −l) : + 2q, H (l, −l) : − q — each at a distance $\text l\sqrt 2$ from O

Let $\text k = \frac{1}{4\pi\varepsilon_0}$ and $\text E_0 = \frac{\text{kq}}{\text l^2}$.

Contribution of the four charges at a distance l : The field of a positive charge at O points away from that charge,

$\vec{\text E}_\text B = \text E_0\hat{\text i}, \quad \vec{\text E}_\text C = -2\text E_0\hat{\text j}, \quad \vec{\text E}_\text G = -2\text E_0\hat{\text i}, \quad \vec{\text E}_\text F = \text E_0\hat{\text j}$

Adding these,

$$\begin{aligned}\vec{\text E}_{(1)} = \left(1 - 2\right)\text E_0\hat{\text i} + \left(-2 + 1\right)\text E_0\hat{\text j} \\[1em] \vec{\text E}_{(1)} = -\text E_0\left(\hat{\text i} + \hat{\text j}\right)\end{aligned}$$

Contribution of the four charges at a distance $\text l\sqrt 2$ : Here the magnitude of the field of a charge Q is $\frac{\text{kQ}}{2\text l^2}$, and the unit vectors are along the diagonals,

$$\begin{aligned}\vec{\text E}_\text A = \frac{\text E_0}{2}\frac{\left(-\hat{\text i} + \hat{\text j}\right)}{\sqrt 2}, \qquad \vec{\text E}_\text D = \frac{\text E_0}{2}\frac{\left(-\hat{\text i} - \hat{\text j}\right)}{\sqrt 2} \\[1em] \vec{\text E}_\text E = \text E_0\frac{\left(\hat{\text i} + \hat{\text j}\right)}{\sqrt 2}, \qquad \vec{\text E}_\text H = \frac{\text E_0}{2}\frac{\left(\hat{\text i} - \hat{\text j}\right)}{\sqrt 2}\end{aligned}$$

Adding the x-components inside the bracket, $-\frac{1}{2} - \frac{1}{2} + 1 + \frac{1}{2} = \frac{1}{2}$, and similarly the y-components, $\frac{1}{2} - \frac{1}{2} + 1 - \frac{1}{2} = \frac{1}{2}$,

$\vec{\text E}_{(2)} = \frac{\text E_0}{2\sqrt 2}\left(\hat{\text i} + \hat{\text j}\right)$

Resultant field :

$$\begin{aligned}\vec{\text E} = \vec{\text E}_{(1)} + \vec{\text E}_{(2)} \\[1em] \vec{\text E} = -\text E_0\left(1 - \frac{1}{2\sqrt 2}\right)\left(\hat{\text i} + \hat{\text j}\right) \\[1em] |\vec{\text E}| = \text E_0\left(1 - \frac{1}{2\sqrt 2}\right)\sqrt 2 \\[1em] |\vec{\text E}| = \text E_0\left(\sqrt 2 - \frac{1}{2}\right) \\[1em] |\vec{\text E}| = \text E_0\frac{\left(2\sqrt 2 - 1\right)}{2} \\[1em] \text E = \frac{1}{4\pi\varepsilon_0}\frac{\text q}{\left(2\text l^2\right)}\left(2\sqrt 2 - 1\right)\end{aligned}$$

Note: The answer key printed with the question gives option 4, which works out to $\sqrt 2\ \text E_0$; the value obtained above from the charges and positions shown in the figure is option 2.

1 s

Reason —

Given,

• Specific charge, $\frac{\text q}{\text m} = 8\ \mu\text C/\text g = \frac{8 \times 10^{-6}}{10^{-3}} = 8 \times 10^{-3}$ C/kg
• Electric field, E = 100 V/m
• Distance from the wall, s = 10 cm = 0.1 m

Acceleration of the body :

$$\begin{aligned}\text a = \frac{\text{qE}}{\text m} \\[1em] \text a = \left(\frac{\text q}{\text m}\right)\text E \\[1em] \text a = \left(8 \times 10^{-3}\right)(100) \\[1em] \text a = 0.8\ \text m/\text s^2\end{aligned}$$

Time to reach the wall : The body starts from rest, so

$$\begin{aligned}\text s = \frac{1}{2}\text a\text t^2 \\[1em] 0.1 = \frac{1}{2}(0.8)\text t^2 \\[1em] \text t^2 = \frac{0.2}{0.8} \\[1em] \text t^2 = 0.25 \\[1em] \text t = 0.5\ \text s\end{aligned}$$

Time period : The collision with the wall is perfectly elastic, so the body rebounds with the same speed and is then retarded by the same electric force. It therefore takes an equal time of 0.5 s to return to its starting point, where it comes momentarily to rest, and the motion repeats.

$$\begin{aligned}\text T = 2\text t \\[1em] \text T = 2 \times 0.5 \\[1em] \text T = 1\ \text s\end{aligned}$$

Note: The printed figure labels the field as 10 V/m, but the question stem gives 100 V/m. The stem value is used; it gives the keyed answer, 1 s.

1.3 s

Reason —

Given,

• Angle of the incline, θ = 30°; height, h = 1 m
• Mass, m = 1 kg; charge, q = 5 mC = 5 × 10^-3 C
• Horizontal electric field, E = 200 N/C
• Coefficient of friction, μ = 0.2; g = 9.8 m/s²

Length of the incline :

$$\begin{aligned}\text L = \frac{\text h}{\sin 30°} \\[1em] \text L = \frac{1}{\frac{1}{2}} \\[1em] \text L = 2\ \text m\end{aligned}$$

Electric force :

$$\begin{aligned}\text F = \text{qE} \\[1em] \text F = \left(5 \times 10^{-3}\right)(200) \\[1em] \text F = 1\ \text N, \quad \text{directed horizontally}\end{aligned}$$

Forces along the incline : The component of the weight down the incline is mg sin θ, while the horizontal electric force has a component qE cos θ directed up the incline, since it opposes the sliding of the body.

Normal reaction : The normal reaction is the sum of the components of the weight and of the electric force perpendicular to the incline,

$$\begin{aligned}\text N = \text{mg}\cos\theta + \text{qE}\sin\theta \\[1em] \text N = (1)(9.8)\left(\frac{\sqrt 3}{2}\right) + (1)\left(\frac{1}{2}\right) \\[1em] \text N = 8.487 + 0.5 \\[1em] \text N = 8.987\ \text N\end{aligned}$$

Force of friction :

$$\begin{aligned}\text f = \mu\text N \\[1em] \text f = (0.2)(8.987) \\[1em] \text f = 1.797\ \text N, \quad \text{up the incline}\end{aligned}$$

Net force down the incline :

$$\begin{aligned}\text F_{\text{net}} = \text{mg}\sin\theta - \text{qE}\cos\theta - \text f \\[1em] = (1)(9.8)\left(\frac{1}{2}\right) - (1)\left(\frac{\sqrt 3}{2}\right) - 1.797 \\[1em] \text F_{\text{net}} = 4.9 - 0.866 - 1.797 \\[1em] \text F_{\text{net}} = 2.237\ \text N\end{aligned}$$

Acceleration :

$$\begin{aligned}\text a = \frac{\text F_{\text{net}}}{\text m} \\[1em] \text a = 2.237\ \text m/\text s^2\end{aligned}$$

Time taken : Starting from rest and covering L = 2 m,

$$\begin{aligned}\text L = \frac{1}{2}\text a\text t^2 \\[1em] \quad 2 = \frac{1}{2}(2.237)\text t^2\\[1em] \text t^2 = \frac{4}{2.237} \\[1em] \text t^2 = 1.788 \\[1em] \text t = 1.3\ \text s\end{aligned}$$

$\frac{\text Q}{2\sqrt 3 \pi\varepsilon_0\text L^2}$

Reason —

Given, a thin wire of length L carries a charge Q, and the point P lies on its perpendicular bisector at a distance $\text a = \frac{\sqrt 3}{2}\text L$ from the centre of the rod.

Linear charge density :

$\lambda = \frac{\text Q}{\text L}$

For a finite line-charge, the components of the field along the length cancel in pairs about the mid-point, and the field at a point on the perpendicular bisector is

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{2\lambda}{\text a}\sin\theta$

where θ is the angle subtended at P by half the rod, so that

$\sin\theta = \frac{\frac{\text L}{2}}{\sqrt{\text a^2 + \frac{\text L^2}{4}}}$

Substituting $\text a = \frac{\sqrt 3}{2}\text L$,

$$\begin{aligned}\sqrt{\text a^2 + \frac{\text L^2}{4}} = \sqrt{\frac{3\text L^2}{4} + \frac{\text L^2}{4}} \\[1em] \sqrt{\text a^2 + \frac{\text L^2}{4}} = \text L \\[1em] \sin\theta = \frac{\text L/2}{\text L} \\[1em] \sin\theta = \frac{1}{2}\end{aligned}$$

Hence

$$\begin{aligned}\text E = \frac{1}{4\pi\varepsilon_0} \times \frac{2\left(\frac{\text Q}{\text L}\right)}{\frac{\sqrt 3}{2}\text L} \times \frac{1}{2} \\[1em] \text E = \frac{1}{4\pi\varepsilon_0}\frac{2\text Q}{\sqrt 3\ \text L^2} \\[1em] \text E = \frac{\text Q}{2\sqrt 3\ \pi\varepsilon_0\text L^2}\end{aligned}$$

$\frac{-2\text Q}{3\sqrt 3 \pi\varepsilon_0\text a^2}\left(\hat{\text x} + \hat{\text y} + \hat{\text z}\right)$

Reason —

Given, a cube of side a with a charge + Q at each of its eight vertices except the origin, where the charge is − Q.

If charges + Q were placed at all the eight vertices, the fields produced at the centre would cancel in pairs, since the charges at the ends of every body diagonal are equidistant from the centre and produce equal and opposite fields. The resultant would then be zero.

The actual arrangement is obtained from this by adding a charge − 2Q at the origin, since (+ Q) + (− 2Q) = − Q. Hence the field at the centre is simply the field due to a single charge − 2Q placed at the origin.

Distance of the centre from the origin : Half the body diagonal of the cube,

$$\begin{aligned}\text r = \frac{\sqrt 3}{2}\text a \\[1em] \text r^2 = \frac{3\text a^2}{4}\end{aligned}$$

Magnitude of the field :

$$\begin{aligned}\text E = \frac{1}{4\pi\varepsilon_0}\frac{2\text Q}{\frac{3\text a^2}{4}} \\[1em] \text E = \frac{8\text Q}{4\pi\varepsilon_0 \times 3\text a^2} \\[1em] \text E = \frac{2\text Q}{3\pi\varepsilon_0\text a^2}\end{aligned}$$

Direction : The field of a negative charge points towards it, that is, from the centre towards the origin. The unit vector along this direction is $-\frac{\left(\hat{\text x} + \hat{\text y} + \hat{\text z}\right)}{\sqrt 3}$.

$$\begin{aligned}\vec{\text E} = \frac{2\text Q}{3\pi\varepsilon_0\text a^2}\times\left(-\frac{\hat{\text x} + \hat{\text y} + \hat{\text z}}{\sqrt 3}\right) \\[1em] \vec{\text E} = \frac{-2\text Q}{3\sqrt 3\ \pi\varepsilon_0\text a^2}\left(\hat{\text x} + \hat{\text y} + \hat{\text z}\right)\end{aligned}$$

$\sqrt{\frac{4\pi^3\varepsilon_0\text{ma}^3}{\text q_0\text Q}}$

Reason —

Given, two charges Q are fixed a distance 2a apart, and a charge q₀ of mass m is placed at the mid-point, that is, at a distance a from each.

Let q₀ be displaced through a small distance x towards the right-hand charge. Its distance from that charge becomes (a − x) and from the charge at A becomes (a + x). The force F₁ due to the nearer charge acts towards the left and the force F₂ due to the charge at A acts towards the right, so, taking the direction of the displacement as positive, the net force on q₀ is

$$\begin{aligned}\text F = \frac{\text Q\text q_0}{4\pi\varepsilon_0}\left[\frac{1}{(\text a + \text x)^2} - \frac{1}{(\text a - \text x)^2}\right] \\[1em] \text F = \frac{\text Q\text q_0}{4\pi\varepsilon_0}\left[\frac{(\text a - \text x)^2 - (\text a + \text x)^2}{\left(\text a^2 - \text x^2\right)^2}\right] \\[1em] \text F = -\frac{\text Q\text q_0}{4\pi\varepsilon_0}\frac{4\text{ax}}{\left(\text a^2 - \text x^2\right)^2}\end{aligned}$$

For a small displacement, x² is negligible compared with a²,

$$\begin{aligned}\text F = -\frac{\text Q\text q_0}{4\pi\varepsilon_0}\frac{4\text{ax}}{\text a^4} \\[1em] \text F = -\frac{\text Q\text q_0\text x}{\pi\varepsilon_0\text a^3}\end{aligned}$$

The negative sign shows that this force is directed towards the mean position, so the motion is simple harmonic. Comparing with F = − mω²x,

$$\begin{aligned}\text m\omega^2 = \frac{\text Q\text q_0}{\pi\varepsilon_0\text a^3} \\[1em] \omega^2 = \frac{\text Q\text q_0}{\pi\varepsilon_0\text{ma}^3}\end{aligned}$$

Time period :

$$\begin{aligned}\text T = \frac{2\pi}{\omega} \\[1em] \text T = 2\pi\sqrt{\frac{\pi\varepsilon_0\text{ma}^3}{\text Q\text q_0}} \\[1em] \text T = \sqrt{\frac{4\pi^2 \times \pi\varepsilon_0\text{ma}^3}{\text q_0\text Q}} \\[1em] \text T = \sqrt{\frac{4\pi^3\varepsilon_0\text{ma}^3}{\text q_0\text Q}}\end{aligned}$$

12 cm

Reason —

Given,

• Mass of each particle, m = 10 g = 10 × 10^-3 kg
• Charge on each, q = 2.0 × 10^-7 C
• Coefficient of friction, μ = 0.25; g = 10 m s^-2

The particles are in limiting equilibrium, so the electrostatic force of repulsion between them is just balanced by the limiting force of friction,

$$\begin{aligned}\frac{1}{4\pi\varepsilon_0}\frac{\text q^2}{\text L^2} = \mu\text{mg} \\[1em] \text L^2 = \frac{1}{4\pi\varepsilon_0}\frac{\text q^2}{\mu\text{mg}}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text L^2 = \frac{\left(9 \times 10^9\right)\left(2.0 \times 10^{-7}\right)^2}{(0.25)\left(10 \times 10^{-3}\right)(10)} \\[1em] \text L^2 = \frac{9 \times 10^9 \times 4 \times 10^{-14}}{2.5 \times 10^{-2}} \\[1em] \text L^2 = \frac{3.6 \times 10^{-4}}{2.5 \times 10^{-2}} \\[1em] \text L^2 = 1.44 \times 10^{-2}\ \text m^2 \\[1em] \text L = 0.12\ \text m \\[1em] \text L = 12\ \text{cm}\end{aligned}$$

tan^-1(2)

Reason —

Given,

• Electric field, $\text E = \frac{8\text m}{\text e}$ V/m
• Length of the plates, L = 1 m
• Speed of entry, u = 2 m/s, parallel to the plates

Acceleration of the electron perpendicular to the plates :

$$\begin{aligned}\text a = \frac{\text{eE}}{\text m} \\[1em] \text a = \frac{\text e}{\text m}\times\frac{8\text m}{\text e} \\[1em] \text a = 8\ \text m/\text s^2\end{aligned}$$

Time spent inside the field :

$$\begin{aligned}\text t = \frac{\text L}{\text u} \\[1em] \text t = \frac{1}{2} \\[1em] \text t = 0.5\ \text s\end{aligned}$$

Velocity components as the electron leaves the field :

$$\begin{aligned}\text v_\text x = \text u \\[1em] \text v_\text x = 2\ \text m/\text s \\[1em] \text v_\text y = \text{at} \\[1em] \text v_\text y = 8 \times 0.5 \\[1em] \text v_\text y = 4\ \text m/\text s\end{aligned}$$

Angle of deviation :

$$\begin{aligned}\tan\theta = \frac{\text v_\text y}{\text v_\text x} \\[1em] \tan\theta = \frac{4}{2} \\[1em] \tan\theta = 2 \\[1em] \theta = \tan^{-1}(2)\end{aligned}$$

$\frac{ρ_0\text r}{4\varepsilon_0}\left(1 - \frac{\text r}{\text R}\right)$

Reason —

Given, $ρ(\text r) = ρ_0\left(\frac{3}{4} - \frac{\text r}{\text R}\right)$ for r ≤ R and zero for r > R.

Charge enclosed within a sphere of radius r : Taking a thin spherical shell of radius r′ and thickness dr′, of volume 4πr′²dr′,

$$\begin{aligned}\text q_{\text{enc}} = \int_0^\text r ρ_0\left(\frac{3}{4} - \frac{\text r'}{\text R}\right)4\pi\text r'^2\ \text{dr}' \\[1em] = 4\piρ_0\int_0^\text r\left(\frac{3\text r'^2}{4} - \frac{\text r'^3}{\text R}\right)\text{dr}'\end{aligned}$$

Carrying out the integration,

$$\begin{aligned}\text q_{\text{enc}} = 4\piρ_0\left[\frac{3}{4}\frac{\text r'^3}{3} - \frac{1}{\text R}\frac{\text r'^4}{4}\right]_0^\text r \\[1em] \text q_{\text{enc}} = 4\piρ_0\left[\frac{\text r^3}{4} - \frac{\text r^4}{4\text R}\right] \\[1em] \text q_{\text{enc}} = \piρ_0\text r^3\left(1 - \frac{\text r}{\text R}\right)\end{aligned}$$

Electric field at P : By the spherical symmetry of the distribution, the charge enclosed behaves as though it were concentrated at the centre,

$$\begin{aligned}\text E = \frac{1}{4\pi\varepsilon_0}\frac{\text q_{\text{enc}}}{\text r^2} \\[1em] \text E = \frac{1}{4\pi\varepsilon_0}\frac{\piρ_0\text r^3\left(1 - \frac{\text r}{\text R}\right)}{\text r^2} \\[1em] \text E = \frac{ρ_0\text r}{4\varepsilon_0}\left(1 - \frac{\text r}{\text R}\right)\end{aligned}$$

$\frac{ρ\text{qR}^2}{4\varepsilon_0}$

Reason —

Given, a long cylinder of radius R carries a uniform volume charge density ρ, and a charge particle q revolves around it in a circular path of radius r (r > R).

Field outside the charged cylinder : For a length L of the cylinder the enclosed charge is ρπR²L, and by the cylindrical symmetry this behaves as a line-charge of linear density

$\lambda = ρ\pi\text R^2$

The field at a distance r from the axis of a line-charge is

$$\begin{aligned}\text E = \frac{\lambda}{2\pi\varepsilon_0\text r} \\[1em] \text E = \frac{ρ\text R^2}{2\varepsilon_0\text r}\end{aligned}$$

Condition for circular motion : The electrostatic force supplies the necessary centripetal force,

$$\begin{aligned}\text{qE} = \frac{\text m\text v^2}{\text r} \\[1em] \text q\frac{ρ\text R^2}{2\varepsilon_0\text r} = \frac{\text m\text v^2}{\text r} \\[1em] \text m\text v^2 = \frac{\text qρ\text R^2}{2\varepsilon_0}\end{aligned}$$

Kinetic energy :

$$\begin{aligned}\text K = \frac{1}{2}\text m\text v^2 \\[1em] \text K = \frac{1}{2}\times\frac{\text qρ\text R^2}{2\varepsilon_0} \\[1em] \text K = \frac{ρ\text{qR}^2}{4\varepsilon_0}\end{aligned}$$

Note that the result is independent of the radius r of the circular path.

$\sqrt 3$ : 1

Reason —

Given, three identical balls each of charge 2 C hang from a common point and form an equilateral triangle of side 1 m.

Force between any two balls :

$$\begin{aligned}\text F = \frac{1}{4\pi\varepsilon_0}\frac{(2)(2)}{(1)^2} \\[1em] \text F = \frac{1}{4\pi\varepsilon_0}\times 4\end{aligned}$$

Net force on one ball : Each ball is repelled by the other two with forces of equal magnitude F, and the angle between these two forces is 60°, since the triangle is equilateral. Hence

$$\begin{aligned}\text F_{\text{net}} = \sqrt{\text F^2 + \text F^2 + 2\text F^2\cos 60°} \\[1em] \text F_{\text{net}} = \sqrt{2\text F^2 + 2\text F^2\left(\frac{1}{2}\right)} \\[1em] \text F_{\text{net}} = \sqrt{3\text F^2} \\[1em] \text F_{\text{net}} = \sqrt 3\ \text F\end{aligned}$$

Required ratio :

$$\begin{aligned}\frac{\text F_{\text{net}}}{\text F} = \frac{\sqrt 3\ \text F}{\text F} \\[1em] \frac{\text F_{\text{net}}}{\text F} = \sqrt 3 : 1\end{aligned}$$

0.5 m

Reason —

Given,

• Mass, m = 100 mg = 100 × 10^-6 kg = 1 × 10^-4 kg
• Charge, q = 40 μC = 40 × 10^-6 C
• Electric field, E = 1 × 10⁵ N C^-1
• Initial velocity, u = 200 m s^-1, opposite to the field

Electric force on the particle :

$$\begin{aligned}\text F = \text{qE} \\[1em] \text F = \left(40 \times 10^{-6}\right)\left(1 \times 10^{5}\right) \\[1em] \text F = 4\ \text N\end{aligned}$$

Since the particle is thrown opposite to the field while the force on the positive charge is along the field, this force retards the particle.

Retardation :

$$\begin{aligned}\text a = \frac{\text F}{\text m} \\[1em] \text a = \frac{4}{1 \times 10^{-4}} \\[1em] \text a = 4 \times 10^{4}\ \text m\ \text s^{-2}\end{aligned}$$

Distance travelled before coming to rest : Using v² = u² − 2as with v = 0,

$$\begin{aligned}0 = (200)^2 - 2\left(4 \times 10^{4}\right)\text s \\[1em] \text s = \frac{4 \times 10^{4}}{8 \times 10^{4}} \\[1em] \text s = 0.5\ \text m\end{aligned}$$

$\text x = \frac{\text d}{2\sqrt 2}$

Reason —

Given,

• Two point-charges, each Q, separated by a distance d
• Third charge q on the perpendicular bisector, at a distance x from the mid-point O

The distance of q from each of the two charges is

$\text r = \sqrt{\text x^2 + \frac{\text d^2}{4}}$

By Coulomb's law, the magnitude of the force on q due to each charge is

$\text F_1 = \text F_2 = \frac{1}{4\pi\varepsilon_0}\frac{\text{Qq}}{\left(\text x^2 + \frac{\text d^2}{4}\right)}$

If θ is the angle each force makes with the bisector, then the components perpendicular to the bisector are equal and opposite and cancel out, while the components along the bisector add up. Hence the resultant force is

$\text F = 2\text F_1\cos\theta, \qquad \text{where}\ \cos\theta = \frac{\text x}{\left(\text x^2 + \frac{\text d^2}{4}\right)^{1/2}}$

Substituting,

$$\begin{aligned}\text F = \frac{2}{4\pi\varepsilon_0}\frac{\text{Qq}}{\left(\text x^2 + \frac{\text d^2}{4}\right)} \times \frac{\text x}{\left(\text x^2 + \frac{\text d^2}{4}\right)^{1/2}} \\[1em] \text F = \frac{\text{Qq}}{2\pi\varepsilon_0}\frac{\text x}{\left(\text x^2 + \frac{\text d^2}{4}\right)^{3/2}}\end{aligned}$$

For F to be maximum, $\frac{\text{dF}}{\text{dx}} = 0$. Differentiating the fraction by the quotient rule,

$\frac{\text{dF}}{\text{dx}} \propto \frac{\left(\text x^2 + \frac{\text d^2}{4}\right)^{3/2}(1) - \text x \times \frac{3}{2}\left(\text x^2 + \frac{\text d^2}{4}\right)^{1/2}(2\text x)}{\left(\text x^2 + \frac{\text d^2}{4}\right)^{3}}$

Putting the numerator equal to zero,

$\left(\text x^2 + \frac{\text d^2}{4}\right)^{3/2} - 3\text x^2\left(\text x^2 + \frac{\text d^2}{4}\right)^{1/2} = 0$

Taking $\left(\text x^2 + \frac{\text d^2}{4}\right)^{1/2}$ common and cancelling it,

$$\begin{aligned}\left(\text x^2 + \frac{\text d^2}{4}\right) - 3\text x^2 = 0 \\[1em] \frac{\text d^2}{4} = 2\text x^2 \\[1em] \text x^2 = \frac{\text d^2}{8} \\[1em] \text x = \frac{\text d}{2\sqrt 2}\end{aligned}$$

$\frac{1}{\text R^3}$

Reason —

Given, the charges − q and + q are placed a distance L apart, so they form an electric dipole of moment

$\text p = \text{qL}$

For a point at a distance R from the dipole, with R >> L, the dipole behaves as a short dipole. The intensity of the electric field is then

$\text E = \frac{1}{4\pi\varepsilon_0}\frac{2\text p}{\text R^3} \quad \text{(axial)}, \qquad \text E = \frac{1}{4\pi\varepsilon_0}\frac{\text p}{\text R^3} \quad \text{(equatorial)}$

In either case

$\text E \propto \frac{1}{\text R^3}$

$\sqrt{\frac{4\text{qE}}{3\text{ml}}}$

Reason —

Given, the positive charge has mass m and the negative charge has mass 2m, the two charges being of magnitude q and separated by a distance l. Hence the dipole moment is

$\text p = \text{ql}$

Moment of inertia about the mid-point of the dipole : Each charge is at a distance $\frac{\text l}{2}$ from the mid-point,

$$\begin{aligned}\text I = \text m\left(\frac{\text l}{2}\right)^2 + 2\text m\left(\frac{\text l}{2}\right)^2 \\[1em] \text I = 3\text m\frac{\text l^2}{4} \\[1em] \text I = \frac{3\text m\text l^2}{4}\end{aligned}$$

Restoring torque : When the dipole axis makes a very small angle θ with the field,

$\tau = -\text{pE}\sin\theta \approx -\text{pE}\theta$

which is of the form τ = − cθ, so the oscillations are angular simple harmonic with

$\omega = \sqrt{\frac{\text{pE}}{\text I}}$

Substituting the values,

$$\begin{aligned}\omega = \sqrt{\frac{\text{qlE}}{\frac{3\text m\text l^2}{4}}} \\[1em] \omega = \sqrt{\frac{4\text{qE}}{3\text{ml}}}\end{aligned}$$

2 mC

Reason —

Given,

• Angle with the field, θ = 30°
• Electric field, E = 2 × 10⁵ N C^-1
• Torque, τ = 4 N-m
• Dipole length, 2l = 2 cm = 2 × 10^-2 m

The torque on a dipole placed in a uniform electric field is

$$\begin{aligned}\tau = \text{pE}\sin\theta \\[1em] \text p = \frac{\tau}{\text E\sin\theta}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text p = \frac{4}{\left(2 \times 10^{5}\right)\left(\frac{1}{2}\right)} \\[1em] \text p = \frac{4}{1 \times 10^{5}} \\[1em] \text p = 4 \times 10^{-5}\ \text{C-m}\end{aligned}$$

Since p = q × 2l,

$$\begin{aligned}\text q = \frac{\text p}{2\text l} \\[1em] \text q = \frac{4 \times 10^{-5}}{2 \times 10^{-2}} \\[1em] \text q = 2 \times 10^{-3}\ \text C \\[1em] \text q = 2\ \text{mC}\end{aligned}$$

q²/(32 πε₀R³m)

Reason —

Given, two beads each of charge q and mass m lie on a frictionless circular hoop of radius R, one of them being glued to the hoop.

Let the angular separation of the two beads, measured at the centre of the hoop, be θ. The straight-line distance between them is the chord

$\text d = 2\text R\sin\frac{\theta}{2}$

Potential energy of the system :

$$\begin{aligned}\text U = \frac{1}{4\pi\varepsilon_0}\frac{\text q^2}{\text d} \\[1em] \text U = \frac{1}{4\pi\varepsilon_0}\frac{\text q^2}{2\text R\sin\frac{\theta}{2}}\end{aligned}$$

U is least when $\sin\frac{\theta}{2}$ is greatest, that is, when θ = π. Hence the equilibrium position is the diametrically opposite point.

Let θ = π + φ, where φ is small. Then $\sin\frac{\theta}{2} = \sin\left(\frac{\pi}{2} + \frac{\phi}{2}\right) = \cos\frac{\phi}{2}$,

$\text U = \frac{\text q^2}{8\pi\varepsilon_0\text R}\sec\frac{\phi}{2} \approx \frac{\text q^2}{8\pi\varepsilon_0\text R}\left(1 + \frac{\phi^2}{8}\right)$

using $\sec\text x \approx 1 + \frac{\text x^2}{2}$ for small x.

Angular simple harmonic motion : The moving bead travels along the hoop, so its moment of inertia about the centre is I = mR². The effective torsional constant is

$$\begin{aligned}\text c = \frac{\text d^2\text U}{\text d\phi^2} \\[1em] \text c = \frac{\text q^2}{8\pi\varepsilon_0\text R}\times\frac{1}{4} \\[1em] \text c = \frac{\text q^2}{32\pi\varepsilon_0\text R} \\[1em] \omega^2 = \frac{\text c}{\text I} \\[1em] \omega^2 = \frac{\text q^2}{32\pi\varepsilon_0\text R \times \text{mR}^2} \\[1em] \omega^2 = \frac{\text q^2}{32\pi\varepsilon_0\text R^3\text m}\end{aligned}$$

Given,

• Mass of the free particle, m = 1 mg = 1 × 10^-6 kg
• The two stationary particles are 2 m apart, so the free particle at the mid-point is at a distance a = 1 m from each
• Each of the three particles carries the same charge q, with q² = 10 C²
• $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$ N-m² C^-2

Let the free particle be displaced through a small distance x from the mid-point, along the line joining the two fixed charges, towards one of them. Its distance from that charge becomes (a − x) and from the other becomes (a + x). The nearer charge pushes it back towards the mid-point with the larger force, so, taking the direction of the displacement as positive, the net force on it is

$$\begin{aligned}\text F = \frac{\text q^2}{4\pi\varepsilon_0}\left[\frac{1}{(\text a + \text x)^2} - \frac{1}{(\text a - \text x)^2}\right] \\[1em] \text F = \frac{\text q^2}{4\pi\varepsilon_0}\left[\frac{(\text a - \text x)^2 - (\text a + \text x)^2}{\left(\text a^2 - \text x^2\right)^2}\right] \\[1em] \text F = -\frac{\text q^2}{4\pi\varepsilon_0}\frac{4\text{ax}}{\left(\text a^2 - \text x^2\right)^2}\end{aligned}$$

Since x << a, the term x² may be neglected in comparison with a²,

$$\begin{aligned}\text F = -\frac{\text q^2}{4\pi\varepsilon_0}\frac{4\text{ax}}{\text a^4} \\[1em] \text F = -\frac{\text q^2}{4\pi\varepsilon_0}\frac{4\text x}{\text a^3}\end{aligned}$$

The force is proportional to the displacement and, being negative, is directed towards the mean position, so the motion is simple harmonic. Comparing with F = − mω²x,

$$\begin{aligned}-\text m\omega^2\text x = -\frac{1}{4\pi\varepsilon_0}\frac{4\text q^2\text x}{\text a^3} \\[1em] \omega^2 = \frac{1}{4\pi\varepsilon_0}\frac{4\text q^2}{\text m\text a^3}\end{aligned}$$

Substituting the values, with a = 1 m,

$$\begin{aligned}\omega^2 = \frac{\left(9 \times 10^9\right)(4)(10)}{\left(1 \times 10^{-6}\right)(1)^3} \\[1em] \omega^2 = \frac{3.6 \times 10^{11}}{1 \times 10^{-6}} \\[1em] \omega^2 = 3.6 \times 10^{17} \\[1em] \omega = \sqrt{3.6 \times 10^{17}} \\[1em] \omega = 6 \times 10^{8}\ \text{rad/s} \\[1em] \omega = 6000 \times 10^{5}\ \text{rad/s}\end{aligned}$$

Hence, the value to be filled in the blank is 6000.

Given,

• Point-charges of 1 μC = 1 × 10^-6 C each, placed on the Y-axis at y = 1 m, 2 m, 4 m, 8 m, ... , the distances forming a geometric progression with common ratio 2
• Test charge at the origin, q₀ = 1 C
• $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$ N-m²/C²

All the charges are of the same sign and lie on the same side of the origin, so all the forces on the charge at the origin are in the same direction and simply add up.

Total force :

$$\begin{aligned}\text F = \frac{1}{4\pi\varepsilon_0}\text q\text q_0\left[\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \dots\right] \\[1em] = \frac{1}{4\pi\varepsilon_0}\text q\text q_0\left[1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots\right]\end{aligned}$$

The series inside the bracket is an infinite geometric progression with first term 1 and common ratio $\frac{1}{4}$. Its sum is

$$\begin{aligned}\text S = \frac{1}{1 - \frac{1}{4}} \\[1em] \text S = \frac{4}{3}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text F = \left(9 \times 10^9\right)\left(1 \times 10^{-6}\right)(1)\times\frac{4}{3} \\[1em] \text F = 9 \times 10^3 \times \frac{4}{3} \\[1em] \text F = 12 \times 10^{3}\ \text N\end{aligned}$$

Comparing with F = x × 10³ N, we get x = 12.

Hence, the value of x to the nearest integer is 12.

Given,

• Mass of each sphere, m = 10 mg = 10 × 10^-6 kg = 1 × 10^-5 kg
• Length of each thread, l = 0.5 m
• Separation of the spheres, x = 0.20 m
• g = 10 m s^-2
• $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$ N-m² C^-2

Let θ be the angle each thread makes with the vertical, and q the charge on each sphere. Resolving the tension into components,

$$\begin{aligned}\text T\sin\theta = \text F, \qquad \text T\cos\theta = \text{mg} \\[1em] \tan\theta = \frac{\text F}{\text{mg}}\end{aligned}$$

Angle of the thread :

$$\begin{aligned}\sin\theta = \frac{\text x/2}{\text l} \\[1em] \sin\theta = \frac{0.10}{0.5} \\[1em] \sin\theta = \frac{1}{5} \\[1em] \cos\theta = \sqrt{1 - \sin^2\theta} \\[1em] \cos\theta = \sqrt{1 - \frac{1}{25}} \\[1em] \cos\theta = \frac{\sqrt{24}}{5} \\[1em] \tan\theta = \frac{\sin\theta}{\cos\theta} \\[1em] \tan\theta = \frac{1}{\sqrt{24}}\end{aligned}$$

Force of repulsion required :

$$\begin{aligned}\text F = \text{mg}\tan\theta \\[1em] \text F = \left(1 \times 10^{-5}\right)(10)\left(\frac{1}{\sqrt{24}}\right) \\[1em] \text F = \frac{10^{-4}}{\sqrt{24}} \\[1em] \text F = 2.041 \times 10^{-5}\ \text N\end{aligned}$$

Charge on each sphere :

$$\begin{aligned}\frac{1}{4\pi\varepsilon_0}\frac{\text q^2}{\text x^2} = \text F \\[1em] \text q^2 = \frac{\text F\text x^2}{9 \times 10^9} \\[1em] \text q^2 = \frac{\left(2.041 \times 10^{-5}\right)(0.20)^2}{9 \times 10^9} \\[1em] \text q^2 = \frac{8.164 \times 10^{-7}}{9 \times 10^9} \\[1em] \text q^2 = 9.071 \times 10^{-17} \\[1em] \text q = 9.524 \times 10^{-9}\ \text C \\[1em] \text q = 0.9524 \times 10^{-8}\ \text C\end{aligned}$$

Comparing with $\text q = \frac{\text a}{21} \times 10^{-8}$ C,

$$\begin{aligned}\text a = 21 \times 0.9524 \\[1em] \text a = 20.00 \approx 20\end{aligned}$$

Hence, the value of a is 20.

Note: The printed key gives 630 because its working treats 10 mg as 10 × 10^-3 kg; using m = 10^-5 kg gives a ≈ 20.

Given,

• Specific charge, $\frac{\text q}{\text m} = 2 \times 10^{11}$ C kg^-1
• Initial velocity, $\vec{\text v}_0 = 3 \times 10^{7}\ \hat{\text i}$ m s^-1
• Electric field, $\vec{\text E} = 1.8\ \hat{\text j}$ kV m^-1 = $1.8 \times 10^{3}\ \hat{\text j}$ V m^-1
• Length of the field region along x, L = 10 cm = 0.1 m

The particles enter the field at right angles to it, so the motion along the x-direction is uniform while that along the y-direction is uniformly accelerated.

Acceleration along the y-direction :

$$\begin{aligned}\text a = \frac{\text{qE}}{\text m} \\[1em] \text a = \left(\frac{\text q}{\text m}\right)\text E \\[1em] \text a = \left(2 \times 10^{11}\right)\left(1.8 \times 10^{3}\right) \\[1em] \text a = 3.6 \times 10^{14}\ \text m\ \text s^{-2}\end{aligned}$$

Time spent inside the field :

$$\begin{aligned}\text t = \frac{\text L}{\text v_0} \\[1em] \text t = \frac{0.1}{3 \times 10^{7}} \\[1em] \text t = 3.33 \times 10^{-9}\ \text s\end{aligned}$$

Deflection along the y-direction :

$$\begin{aligned}\text y = \frac{1}{2}\text a\text t^2 \\[1em] \text y = \frac{1}{2}\left(3.6 \times 10^{14}\right)\left(3.33 \times 10^{-9}\right)^2 \\[1em] \text y = \frac{1}{2}\left(3.6 \times 10^{14}\right)\left(1.111 \times 10^{-17}\right) \\[1em] \text y = \frac{1}{2}\left(4 \times 10^{-3}\right) \\[1em] \text y = 2 \times 10^{-3}\ \text m \\[1em] \text y = 2\ \text{mm}\end{aligned}$$

Hence, the deflection of the charge particles in the y direction is 2 mm.

Given,

• The angle between the two strings is 37°, so each string makes an angle $\theta = \frac{37°}{2}$ with the vertical
• Density of the liquid, σ = 0.7 g/cm³
• Density of the material of the spheres, ρ = 1.4 g/cm³
• The angle remains the same in air and in the liquid

In air : If F is the force of repulsion and m the mass of each sphere,

$\tan\theta = \frac{\text F}{\text{mg}} \qquad \dots(\text i)$

In the liquid : Two changes occur.

(a) The force of repulsion is reduced K times, where K is the dielectric constant of the liquid,

$\text F' = \frac{\text F}{\text K}$

(b) The effective weight is reduced by the upthrust. If V is the volume of a sphere, the weight is Vρg and the upthrust is Vσg, so the apparent weight is

$$\begin{aligned}\text W' = \text V\rho\text g - \text V\sigma\text g \\[1em] \text W' = \text{mg}\left(1 - \frac{\sigma}{\rho}\right) \\[1em] \tan\theta = \frac{\text F'}{\text W'} \\[1em] \tan\theta = \frac{\text F}{\text K\ \text{mg}\left(1 - \frac{\sigma}{\rho}\right)} \qquad \dots(\text{ii})\end{aligned}$$

Equating (i) and (ii), since the angle is unchanged,

$$\begin{aligned}\frac{\text F}{\text{mg}} = \frac{\text F}{\text K\ \text{mg}\left(1 - \frac{\sigma}{\rho}\right)} \\[1em] \text K\left(1 - \frac{\sigma}{\rho}\right) = 1 \\[1em] \text K = \frac{\rho}{\rho - \sigma}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text K = \frac{1.4}{1.4 - 0.7} \\[1em] \text K = \frac{1.4}{0.7} \\[1em] \text K = 2\end{aligned}$$

Hence, the dielectric constant of the liquid is 2.

Given,

• Radius of the ring, R = 30 cm = 0.3 m
• Charge distributed over the ring, Q = 2π C
• Charge at the centre, q₀ = 30 pC = 30 × 10^-12 C
• $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$ SI units

Linear charge density of the ring :

$$\begin{aligned}\lambda = \frac{\text Q}{2\pi\text R} \\[1em] \lambda = \frac{2\pi}{2\pi\text R} \\[1em] \lambda = \frac{1}{\text R}\ \text C\ \text m^{-1}\end{aligned}$$

Outward force on a small element of the ring : Consider an element subtending a small angle dθ at the centre. Its length is R dθ and its charge is dQ = λR dθ. The central charge q₀ repels it radially outwards with a force

$$\begin{aligned}\text{dF} = \frac{1}{4\pi\varepsilon_0}\frac{\text q_0\ \lambda\text R\ \text{d}\theta}{\text R^2} \\[1em] \text{dF} = \frac{1}{4\pi\varepsilon_0}\frac{\text q_0\lambda\ \text{d}\theta}{\text R}\end{aligned}$$

Condition of equilibrium of the element : The tension T acts along the ring at both ends of the element. Their resultant, directed inwards, is

$2\text T\sin\frac{\text{d}\theta}{2} \approx \text T\ \text{d}\theta$

Equating this to the outward force,

$$\begin{aligned}\text T\ \text{d}\theta = \frac{1}{4\pi\varepsilon_0}\frac{\text q_0\lambda\ \text{d}\theta}{\text R} \\[1em] \text T = \frac{1}{4\pi\varepsilon_0}\frac{\text q_0\lambda}{\text R} \\[1em] \text T = \frac{1}{4\pi\varepsilon_0}\frac{\text q_0\text Q}{2\pi\text R^2}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text T = \frac{\left(9 \times 10^9\right)\left(30 \times 10^{-12}\right)(2\pi)}{2\pi(0.3)^2} \\[1em] \text T = \frac{\left(9 \times 10^9\right)\left(30 \times 10^{-12}\right)}{0.09} \\[1em] \text T = \frac{0.27}{0.09} \\[1em] \text T = 3\ \text N\end{aligned}$$

Hence, the tension in the ring is 3 N.