Chapter 1 Electric Charges and Fields Class 12 Physics

Question 1.1

What is the force between two small charged spheres having charges of $2 \times 10^{-7} \text{ C}$ and $3 \times 10^{-7} \text{ C}$ placed 30 cm apart in air?

Calculation

Using Coulomb’s Law: $F = \frac{k q_1 q_2}{r^2}$

$$F = \frac{9 \times 10^9 \times (2 \times 10^{-7}) \times (3 \times 10^{-7})}{(0.3)^2} = 6 \times 10^{-3} \text{ N}$$

Result: $6 \times 10^{-3} \text{ N}$ (Repulsive)

Question 1.2

The electrostatic force on a small sphere of charge $0.4 \mu\text{C}$ due to another small sphere of charge $-0.8 \mu\text{C}$ in air is 0.2 N. Find (a) distance (b) force on second sphere.

(a) Distance

$$r^2 = \frac{k q_1 q_2}{F} = \frac{9 \times 10^9 \times (0.4 \times 10^{-6})(0.8 \times 10^{-6})}{0.2} \implies r = 0.12 \text{ m}$$

(b) Force on 2nd Sphere

By Newton’s 3rd Law, force is equal and opposite.

0.2 N (Attractive)

Question 1.3

Check that the ratio $k e^2 / G m_e m_p$ is dimensionless. Determine its value.

Analysis

This is the ratio of Electric Force to Gravitational Force. Since it’s a ratio of forces, it is dimensionless.

$$\text{Value} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.67 \times 10^{-27}} \approx 2.3 \times 10^{39}$$

Question 1.4

(a) Explain ‘electric charge is quantised’. (b) Why ignore it for macroscopic charges?

(a) Charge exists in discrete packets $q = ne$, where $n$ is an integer.

(b) At macroscopic scales, the elementary charge $e$ is so small relative to the total charge that the discrete nature is not observable (continuous distribution approximation).

Question 1.5

Explain consistency of rubbing glass/silk with conservation of charge.

Rubbing causes electron transfer. Glass loses electrons (becomes positive), Silk gains them (becomes negative). The net charge of the system remains zero, conserving charge.

Question 1.6

Four charges $2\mu\text{C}, -5\mu\text{C}, 2\mu\text{C}, -5\mu\text{C}$ on a square. Force on $1\mu\text{C}$ at center?

Symmetry

Charges at opposite corners (A & C) are equal and positive; their forces cancel. Charges at (B & D) are equal and negative; their forces cancel.

Net Force = 0 N

Question 1.7

(a) Why are field lines continuous? (b) Why don’t they cross?

(a) Charges experience continuous force; they don’t jump. Thus, field lines are continuous.

(b) Intersection would imply two directions of electric field at a single point, which is physically impossible.

Question 1.8

$q_A = 3 \mu\text{C}, q_B = -3 \mu\text{C}$ separated by 20 cm. (a) Field at midpoint? (b) Force on $-1.5 \text{ nC}$?

(a) Electric Field

Fields add up (both point to B).

$$E = \frac{2kq}{(d/2)^2} = \frac{2 \times 9 \times 10^9 \times 3 \times 10^{-6}}{0.01} = 5.4 \times 10^6 \text{ N/C}$$

(b) Force

$$F = qE = 1.5 \times 10^{-9} \times 5.4 \times 10^6 = 8.1 \times 10^{-3} \text{ N}$$

Force towards A (Opposite to field).

Question 1.9

Total charge and dipole moment for $q = \pm 2.5 \times 10^{-7} \text{ C}$ at $z = \mp 15 \text{ cm}$.

Total Charge: $+q + (-q) = 0$.

Dipole Moment: $p = q \times 2a = 2.5 \times 10^{-7} \times 0.3 = 7.5 \times 10^{-8} \text{ Cm}$.

Direction: Along negative Z-axis.

Question 1.10

Torque on dipole $p = 4 \times 10^{-9} \text{ Cm}$ at $30^\circ$ to $E = 5 \times 10^4 \text{ N/C}$.

$$\tau = pE \sin 30^\circ = (4 \times 10^{-9})(5 \times 10^4)(0.5) = 10^{-4} \text{ Nm}$$

Question 1.11

Polythene rubbed with wool has charge $-3 \times 10^{-7} \text{ C}$. (a) Electron transfer? (b) Mass transfer?

(a) $n = \frac{q}{e} = \frac{3 \times 10^{-7}}{1.6 \times 10^{-19}} \approx 1.87 \times 10^{12}$ electrons.

(b) Yes. Mass $m = n \times m_e \approx 1.7 \times 10^{-18} \text{ kg}$.

Question 1.12

(a) Force spheres $q=6.5 \times 10^{-7} \text{ C}, r=50 \text{ cm}$. (b) Force if $q \to 2q, r \to r/2$?

(a) Initial

$$F = \frac{9 \times 10^9 \times (6.5 \times 10^{-7})^2}{0.5^2} \approx 1.52 \times 10^{-2} \text{ N}$$

(b) Modified

Factor = $(2 \times 2) / (1/2)^2 = 16$. Force $F’ = 16F \approx 0.24 \text{ N}$.

Question 1.13

Uncharged sphere C touches A, then B, then removed. New force A-B?

Charge sharing: A becomes $q/2$. B becomes $3q/4$.

$$F_{new} = k \frac{(q/2)(3q/4)}{r^2} = \frac{3}{8} F_{old} \approx 5.7 \times 10^{-3} \text{ N}$$

Question 1.14

Identify charge signs from tracks. Which has highest charge-to-mass ratio?

Signs: Deflection towards positive plate = Negative (1, 2). Towards negative plate = Positive (3).

Ratio: Particle 3 deflects most, so highest $q/m$ ratio.

Question 1.15

Flux of $E = 3 \times 10^3 \hat{i}$ through 10 cm square (a) parallel to yz (b) normal at $60^\circ$ to x-axis.

(a) $\Phi = EA = 30 \text{ Nm}^2/\text{C}$.

(b) $\Phi = EA \cos 60^\circ = 15 \text{ Nm}^2/\text{C}$.

Question 1.16

Net flux of uniform field through a cube?

Zero (Flux In = Flux Out).

Question 1.17

Net outward flux $8.0 \times 10^3 \text{ Nm}^2/\text{C}$. (a) Charge? (b) Does zero flux mean no charge?

(a) $q = \epsilon_0 \Phi \approx 0.07 \mu\text{C}$.

(b) No. It means net charge is zero (e.g., dipoles).

Question 1.18

Flux through square from point charge 5 cm above center.

[Image of Cube Symmetry]

Consider the square as one face of a cube. Total flux $q/\epsilon_0$. Flux through one face is $1/6$.

$$\Phi = \frac{1}{6} \frac{10 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 1.88 \times 10^5 \text{ Nm}^2/\text{C}$$

Question 1.19

Net flux from $2.0 \mu\text{C}$ inside a cube.

$$\Phi = \frac{q}{\epsilon_0} = \frac{2 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 2.26 \times 10^5 \text{ Nm}^2/\text{C}$$

Question 1.20

Flux $-10^3$ through sphere. (a) Flux if radius doubled? (b) Charge?

(a) Unchanged (Flux depends only on enclosed charge).

(b) $q = \epsilon_0 \Phi = -8.85 \text{ nC}$.

Question 1.21

Sphere radius 10 cm. E-field 1.5 kN/C inward at 20 cm. Net charge?

$$q = \frac{E r^2}{k} = \frac{1.5 \times 10^3 \times (0.2)^2}{9 \times 10^9} \approx 6.67 \text{ nC}$$

Field is inward, so charge is Negative.

Question 1.22

Sphere 2.4 m diameter, $\sigma = 80 \mu\text{C/m}^2$. (a) Charge? (b) Flux?

(a) $Q = \sigma \times 4\pi R^2 \approx 1.45 \times 10^{-3} \text{ C}$.

(b) $\Phi = Q/\epsilon_0 \approx 1.6 \times 10^8 \text{ Nm}^2/\text{C}$.

Question 1.23

Infinite line charge. Field $9 \times 10^4$ N/C at 2 cm. Linear charge density?

$$\lambda = \frac{E r}{2k} = \frac{9 \times 10^4 \times 0.02}{18 \times 10^9} = 0.1 \mu\text{C/m}$$

Question 1.24

Two plates with $\pm \sigma$. Field in regions?

Outside Plates

Fields cancel out. $E = 0$.

Between Plates

Fields add up.

$$E = \frac{\sigma}{\epsilon_0} = \frac{17 \times 10^{-22}}{8.85 \times 10^{-12}} \approx 1.92 \times 10^{-10} \text{ N/C}$$

Question 1.25

Millikan oil drop (12 electrons). Find radius.

Equilibrium

Electric Force $qE$ = Weight $mg$.

$$r = \left[ \frac{3neE}{4\pi \rho g} \right]^{1/3} \approx 9.81 \times 10^{-7} \text{ m}$$

Question 1.26

Which field line diagrams are invalid?

(a) Invalid (Not normal to surface).
(b) Invalid (Starts from negative).
(d) Invalid (Lines intersect).
(e) Invalid (Closed loops).
(c) Valid.

Question 1.27

Force and Torque in non-uniform field along z-axis.

Force: $F = p \frac{dE}{dz} = (-10^{-7})(10^5) = -10^{-2} \text{ N}$ (Downwards).

Torque: Zero (Dipole is antiparallel to field).

Question 1.28

Conductor with cavity. (a) Charge Q on conductor. (b) Charge q inside cavity.

(a) Charge Q resides on outer surface.

(b) Charge $q$ induces $-q$ on inner surface. To conserve charge, $+q$ moves to outer surface. Total outer charge: $Q+q$.

Question 1.29

Field in a tiny hole of conductor.

Field just outside is $\sigma/\epsilon_0$. This is superposition of field from hole ($\sigma/2\epsilon_0$) and rest of conductor. Thus, field due to rest of conductor is $\sigma/2\epsilon_0$.

Question 1.30

Derive field for line charge without Gauss Law.

Integrate perpendicular components of electric field using Coulomb’s law:

$$E = \int dE \cos\theta = \frac{\lambda}{2\pi\epsilon_0 r}$$

Question 1.31

Quark composition of Proton and Neutron.

Proton (+e): 2 up (+4/3) + 1 down (-1/3) = +1. (uud)

Neutron (0): 1 up (+2/3) + 2 down (-2/3) = 0. (udd)

Question 1.32

Show equilibrium of test charge at null point is unstable.

Stable equilibrium requires restoring force (inward field lines). By Gauss Law, inward lines imply a charge exists at that point. Since null point is free space (no charge), stable equilibrium is impossible (Earnshaw’s Theorem).

Question 1.33

Show deflection $y = qEL^2/(2mv_x^2)$.

[Image of Electron Deflection]

Derivation

Time $t = L/v_x$. Vertical acceleration $a = qE/m$.

$$y = \frac{1}{2} a t^2 = \frac{1}{2} \left( \frac{qE}{m} \right) \left( \frac{L}{v_x} \right)^2 = \frac{qEL^2}{2mv_x^2}$$

Question 1.34

Electron trajectory calculation from Q1.33.

Strikes when $y = d/2 = 0.25 \text{ cm}$. Solve for $x$:

$$x = \sqrt{\frac{2m v_x^2 y}{eE}} \approx 1.12 \text{ cm}$$

The electron strikes the upper plate at 1.12 cm.

Physics Solutions