Class 12 Physics Chapter 1: Electric Charges and Fields — Notes & Solutions

01 / Chapter Theory

Electric Charges & Fields

Complete official NCERT textbook coverage: Electric Charge, Conductors & Insulators, Basic Properties, Coulomb's Law, Superposition, Electric Field & Field Lines, Electric Flux, Dipole & Torque, Continuous Charge Distributions, and Gauss's Law Applications.

14 SECTIONS

Chapter Navigation

1.1Introduction 1.2Electric Charge 1.3Conductors & Insulators 1.4Basic Properties of Charge 1.5Coulomb's Law 1.6Forces Between Multiple Charges 1.7Electric Field 1.8Electric Field Lines 1.9Electric Flux 1.10Electric Dipole 1.11Dipole in Uniform Field 1.12Continuous Charge Distribution 1.13Gauss's Law 1.14Applications of Gauss's Law

1.1

NCERT Section

Introduction

Everyday Observations & Electrostatics

1.1.1 Static Electricity in Everyday Life

All of us have the experience of seeing a spark or hearing a crackle when we take off our synthetic clothes or sweater, particularly in dry weather. Have you ever tried to find any explanation for this phenomenon? Another common example of electric discharge is the lightning that we see in the sky during thunderstorms. We also experience a sensation of an electric shock either while opening the door of a car or holding the iron bar of a bus after sliding from our seat.

The reason for these experiences is the discharge of electric charges through our body, which were accumulated due to rubbing of insulating surfaces. This is due to the generation of static electricity.

Static: Means anything that does not move or change with time.
Electrostatics: Deals with the study of forces, fields, and potentials arising from static charges.

Core Definition: Electrostatics is the branch of physics that investigates the interactions, forces, fields, and potentials produced by static (stationary) electric charges.

1.2

NCERT Section

Electric Charge

Historical Discovery & Types of Charge

1.2.1 Discovery & Polarity of Charge

Historically, the credit for discovering that amber rubbed with wool or silk cloth attracts light objects goes to Thales of Miletus, Greece (around 600 BC). The word electricity is coined from the Greek word elektron meaning amber.

Many such pairs of materials were known which on rubbing could attract light objects like straw, pith balls, and bits of paper.

Two glass rods rubbed with wool or silk cloth repel each other [Fig. 1.1(a)]. The two pieces of silk or wool also repel each other. However, glass rod and silk attract each other.
Two plastic rods rubbed with cat's fur repel each other [Fig. 1.1(b)] but attract the fur.
Plastic rod attracts the glass rod [Fig. 1.1(c)] and repels the silk/wool. The glass rod repels the fur.

Figure 1.1: Experimental demonstration of electrification — (a) Two glass rods rubbed with silk repel, (b) Two plastic rods rubbed with fur repel, (c) Glass rod attracts plastic rod.

American scientist Benjamin Franklin named the two kinds of charges positive (+) and negative (−).

By Convention: Charge on a glass rod (or cat's fur) is positive (+); charge on silk (or plastic rod) is termed negative (−).
Polarity of Charge: The property that differentiates the two kinds of charges.
Neutralization: When an electrified glass rod is brought in contact with the silk with which it was rubbed, they neutralize each other and no longer attract or repel other light objects.
Origin of Charge: All matter consists of atoms/molecules. Normally materials are electrically neutral with balanced positive and negative charges. Transfer of loosely bound electrons electrifies a body. Losing electrons leaves an excess of positive charge; gaining electrons creates an excess of negative charge.

Figure 1.2: Gold-leaf electroscope — charge transferred via the metal knob flows down the vertical metal rod into the gold leaves, causing them to diverge proportionally to the charge.

1.3

NCERT Section

Conductors and Insulators

Material Classification & Grounding

1.3.1 Conductors, Insulators, Semiconductors & Earthing

Conductors: Substances that readily allow the passage of electricity through them (metals, human and animal bodies, Earth, electrolyte solutions). They possess free mobile electrons. When charge is transferred to a conductor, it distributes over the entire outer surface.
Insulators: Substances that offer very high resistance to the flow of electric charge (glass, porcelain, plastic, nylon, rubber, dry wood). When charge is placed on an insulator, it stays confined to the local rubbed spot.
Semiconductors: Materials (e.g. Silicon, Germanium) offering resistance intermediate between conductors and insulators.
Why a Metal Spoon Cannot Be Charged by Hand: The charges on metal leak through our conducting body to the ground. However, if a metal rod with a plastic/wooden handle is rubbed without touching the metal part, it shows clear signs of charging.
Earthing / Grounding: When a charged body is brought into contact with the Earth, all excess charge disappears by causing a momentary current to pass to the ground through the conductor. In home electrical wiring, the Earth wire protects lives from appliance insulation failures.

1.4

NCERT Section

Basic Properties of Electric Charge

Fundamental Postulates

1.4.1 Point Charges & 3 Basic Properties

When the physical sizes of charged bodies are very small compared to the distances between them, they are treated as point charges, with all charge content assumed concentrated at a point in space.

1. Additivity of Charges (Section 1.4.1)

Electric charges add algebraically like real numbers or scalars (similar to mass, but charges can be positive or negative):

$$\color{#9333ea}{q_{\text{total}}} = \color{#9333ea}{q_1} + \color{#9333ea}{q_2} + \color{#9333ea}{q_3} + \dots + \color{#9333ea}{q_n}$$

Example: A system with five charges $+1, +2, -3, +4, -5$ (in arbitrary units) has total charge $(+1)+(+2)+(-3)+(+4)+(-5) = -1$ unit.

2. Charge is Conserved (Section 1.4.2)

Within an isolated system, the total algebraic sum of electric charges remains strictly invariant with time. While charge-carrying particles may be created or destroyed (e.g. neutron decaying into a proton and electron: $n \to p + e^- + \bar{\nu}_e$), the net charge before and after remains zero.

3. Quantisation of Charge (Section 1.4.3)

All free observable charges in the universe are integral multiples of a basic fundamental unit of charge denoted by $e$ ($e = 1.602192 \times 10^{-19}\text{ C}$):

$$\color{#9333ea}{q} = n \color{#9333ea}{e} \quad (n = 0, \pm 1, \pm 2, \pm 3, \dots)$$

History: First suggested by Faraday's laws of electrolysis; experimentally demonstrated by Millikan in 1912.
Proton & Electron: Proton charge is $+e$; electron charge is $-e$. A body with $n_1$ electrons and $n_2$ protons has total charge $(n_2 - n_1)e$.
SI Unit (Coulomb): $1\text{ C}$ is defined as the charge flowing through a wire in $1\text{ s}$ when current is $1\text{ A}$. $1\text{ C}$ contains $\approx 6 \times 10^{18}$ electrons.
Macroscopic vs Microscopic Scale: For macroscopic charges ($\mu\text{C}$ containing $\sim 10^{13}$ electrons), charge appears smooth and continuous, so quantisation can be ignored. At microscopic subatomic scales, the discrete graininess of $e$ is essential.

Example 1.1

Time to Accumulate 1 Coulomb Charge

If $10^9$ electrons move out of a body to another body every second, how much time is required to get a total charge of $1\text{ C}$ on the other body?

Step 1: Charge transferred per second
$\Delta q = 10^9 \times e = 10^9 \times 1.6 \times 10^{-19}\text{ C} = 1.6 \times 10^{-10}\text{ C s}^{-1}$.

Step 2: Total time calculation
$t = \frac{1\text{ C}}{1.6 \times 10^{-10}\text{ C s}^{-1}} = 6.25 \times 10^9\text{ s}$.

Step 3: Conversion to years
$t = \frac{6.25 \times 10^9}{365 \times 24 \times 3600}\text{ years} \approx 198.18\text{ years} \approx 200\text{ years}$.

Final Answer: Approximately 198–200 years (demonstrating that 1 Coulomb is a very large unit for practical electrostatic phenomena). Note: $1\text{ cm}^3$ of copper contains about $2.5 \times 10^{24}$ electrons.

Example 1.2

Total Positive & Negative Charge in a Cup of Water

How much positive and negative charge is there in a cup ($250\text{ g}$) of water?

Step 1: Molar mass & number of molecules
Molecular mass of $\text{H}_2\text{O} = 18\text{ g}$.
Number of moles in $250\text{ g} = \frac{250}{18}\text{ mol}$.
Number of molecules $N = \left(\frac{250}{18}\right) \times 6.02 \times 10^{23} = 8.36 \times 10^{24}$ molecules.

Step 2: Protons and electrons per molecule
Each $\text{H}_2\text{O}$ molecule contains $2(1) + 8 = 10$ protons and $10$ electrons.

Step 3: Total magnitude of charge
$Q = N \times 10 \times e = \left(\frac{250}{18}\right) \times 6.02 \times 10^{23} \times 10 \times 1.6 \times 10^{-19}\text{ C} = 1.34 \times 10^7\text{ C}$.

Final Answer: Total positive charge = $+1.34 \times 10^7\text{ C}$, Total negative charge = $-1.34 \times 10^7\text{ C}$.

1.5

NCERT Section

Coulomb's Law

Fundamental Quantitative Law

1.5.1 Statement, Torsion Balance & Vector Form

Coulomb's law states that the electrostatic force ($F$) between two stationary point charges ($q_1, q_2$) is inversely proportional to the square of the distance ($r$) between them and directly proportional to the product of the magnitudes of the two charges, acting along the line joining them.

$$\color{#059669}{F} = \color{#0891b2}{k} \frac{|\color{#9333ea}{q_1} \color{#9333ea}{q_2}|}{\color{#d97706}{r}^2} = \frac{1}{4\pi\color{#0891b2}{\varepsilon_0}} \frac{|\color{#9333ea}{q_1} \color{#9333ea}{q_2}|}{\color{#d97706}{r}^2} \quad \text{[Eq. 1.1, 1.2]}$$

Charles Augustin de Coulomb (1736–1806): Arrived at this law in 1785 using a sensitive torsion balance. By touching a charged sphere with an identical uncharged sphere, charge divides into $q/2, q/4$, etc. by symmetry.
Electrostatic Constant ($k$): $\color{#0891b2}{k} = \frac{1}{4\pi\color{#0891b2}{\varepsilon_0}} \approx 8.987 \times 10^9 \approx 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$.
Permittivity of Free Space ($\varepsilon_0$): $\color{#0891b2}{\varepsilon_0} = 8.854 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}$.
Definition of 1 Coulomb ($1\text{ C}$): That charge which when placed at a distance of $1\text{ m}$ from another identical charge in vacuum experiences an electrostatic repulsion force of $9 \times 10^9\text{ N}$.

Figure 1.3: Vector geometry of Coulomb's Law: $\mathbf{F}_{21} = -\mathbf{F}_{12}$ agreeing identically with Newton's third law.

Vector Form of Coulomb's Law (Eq. 1.3)

Let $\mathbf{r}_1, \mathbf{r}_2$ be position vectors of $q_1, q_2$. The displacement vector leading from $1$ to $2$ is $\mathbf{r}_{21} = \mathbf{r}_2 - \mathbf{r}_1$, with unit vector $\hat{\mathbf{r}}_{21} = \frac{\mathbf{r}_{21}}{r_{21}}$. Force $\mathbf{F}_{21}$ on $q_2$ due to $q_1$ is:

$$\color{#059669}{\mathbf{F}_{21}} = \frac{1}{4\pi\color{#0891b2}{\varepsilon_0}} \frac{\color{#9333ea}{q_1 q_2}}{\color{#d97706}{r_{21}}^2} \color{#d97706}{\hat{\mathbf{r}}_{21}} \quad \text{[Eq. 1.3]}$$

Automatic Sign Handling: If $q_1, q_2$ have like signs ($q_1 q_2 > 0$), $\mathbf{F}_{21}$ is along $\hat{\mathbf{r}}_{21}$ (repulsion). If opposite signs ($q_1 q_2 < 0$), $\mathbf{F}_{21}$ is along $-\hat{\mathbf{r}}_{21} = \hat{\mathbf{r}}_{12}$ (attraction).
Newton's Third Law Agreement: Interchanging indices gives $\mathbf{F}_{12} = -\mathbf{F}_{21}$.

Example 1.3

Comparison of Electrostatic & Gravitational Forces

(a) Compare the strength of electrostatic and gravitational forces: (i) for an electron and a proton, (ii) for two protons. (b) Estimate the accelerations of an electron and a proton due to mutual electrical attraction at separation $1\text{ \AA} = 10^{-10}\text{ m}$. ($m_p = 1.67 \times 10^{-27}\text{ kg}, m_e = 9.11 \times 10^{-31}\text{ kg}$)

(a)(i) Ratio for Electron & Proton:
$F_e = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2}$, $F_G = G \frac{m_p m_e}{r^2}$.
$\frac{F_e}{F_G} = \frac{e^2}{4\pi\varepsilon_0 G m_p m_e} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times 9.11 \times 10^{-31}} \approx 2.4 \times 10^{39}$.

(a)(ii) Ratio for Two Protons:
$\frac{F_e}{F_G} = \frac{e^2}{4\pi\varepsilon_0 G m_p^2} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2} \approx 1.3 \times 10^{36}$.
Inside nucleus ($r \sim 10^{-15}\text{ m}$): $F_e \approx 230\text{ N}$, while $F_G \approx 1.9 \times 10^{-34}\text{ N}$.

(b) Accelerations at $r = 10^{-10}\text{ m}$:
$|F| = \frac{8.987 \times 10^9 \times (1.6 \times 10^{-19})^2}{(10^{-10})^2} = 2.3 \times 10^{-8}\text{ N}$.
$a_e = \frac{F}{m_e} = \frac{2.3 \times 10^{-8}\text{ N}}{9.11 \times 10^{-31}\text{ kg}} = 2.5 \times 10^{22}\text{ m s}^{-2}$.
$a_p = \frac{F}{m_p} = \frac{2.3 \times 10^{-8}\text{ N}}{1.67 \times 10^{-27}\text{ kg}} = 1.4 \times 10^{19}\text{ m s}^{-2}$.

Summary: Electrostatic forces are $\sim 10^{39}$ times stronger than gravity. Gravitational fields are negligible on electron and proton trajectories.

Example 1.4

Redistribution of Charge on Identical Spheres

Two identical charged spheres A ($q$) and B ($q'$) at distance $r$ experience repulsion $F$. Identical uncharged spheres C and D touch A and B respectively and are removed. If separation is halved to $r/2$, what is the new force $F'$?

Step 1: Initial force
$F = \frac{1}{4\pi\varepsilon_0} \frac{q q'}{r^2}$.

Step 2: Charge redistribution
By symmetry, after contact with uncharged identical spheres: $q_A = q/2$ and $q_B = q'/2$.

Step 3: New force at separation $r/2$
$F' = \frac{1}{4\pi\varepsilon_0} \frac{(q/2)(q'/2)}{(r/2)^2} = \frac{1}{4\pi\varepsilon_0} \frac{(q q'/4)}{(r^2/4)} = \frac{1}{4\pi\varepsilon_0} \frac{q q'}{r^2} = F$.

Final Answer: The electrostatic repulsion remains unaltered ($F' = F$).

1.6

NCERT Section

Forces Between Multiple Charges

Superposition Theorem

1.6.1 The Principle of Superposition

Coulomb's law gives the mutual electric force between two charges. To calculate force in a system of $n$ charges $q_1, q_2, \dots, q_n$, we use the Principle of Superposition: force on any charge is the vector sum of all forces exerted on that charge due to the other charges taken one at a time. The individual pairwise forces are unaffected by other charges.

$$\color{#059669}{\mathbf{F}_1} = \color{#059669}{\mathbf{F}_{12}} + \color{#059669}{\mathbf{F}_{13}} + \dots + \color{#059669}{\mathbf{F}_{1n}} = \frac{\color{#9333ea}{q_1}}{4\pi\color{#0891b2}{\varepsilon_0}} \sum_{i=2}^{n} \frac{\color{#9333ea}{q_i}}{\color{#d97706}{r_{1i}}^2} \color{#d97706}{\hat{\mathbf{r}}_{1i}} \quad \text{[Eq. 1.4, 1.5]}$$

Example 1.5

Net Force on Centroid Charge of Equilateral Triangle

Three equal charges $q_1 = q_2 = q_3 = q$ are placed at the vertices of an equilateral triangle of side $l$. What is the net electrostatic force on a charge $Q$ (same sign as $q$) placed at the centroid $O$?

Step 1: Distance from vertex to centroid
In equilateral triangle ABC: $AD = AC \cos 30^\circ = \frac{\sqrt{3}}{2}l$. Distance $AO = BO = CO = \frac{2}{3} AD = \frac{l}{\sqrt{3}}$.

Step 2: Magnitude of individual forces
$\mathbf{F}_1$ (due to A) $= \frac{3Qq}{4\pi\varepsilon_0 l^2}$ along $AO$.
$\mathbf{F}_2$ (due to B) $= \frac{3Qq}{4\pi\varepsilon_0 l^2}$ along $BO$.
$\mathbf{F}_3$ (due to C) $= \frac{3Qq}{4\pi\varepsilon_0 l^2}$ along $CO$.

Step 3: Vector resultant
By parallelogram law, resultant of $\mathbf{F}_2$ and $\mathbf{F}_3$ (at $120^\circ$) has magnitude $\frac{3Qq}{4\pi\varepsilon_0 l^2}$ along $OA$.
$\mathbf{F}_{\text{total}} = \frac{3Qq}{4\pi\varepsilon_0 l^2} (\hat{\mathbf{r}} - \hat{\mathbf{r}}) = \mathbf{0}$.

Final Answer: Total force on centroid charge $Q$ is zero ($\mathbf{F} = \mathbf{0}$).

Example 1.6

Forces on Vertices with Charges $q, q, -q$

Charges $q, q, -q$ are placed at vertices A, B, C of an equilateral triangle of side $l$. Determine the resultant force on each charge.

Step 1: Pairwise magnitude $F$
$F = \frac{q^2}{4\pi\varepsilon_0 l^2}$.

Step 2: Force on charge $q$ at A ($\mathbf{F}_1$)
Repulsion from B ($\mathbf{F}_{12}$ along $BA$) and attraction to C ($\mathbf{F}_{13}$ along $AC$) at angle $120^\circ$.
$F_1 = \sqrt{F^2 + F^2 + 2F^2\cos 120^\circ} = F = \frac{q^2}{4\pi\varepsilon_0 l^2}$, directed along unit vector $\hat{\mathbf{r}}_1$ parallel to $BC$.

Step 3: Force on charge $q$ at B ($\mathbf{F}_2$)
$F_2 = F = \frac{q^2}{4\pi\varepsilon_0 l^2}$, directed along unit vector $\hat{\mathbf{r}}_2$ parallel to $AC$.

Step 4: Force on charge $-q$ at C ($\mathbf{F}_3$)
Attracted to both A and B at angle $60^\circ$.
$F_3 = \sqrt{F^2 + F^2 + 2F^2\cos 60^\circ} = \sqrt{3} F = \frac{\sqrt{3} q^2}{4\pi\varepsilon_0 l^2}$, directed along unit vector $\hat{\mathbf{n}}$ bisecting $\angle BCA$.

Note: Vector sum $\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}$, consistent with Newton's third law.

1.7

NCERT Section

Electric Field

Field Theory & Physical Significance

1.7.1 Electric Field Definition & Four Key Remarks

A source charge ($Q$) produces an electric field ($\mathbf{E}$) everywhere in its surroundings. When a test charge ($q$) is brought to point $\mathbf{r}$, the field acts on it producing force $\color{#059669}{\mathbf{F}}(\mathbf{r}) = \color{#db2777}{q}\color{#4f46e5}{\mathbf{E}}(\mathbf{r})$.

$$\color{#4f46e5}{\mathbf{E}}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{\color{#9333ea}{Q}}{\color{#d97706}{r}^2} \hat{\mathbf{r}} \quad \text{[Eq. 1.6]} \qquad \text{and} \qquad \color{#059669}{\mathbf{F}}(\mathbf{r}) = \color{#db2777}{q}\color{#4f46e5}{\mathbf{E}}(\mathbf{r}) \quad \text{[Eq. 1.8]}$$

Test Charge Limit (Eq. 1.9): $\color{#4f46e5}{\mathbf{E}} = \lim_{\color{#db2777}{q} \to 0} \frac{\color{#059669}{\mathbf{F}}}{\color{#db2777}{q}}$ (so the test charge $\color{#db2777}{q}$ does not disturb the source charge $\color{#9333ea}{Q}$).
Independence of $q$: Ratio $\color{#059669}{\mathbf{F}}/\color{#db2777}{q}$ is independent of the magnitude of test charge $q$.
Direction: Radially outwards for positive source ($\color{#9333ea}{Q} > 0$); radially inwards for negative source ($\color{#9333ea}{Q} < 0$).
Spherical Symmetry: At equal distances ($r$) from point charge $Q$, $|\color{#4f46e5}{\mathbf{E}}|$ is identical on the sphere centered at $Q$.
Superposition for System of Charges (Eq. 1.10): $\color{#4f46e5}{\mathbf{E}}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^{n} \frac{\color{#9333ea}{q_i}}{\color{#d97706}{r_{iP}}^2} \hat{\mathbf{r}}_{iP}$.
Physical Significance (Section 1.7.2): In electrostatics, $\color{#4f46e5}{\mathbf{E}}$ is convenient. In time-dependent electromagnetism, accelerated charges produce electromagnetic waves propagating at speed $c$, transporting energy and momentum. Electric and magnetic fields are physical entities with independent dynamics.

Example 1.7

Time of Fall: Electron vs Proton in Uniform Field

An electron falls through $h = 1.5\text{ cm}$ in uniform $E = 2.0 \times 10^4\text{ N C}^{-1}$ (upward). The field direction is reversed (downward) and a proton falls through the same distance. Compute the time of fall in each case and contrast with 'free fall under gravity'.

Step 1: Formula for time of fall from rest
$a = \frac{eE}{m} \implies h = \frac{1}{2} a t^2 \implies t = \sqrt{\frac{2hm}{eE}}$.

Step 2: Electron fall time ($t_e$)
$t_e = \sqrt{\frac{2 \times 1.5 \times 10^{-2}\text{ m} \times 9.11 \times 10^{-31}\text{ kg}}{1.6 \times 10^{-19}\text{ C} \times 2.0 \times 10^4\text{ N C}^{-1}}} \approx 2.9 \times 10^{-9}\text{ s} = 2.9\text{ ns}$.

Step 3: Proton fall time ($t_p$)
$t_p = \sqrt{\frac{2 \times 1.5 \times 10^{-2}\text{ m} \times 1.67 \times 10^{-27}\text{ kg}}{1.6 \times 10^{-19}\text{ C} \times 2.0 \times 10^4\text{ N C}^{-1}}} \approx 1.3 \times 10^{-7}\text{ s} = 130\text{ ns}$.

Step 4: Comparison with gravity
Proton acceleration $a_p = \frac{eE}{m_p} = \frac{1.6 \times 10^{-19} \times 2.0 \times 10^4}{1.67 \times 10^{-27}} = 1.9 \times 10^{12}\text{ m s}^{-2} \gg g$ ($9.8\text{ m s}^{-2}$). Thus gravity is negligible. Unlike free fall where time is mass-independent, here heavier proton takes greater time ($t \propto \sqrt{m}$).

Answers: $t_e = 2.9\text{ ns}$, $t_p = 130\text{ ns}$.

Example 1.8

Electric Field of Pair of Opposite Charges at Key Points

Two point charges $q_1 = +10^{-8}\text{ C}$ and $q_2 = -10^{-8}\text{ C}$ are placed $0.1\text{ m}$ apart. Calculate electric fields at point A (midpoint), point B ($0.05\text{ m}$ left of $q_1$), and point C (vertex of equilateral triangle of side $0.10\text{ m}$).

(a) Field at midpoint A ($r = 0.05\text{ m}$):
$E_{1A} = \frac{9 \times 10^9 \times 10^{-8}}{(0.05)^2} = 3.6 \times 10^4\text{ N C}^{-1}$ (towards right).
$E_{2A} = \frac{9 \times 10^9 \times 10^{-8}}{(0.05)^2} = 3.6 \times 10^4\text{ N C}^{-1}$ (towards right).
$E_A = E_{1A} + E_{2A} = 7.2 \times 10^4\text{ N C}^{-1}$ (directed right).

(b) Field at point B:
$E_{1B} = \frac{9 \times 10^9 \times 10^{-8}}{(0.05)^2} = 3.6 \times 10^4\text{ N C}^{-1}$ (left).
$E_{2B} = \frac{9 \times 10^9 \times 10^{-8}}{(0.15)^2} = 4 \times 10^3\text{ N C}^{-1}$ (right).
$E_B = E_{1B} - E_{2B} = 3.2 \times 10^4\text{ N C}^{-1}$ (directed left).

(c) Field at point C ($r = 0.10\text{ m}$):
$E_{1C} = E_{2C} = \frac{9 \times 10^9 \times 10^{-8}}{(0.10)^2} = 9 \times 10^3\text{ N C}^{-1}$ at angle $120^\circ$.
$E_C = 2 E_{1C} \cos\left(\frac{\pi}{3}\right) = 2 \times (9 \times 10^3) \times \frac{1}{2} = 9 \times 10^3\text{ N C}^{-1}$ (directed right).

Answers: $E_A = 7.2 \times 10^4\text{ N C}^{-1}$ (right), $E_B = 3.2 \times 10^4\text{ N C}^{-1}$ (left), $E_C = 9 \times 10^3\text{ N C}^{-1}$ (right).

1.8

NCERT Section

Electric Field Lines

Pictorial Mapping & Solid Angle

1.8.1 Concept & 4 General Properties of Field Lines

Invented by Michael Faraday (who originally called them 'lines of force'), field lines pictorially map electric fields around charge configurations. An electric field line is a curve whose tangent at each point gives the direction of the net electric field at that point.

Figure 1.14: Electric field line patterns for isolated positive charge, electric dipole, and pair of equal positive charges.

Solid Angle & $1/r^2$ Dependence

Solid angle $\Delta\Omega = \Delta S/r^2$. In a given solid angle, the number of radial field lines $n$ is constant. The number of lines per unit area element at distance $r$ is $\frac{n}{r^2 \Delta\Omega}$, explaining the geometric origin of the $1/r^2$ law.

4 General Properties:

Origin & Termination: Field lines start from positive charges and end at negative charges (or extend to infinity for isolated single charges).
Continuity: In a charge-free region, field lines are continuous curves without sudden breaks.
No Intersection: Two field lines can never cross each other. If they did, the field at the intersection point would have two different tangents and thus two unique directions, which is absurd.
No Closed Loops: Electrostatic field lines do not form closed loops, following from the conservative nature of electrostatic fields.

1.9

NCERT Section

Electric Flux

Dot Product Formulation

1.9.1 Definition & Flux Calculation

Analogous to liquid flow rate through area $\Delta S$, electric flux ($\Delta\Phi$) through area element $\Delta\mathbf{S}$ measures the number of electric field lines crossing $\Delta\mathbf{S}$:

$$\color{#0f766e}{\Delta\Phi} = \color{#4f46e5}{\mathbf{E}}\cdot\color{#0f766e}{\Delta\mathbf{S}} = \color{#4f46e5}{E}\,\color{#0f766e}{\Delta S}\cos\theta \quad \text{[Eq. 1.11]} \qquad \text{and} \qquad \color{#0f766e}{\Phi} = \oint_S \color{#4f46e5}{\mathbf{E}}\cdot \color{#0f766e}{d\mathbf{S}} \quad \text{[Eq. 1.12]}$$

Area Vector ($\Delta\mathbf{S}$): $\Delta\mathbf{S} = \Delta S\,\hat{\mathbf{n}}$, where $\hat{\mathbf{n}}$ is the unit normal vector. For closed surfaces, $\hat{\mathbf{n}}$ is always chosen as the outward normal.
Angle $\theta$: Angle between electric field $\mathbf{E}$ and outward normal $\hat{\mathbf{n}}$. When $\theta = 90^\circ$, field lines are parallel to surface and flux is zero ($\Delta\Phi = 0$).
SI Unit: $\text{N C}^{-1}\text{ m}^2$ (or $\text{V m}$).

1.10

NCERT Section

Electric Dipole

Dipole Moment & Field Derivations

1.10.1 Axial, Equatorial Fields & Polar Molecules

An electric dipole is a pair of equal and opposite point charges $+q$ and $-q$ separated by a distance $2a$. Total charge is zero, but electric fields do not cancel completely.

Electric Dipole Moment ($\mathbf{p}$):

$$\color{#db2777}{\mathbf{p}} = \color{#9333ea}{q}(2\color{#d97706}{a})\,\color{#db2777}{\hat{\mathbf{p}}} \quad \text{[Eq. 1.19]}$$
(Vector directed along dipole axis from negative charge $-q$ to positive charge $+q$; SI Unit: $\text{C m}$)

Figure 1.17: Electric field of a dipole: (a) Axial point field $\mathbf{E}_{\text{axial}} \parallel \mathbf{p}$, (b) Equatorial point field $\mathbf{E}_{\text{eq}} \parallel -\mathbf{p}$.

1. Field on Dipole Axis (Eq. 1.14 & 1.20)

$\mathbf{E} = \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{(r-a)^2} - \frac{1}{(r+a)^2}\right]\hat{\mathbf{p}} = \frac{4qar}{4\pi\varepsilon_0(r^2-a^2)^2}\hat{\mathbf{p}}$. For $r \gg a$:

$$\color{#4f46e5}{\mathbf{E}_{\text{axial}}} = \frac{1}{4\pi\color{#0891b2}{\varepsilon_0}} \frac{2\color{#db2777}{\mathbf{p}}}{\color{#d97706}{r}^3} \quad (\color{#d97706}{r} \gg \color{#d97706}{a}) \quad \text{[Eq. 1.20]}$$

2. Field on Equatorial Plane (Eq. 1.17 & 1.21)

$\mathbf{E} = -2 E_{+q} \cos\theta\,\hat{\mathbf{p}} = -\frac{2qa}{4\pi\varepsilon_0(r^2+a^2)^{3/2}}\hat{\mathbf{p}}$. For $r \gg a$:

$$\color{#4f46e5}{\mathbf{E}_{\text{equatorial}}} = -\frac{1}{4\pi\color{#0891b2}{\varepsilon_0}} \frac{\color{#db2777}{\mathbf{p}}}{\color{#d97706}{r}^3} \quad (\color{#d97706}{r} \gg \color{#d97706}{a}) \quad \text{[Eq. 1.21]}$$

Physical Significance (Section 1.10.2):

Non-polar molecules: Centers of positive and negative charges coincide; dipole moment is zero in absence of field (e.g. $\text{CO}_2, \text{CH}_4$).
Polar molecules: Centers of positive and negative charges do not coincide; possess permanent dipole moment even without external field (e.g. $\text{H}_2\text{O}$).
Point Dipole: Ideal limit as $2a \to 0, q \to \infty$ such that $\mathbf{p} = q(2\mathbf{a})$ remains finite. Equations 1.20 and 1.21 are exact for point dipoles.

Example 1.9

Axial and Equatorial Fields of a Dipole

Two charges $\pm 10\,\mu\text{C}$ are placed $5.0\text{ mm}$ apart ($2a = 5\times 10^{-3}\text{ m}$). Determine the electric field at (a) point P on the axis $15\text{ cm}$ away from center O, and (b) point Q $15\text{ cm}$ away on the line normal to the dipole axis.

Step 1: Calculate Dipole Moment ($p$)
$p = q(2a) = 10 \times 10^{-6}\text{ C} \times 5.0 \times 10^{-3}\text{ m} = 5.0 \times 10^{-8}\text{ C m}$.

(a) Field at Axial Point P ($r = 0.15\text{ m}$):
Exact: $E_{+10\mu\text{C}} = \frac{10^{-5}}{4\pi(8.854\times 10^{-12})(15-0.25)^2\times 10^{-4}} = 4.13 \times 10^6\text{ N C}^{-1}$ along $BP$.
$E_{-10\mu\text{C}} = \frac{10^{-5}}{4\pi(8.854\times 10^{-12})(15+0.25)^2\times 10^{-4}} = 3.86 \times 10^6\text{ N C}^{-1}$ along $PA$.
$E_{\text{net}} = 4.13 \times 10^6 - 3.86 \times 10^6 = 2.7 \times 10^5\text{ N C}^{-1}$ along $BP$.
Using formula ($r/a = 60 \gg 1$): $E = \frac{2p}{4\pi\varepsilon_0 r^3} = \frac{2 \times (5 \times 10^{-8})}{4\pi(8.854\times 10^{-12})(0.15)^3} \approx 2.6 \times 10^5\text{ N C}^{-1}$ along $\mathbf{p}$.

(b) Field at Equatorial Point Q ($r = 0.15\text{ m}$):
$E = \frac{p}{4\pi\varepsilon_0 r^3} = \frac{5 \times 10^{-8}}{4\pi(8.854\times 10^{-12})(0.15)^3} = 1.33 \times 10^5\text{ N C}^{-1}$ along $BA$ (opposite to dipole moment direction).

Answers: (a) $2.7 \times 10^5\text{ N C}^{-1}$ along $\mathbf{p}$, (b) $1.33 \times 10^5\text{ N C}^{-1}$ opposite to $\mathbf{p}$.

1.11

NCERT Section

Dipole in a Uniform External Field

Torque & Potential Energy

1.11.1 Torque on Dipole in Uniform & Non-Uniform Fields

Consider a permanent dipole $\mathbf{p}$ in a uniform external field $\mathbf{E}$ (Fig. 1.19):

Net Force: $\mathbf{F}_{\text{net}} = q\mathbf{E} + (-q\mathbf{E}) = \mathbf{0}$ (no translational acceleration).
Torque ($\boldsymbol{\tau}$): Magnitude of couple $= qE \times 2a\sin\theta = 2qa E\sin\theta = pE\sin\theta$.

$$\color{#b45309}{\boldsymbol{\tau}} = \color{#db2777}{\mathbf{p}} \times \color{#4f46e5}{\mathbf{E}} \quad \text{[Eq. 1.22]}$$

$\theta = 0^\circ$: Stable equilibrium ($\tau = 0$, $\mathbf{p} \parallel \mathbf{E}$).
$\theta = 180^\circ$: Unstable equilibrium ($\tau = 0$, $\mathbf{p}$ antiparallel to $\mathbf{E}$).
$\theta = 90^\circ$: Maximum torque ($\tau_{\max} = pE$).

Non-Uniform Field (Fig. 1.20): When field is non-uniform, net force is non-zero in addition to torque. When $\mathbf{p} \parallel \mathbf{E}$, net force is in the direction of increasing field. When $\mathbf{p}$ is antiparallel, net force is in direction of decreasing field. This explains why a charged comb polarizes and attracts neutral bits of paper!

1.12

NCERT Section

Continuous Charge Distribution

Macroscopic Densities & Superposition Integral

1.12.1 Linear, Surface & Volume Densities

On macroscopic surfaces, discrete charge constituents are smoothed into continuous charge density functions:

Type	Symbol	Definition	SI Unit	Total Field Contribution
Linear Charge Density	λ	$\lambda = \frac{\Delta Q}{\Delta l}$ [Eq. 1.24]	C m⁻¹	$\mathbf{E} = \frac{1}{4\pi\varepsilon_0}\int \frac{\lambda\,dl}{r'^2}\hat{\mathbf{r}}'$
Surface Charge Density	σ	$\sigma = \frac{\Delta Q}{\Delta S}$ [Eq. 1.23]	C m⁻²	$\mathbf{E} = \frac{1}{4\pi\varepsilon_0}\int \frac{\sigma\,dS}{r'^2}\hat{\mathbf{r}}'$
Volume Charge Density	ρ	$\rho = \frac{\Delta Q}{\Delta V}$ [Eq. 1.25]	C m⁻³	$\mathbf{E} \cong \frac{1}{4\pi\varepsilon_0}\sum_{\text{all }\Delta V} \frac{\rho\Delta V}{r'^2}\hat{\mathbf{r}}'$ [Eq. 1.27]

1.13

NCERT Section

Gauss's Law

Fundamental Theorem

1.13.1 Statement & Six Key Remarks

For a point charge $q$ at the center of a sphere of radius $r$, flux through element $\Delta S$ is $\Delta\Phi = \mathbf{E}\cdot\Delta\mathbf{S} = \frac{q}{4\pi\varepsilon_0 r^2}\Delta S$. Summing over the whole sphere ($\sum \Delta S = 4\pi r^2$): $\Phi = \frac{q}{4\pi\varepsilon_0 r^2}(4\pi r^2) = \frac{q}{\varepsilon_0}$.

$$\color{#0f766e}{\Phi} = \oint_S \color{#4f46e5}{\mathbf{E}}\cdot \color{#0f766e}{d\mathbf{S}} = \frac{\color{#9333ea}{q_{\text{enclosed}}}}{\color{#0891b2}{\varepsilon_0}} \quad \text{[Eq. 1.31]}$$

6 Important Points Regarding Gauss's Law:

True for any closed surface regardless of its shape or geometric size.
$q$ represents the algebraic sum of all charges enclosed inside the surface.
$\mathbf{E}$ in the flux integral is the resultant field due to all charges both inside and outside $S$.
The chosen Gaussian surface must not pass through discrete point charges (where $\mathbf{E}$ is unbounded), but can pass through continuous distributions.
Gauss's law provides convenient calculation of electrostatic fields when the charge configuration has high symmetry.
Gauss's law is directly based on the inverse square ($1/r^2$) dependence in Coulomb's law. Any departure indicates deviation from inverse square law.

Example 1.10

Flux Through Cube in Non-Uniform Field & Charge Enclosed

Electric field components in Fig. 1.24 are $E_x = \alpha x^{1/2}, E_y = E_z = 0$, where $\alpha = 800\text{ N C}^{-1}\text{ m}^{-1/2}$. Calculate: (a) flux through the cube, and (b) charge within the cube ($a = 0.1\text{ m}$).

(a) Flux Calculation:
Only left ($x = a$) and right ($x = 2a$) faces perpendicular to x-direction have non-zero flux.
$\Phi_L = -E_L a^2 = -\alpha a^{1/2} a^2 = -\alpha a^{5/2}$.
$\Phi_R = +E_R a^2 = +\alpha (2a)^{1/2} a^2 = \sqrt{2}\alpha a^{5/2}$.
$\Phi_{\text{net}} = \Phi_R + \Phi_L = \alpha a^{5/2}(\sqrt{2}-1) = 800 \times (0.1)^{5/2} \times (1.414 - 1) = 1.05\text{ N m}^2\text{ C}^{-1}$.

(b) Charge inside cube ($q$):
$q = \varepsilon_0 \Phi = (8.854 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}) \times (1.05\text{ N m}^2\text{ C}^{-1}) = 9.27 \times 10^{-12}\text{ C}$.

Answers: (a) $\Phi = 1.05\text{ N m}^2\text{ C}^{-1}$, (b) $q = 9.27 \times 10^{-12}\text{ C} = 9.27\text{ pC}$.

Example 1.11

Cylindrical Gaussian Surface in Uniform Step Field

$\mathbf{E} = +200\,\hat{\mathbf{i}}\text{ N/C}$ for $x > 0$ and $\mathbf{E} = -200\,\hat{\mathbf{i}}\text{ N/C}$ for $x < 0$. A right circular cylinder of length $20\text{ cm}$ and radius $5\text{ cm}$ has its center at the origin and axis along the x-axis. Find: (a) outward flux through each flat face, (b) flux through curved side, (c) net outward flux, (d) net charge inside.

(a) Flat faces:
Left face: $\Phi_L = (-200\hat{\mathbf{i}})\cdot(-\Delta S\hat{\mathbf{i}}) = +200 \times \pi(0.05)^2 = +1.57\text{ N m}^2\text{ C}^{-1}$.
Right face: $\Phi_R = (+200\hat{\mathbf{i}})\cdot(+\Delta S\hat{\mathbf{i}}) = +200 \times \pi(0.05)^2 = +1.57\text{ N m}^2\text{ C}^{-1}$.

(b) Curved side: $\mathbf{E} \perp \Delta\mathbf{S} \implies \Phi_{\text{side}} = 0$.

(c) Net outward flux: $\Phi = 1.57 + 1.57 + 0 = 3.14\text{ N m}^2\text{ C}^{-1}$.

(d) Net charge: $q = \varepsilon_0 \Phi = 8.854 \times 10^{-12} \times 3.14 = 2.78 \times 10^{-11}\text{ C}$.

Answers: (a) $+1.57\text{ N m}^2\text{ C}^{-1}$ each, (b) $0$, (c) $3.14\text{ N m}^2\text{ C}^{-1}$, (d) $2.78 \times 10^{-11}\text{ C}$.

1.14

NCERT Section

Applications of Gauss's Law

3 Classical Symmetry Derivations

1.14.1 Infinitely Long Uniformly Charged Straight Wire

For an infinitely long thin wire with linear charge density $\lambda$, choose a co-axial cylindrical Gaussian surface of radius $r$ and length $l$ (Fig. 1.26):

Flux through circular flat ends = $0$ (since $\mathbf{E} \parallel$ flat caps).
Flux through curved cylinder $= E(2\pi r l)$.
Enclosed charge $= \lambda l$.

$$\color{#4f46e5}{E}(2\pi \color{#d97706}{r} l) = \frac{\color{#9333ea}{\lambda} l}{\color{#0891b2}{\varepsilon_0}} \implies \color{#4f46e5}{\mathbf{E}} = \frac{\color{#9333ea}{\lambda}}{2\pi\color{#0891b2}{\varepsilon_0} \color{#d97706}{r}}\,\hat{\mathbf{n}} \quad \text{[Eq. 1.32]}$$

1.14.2 Uniformly Charged Infinite Plane Sheet

For an infinite plane sheet of uniform surface charge density $\sigma$, choose a rectangular pillbox / cylinder of cross-section $A$ cutting through the sheet (Fig. 1.27):

Flux through 2 end faces $= EA + EA = 2EA$.
Curved sides parallel to $\mathbf{E} \implies$ flux is $0$.
Enclosed charge $= \sigma A$.

$$2\color{#4f46e5}{E}A = \frac{\color{#9333ea}{\sigma} A}{\color{#0891b2}{\varepsilon_0}} \implies \color{#4f46e5}{\mathbf{E}} = \frac{\color{#9333ea}{\sigma}}{2\color{#0891b2}{\varepsilon_0}}\,\hat{\mathbf{n}} \quad \text{[Eq. 1.33]}$$
(Note: $\mathbf{E}$ is directed away from sheet if $\sigma > 0$ and is completely independent of distance from the sheet!)

1.14.3 Uniformly Charged Thin Spherical Shell

For a spherical shell of radius $R$, surface density $\sigma$, total charge $q = 4\pi R^2 \sigma$ (Fig. 1.28):

(i) Outside the shell ($r \ge R$):

Spherical Gaussian surface of radius $r > R$ encloses charge $q$:

$$\color{#4f46e5}{E}(4\pi \color{#d97706}{r}^2) = \frac{\color{#9333ea}{q}}{\color{#0891b2}{\varepsilon_0}} \implies \color{#4f46e5}{\mathbf{E}} = \frac{1}{4\pi\color{#0891b2}{\varepsilon_0}}\frac{\color{#9333ea}{q}}{\color{#d97706}{r}^2}\,\color{#d97706}{\hat{\mathbf{r}}} \quad \text{[Eq. 1.34]}$$

Interpretation: The electric field outside is as if the entire charge $q$ were concentrated at the center $O$.

(ii) Inside the shell ($r < R$):

Gaussian surface of radius $r < R$ encloses no charge ($q_{\text{enc}} = 0$):

$$\color{#4f46e5}{E}(4\pi \color{#d97706}{r}^2) = 0 \implies \color{#4f46e5}{\mathbf{E}} = \mathbf{0} \quad (\color{#d97706}{r} < \color{#d97706}{R}) \quad \text{[Eq. 1.35]}$$

Core Result: The electric field due to a uniformly charged thin spherical shell is zero at all points inside the shell.

Example 1.12

Electric Field in Early Atomic Model

An early atom model has a positive point nucleus $+Ze$ surrounded by uniform negative charge density $\rho$ up to radius $R$ (atom is neutral). Find electric field $E(r)$ for: (i) $r < R$, (ii) $r > R$.

Step 1: Negative charge density ($\rho$)
Total negative charge in sphere $= -Ze = \frac{4}{3}\pi R^3 \rho \implies \rho = -\frac{3Ze}{4\pi R^3}$.

(i) Inside atom ($r < R$):
Enclosed charge $q_{\text{enc}} = +Ze + \frac{4}{3}\pi r^3 \rho = Ze - Ze\frac{r^3}{R^3} = Ze\left(1 - \frac{r^3}{R^3}\right)$.
By Gauss's law: $E(4\pi r^2) = \frac{Ze}{\varepsilon_0}\left(1 - \frac{r^3}{R^3}\right) \implies E(r) = \frac{Ze}{4\pi\varepsilon_0}\left(\frac{1}{r^2} - \frac{r}{R^3}\right)$ (radially outward).

(ii) Outside atom ($r > R$):
Total enclosed charge $= +Ze - Ze = 0 \implies E(r) = 0$. (At $r = R$, both expressions give $E = 0$).

Final Answer: Inside ($r < R$): $E(r) = \frac{Ze}{4\pi\varepsilon_0}\left(\frac{1}{r^2} - \frac{r}{R^3}\right)$; Outside ($r > R$): $E(r) = 0$.

02 / Interactive Assessment

Concept Check & Board Mastery Quiz

Targeted multiple choice questions covering properties of charge, Coulomb's law, superposition, field lines, dipoles, and Gauss's law with detailed step-by-step explanations for every option.

15 MCQS

0 / 15 Answered

Score: 0 (0%)

Q1 UNANSWERED

When a polythene piece is rubbed with wool, a negative charge of $3.2 \times 10^{-7}\text{ C}$ is developed on polythene. The number of electrons transferred is:

Option A is correct. $n = \frac{q}{e} = \frac{3.2 \times 10^{-7}}{1.6 \times 10^{-19}} = 2 \times 10^{12}$. Since polythene becomes negatively charged, electrons transfer from wool to polythene.

Option B is incorrect. If electrons moved from polythene, polythene would become positively charged.

Option C is incorrect. Calculation error in dividing $3.2/1.6$.

Option D is incorrect. $6.25 \times 10^{18}$ is the number of electrons in $1\text{ C}$, not $3.2 \times 10^{-7}\text{ C}$.

Core Rule

$q = ne \implies n = q/e$. Gaining electrons yields negative charge.

Q2 UNANSWERED

Two point charges placed in vacuum repel with force $F$. If a dielectric slab of dielectric constant $K = 4$ is introduced between them keeping separation unchanged, the new force is:

Option A is incorrect. Dielectric medium reduces electrostatic attraction/repulsion.

Option B is correct. Force in a medium of dielectric constant $K$ is $F_{\text{med}} = \frac{F_{\text{vac}}}{K} = \frac{F}{4}$.

Option C is incorrect. The force scales as $1/K$, not $K^2$.

Option D is incorrect. The dielectric permittivity $\varepsilon = K\varepsilon_0$ directly weakens the electric force.

Core Rule

$F_{\text{medium}} = \frac{F_0}{K} = \frac{F_0}{\varepsilon_r}$.

Q3 UNANSWERED

Which of the following statements about electrostatic field lines is INCORRECT?

Option A is a correct property. Field lines emanate from $+q$ and sink into $-q$.

Option B is a correct property. By definition, the tangent defines field vector direction.

Option C is the INCORRECT statement (hence the correct answer). Electrostatic field lines NEVER form closed loops because electrostatic fields are conservative ($\oint \mathbf{E}\cdot d\mathbf{l} = 0$).

Option D is a correct property. Intersection would imply two unique directions of $\mathbf{E}$ at one point.

Core Rule

Electrostatic field lines never form closed loops due to conservative nature of electrostatic forces.

Q4 UNANSWERED

At a large distance $r$ from an electric dipole, the electric field falls off as:

Option A is incorrect. $1/r$ is the field of an infinite straight line charge.

Option B is incorrect. $1/r^2$ is the field of a single isolated point charge.

Option C is correct. Due to partial cancellation of opposite charges $+q$ and $-q$, dipole field decays as $E \propto 1/r^3$.

Option D is incorrect. $1/r^4$ applies to electric quadrupoles.

Core Rule

Point charge: $E \propto 1/r^2$. Dipole: $E \propto 1/r^3$. Line charge: $E \propto 1/r$.

Q5 UNANSWERED

An electric dipole of moment $\mathbf{p}$ placed in a uniform electric field $\mathbf{E}$ experiences:

Option A is correct. In uniform $\mathbf{E}$, $\mathbf{F}_{\text{net}} = q\mathbf{E} - q\mathbf{E} = 0$, but torque $\boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}$ acts to align dipole with the field.

Option B is incorrect. Non-zero net force occurs ONLY in non-uniform electric fields.

Option C is incorrect. Net force is strictly zero in uniform $\mathbf{E}$.

Option D is incorrect. Torque is zero only when dipole is parallel/antiparallel to $\mathbf{E}$ ($\theta = 0^\circ, 180^\circ$).

Core Rule

Uniform field: $\mathbf{F}_{\text{net}} = 0, \boldsymbol{\tau} = \mathbf{p}\times\mathbf{E}$. Non-uniform field: $\mathbf{F}_{\text{net}} \ne 0, \boldsymbol{\tau} \ne 0$.

Q6 UNANSWERED

If the radius of a spherical Gaussian surface enclosing a point charge $q$ is doubled, the outward electric flux through the surface will:

Option A is incorrect. Flux depends solely on enclosed charge, not surface radius.

Option B is incorrect. Electric field decreases by $1/4$ while surface area increases by $4$, keeping flux constant.

Option C is correct. By Gauss's Law, $\Phi = q/\varepsilon_0$, which is completely independent of the radius $r$ of the Gaussian sphere.

Option D is incorrect. Total number of field lines leaving the enclosed charge remains invariant.

Core Rule

Total electric flux depends strictly on enclosed charge ($\Phi = q_{\text{enc}}/\varepsilon_0$) and is independent of geometry or size.

Q7 UNANSWERED

The electric field inside a thin, uniformly charged spherical conducting shell of radius $R$ at distance $r < R$ from the center is:

Option A is correct. Gaussian surface inside the shell ($r < R$) encloses zero charge ($q_{\text{enc}} = 0$), hence $E = 0$.

Option B is incorrect. This holds only for outside points ($r \ge R$).

Option C is incorrect. This is the surface field value at $r = R$.

Option D is incorrect. Linear increase occurs in uniformly charged solid dielectric spheres, not thin shells.

Core Rule

Thin spherical shell: $E_{\text{inside}} = 0$, $E_{\text{outside}} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$.

Q8 UNANSWERED

Electric field at distance $r$ from an infinite plane sheet of uniform surface charge density $\sigma$ is:

Option A is correct. Derived from Gauss's law pillbox: $2EA = \frac{\sigma A}{\varepsilon_0} \implies E = \frac{\sigma}{2\varepsilon_0}$.

Option B is incorrect. $\sigma/\varepsilon_0$ is the field between two oppositely charged plates or outside a charged conductor.

Option C is incorrect. The field does not decay with $1/r$.

Option D is incorrect. $1/r^2$ is for point charges.

Core Rule

Infinite plane sheet field is uniform and distance-independent: $E = \sigma / (2\varepsilon_0)$.

Q9 UNANSWERED

The SI unit of electric flux is:

Option A is correct. $\Phi = E S \implies (\text{N/C})\times\text{m}^2 = \text{N m}^2\text{ C}^{-1} = \text{V m}$.

Option B is incorrect. $\text{N C}^{-1}$ is the unit of electric field.

Option C is incorrect. $\text{C m}^{-2}$ is surface charge density ($\sigma$).

Option D is incorrect. $\text{C m}$ is dipole moment ($p$).

Core Rule

Electric flux SI unit: $\text{N m}^2\text{ C}^{-1} \equiv \text{V m}$.

Q10 UNANSWERED

Four charges $+q, -q, +q, -q$ are placed at four corners A, B, C, D of a square of side $a$. The net electric field at the center of the square is:

Option A is incorrect. Fields along opposite diagonals cancel out.

Option B is correct. Diametrically opposite corners A and C are both $+q$, so their fields at center point in opposite directions and cancel. Likewise B and D are both $-q$, cancelling each other. Hence net field is identically Zero.

Option C is incorrect. This would only occur if charges had an asymmetric configuration.

Option D is incorrect. Distance to center is finite ($a/\sqrt{2}$).

Core Rule

By symmetry, identical opposite charges across center diagonals cancel out pairwise.

03 / Textbook Solutions

NCERT Exercises 1.1 – 1.23

Complete stepwise solutions for every textbook exercise question in NCERT Class 12 Physics Chapter 1 Reprint 2026-27.

23 QUESTIONS

Ex 1.1 Coulomb Force Between Spheres

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 $30\text{ cm}$ apart in air?

Prerequisite: ↗ 1.5 Coulomb's Law

Force $F = 6 \times 10^{-3}\text{ N}$ (Repulsive).

Given:
$q_1 = 2 \times 10^{-7}\text{ C}$, $q_2 = 3 \times 10^{-7}\text{ C}$, $r = 30\text{ cm} = 0.3\text{ m}$, $k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$.

Formula:
$F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} = \frac{9 \times 10^9 \times (2 \times 10^{-7}) \times (3 \times 10^{-7})}{(0.3)^2}$.

Calculation:
$F = \frac{54 \times 10^{-5}}{0.09} = 6 \times 10^{-3}\text{ N}$.

Nature of force: Since both charges are positive (like charges), the force is repulsive.

Ex 1.2 Distance and Mutual Force

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 $0.2\text{ N}$. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

(a) Distance $r = 0.12\text{ m} = 12\text{ cm}$ • (b) Force on 2nd sphere = $0.2\text{ N}$ (Attractive).

(a) Separation $r$:
$|F| = \frac{1}{4\pi\varepsilon_0}\frac{|q_1 q_2|}{r^2} \implies r^2 = \frac{9 \times 10^9 \times (0.4 \times 10^{-6}) \times (0.8 \times 10^{-6})}{0.2} = \frac{2.88 \times 10^{-3}}{0.2} = 1.44 \times 10^{-2}\text{ m}^2$.
$r = \sqrt{1.44 \times 10^{-2}} = 0.12\text{ m} = 12\text{ cm}$.

(b) Force on second sphere:
By Newton's third law ($\mathbf{F}_{21} = -\mathbf{F}_{12}$), the force on the second sphere due to the first is equal in magnitude ($0.2\text{ N}$) and attractive (towards the first sphere).

Ex 1.3 Dimensionless Ratio of Forces

1.3 Check that the ratio $\frac{k e^2}{G m_e m_p}$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Dimensions = $[M^0 L^0 T^0]$ (Dimensionless) • Ratio value $\approx 2.4 \times 10^{39}$.

1. Dimensional Analysis:
$[k] = [\text{N m}^2\text{ C}^{-2}] = [M L^3 T^{-4} A^{-2}]$, $[e^2] = [A^2 T^2]$, $[G] = [M^{-1} L^3 T^{-2}]$, $[m_e m_p] = [M^2]$.
$\left[\frac{k e^2}{G m_e m_p}\right] = \frac{[M L^3 T^{-2}]}{[M L^3 T^{-2}]} = [M^0 L^0 T^0 A^0]$.

2. Numerical Calculation:
$\text{Ratio} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{6.67 \times 10^{-11} \times 9.11 \times 10^{-31} \times 1.67 \times 10^{-27}} \approx 2.4 \times 10^{39}$.

3. Physical Significance:
It signifies the ratio of electrostatic force to gravitational force between an electron and a proton. Electrostatic forces are overwhelmingly stronger ($\sim 10^{39}$ times) than gravitational forces at subatomic scales.

Ex 1.4 Quantisation Conceptual

1.4 (a) Explain the meaning of the statement 'electric charge of a body is quantised'. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

(a) Meaning of Quantisation:
Electric charge is not continuous; it exists in discrete packets. Any observable charge $q$ on a body is always an integral multiple of the basic elementary unit $e = 1.6 \times 10^{-19}\text{ C}$ ($q = \pm ne$, $n \in \mathbb{Z}$).

(b) Why ignored macroscopically:
At macroscopic levels (e.g. $1\,\mu\text{C}$), the number of electrons involved is enormous ($n \approx 10^{13}$). The discrete addition or removal of a few electrons changes charge by an imperceptibly tiny step ($10^{-19}\text{ C}$), making charge appear continuous—analogous to a dotted line appearing solid from a distance.

Ex 1.5 Frictional Charging & Conservation

1.5 When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Explanation:
Rubbing simply transfers loosely bound electrons from the glass rod to the silk cloth. The number of electrons lost by the glass rod equals the number of electrons gained by the silk cloth. Thus, the glass rod acquires a charge $+q$ and the silk cloth acquires an equal and opposite charge $-q$. The total net charge of the isolated system before and after rubbing remains identically zero ($q_{\text{net}} = +q + (-q) = 0$). This completely conforms to the law of conservation of charge.

Ex 1.6 Force at Center of Square

1.6 Four point charges $q_A = 2\,\mu\text{C}, q_B = -5\,\mu\text{C}, q_C = 2\,\mu\text{C}$, and $q_D = -5\,\mu\text{C}$ are located at the corners of a square ABCD of side $10\text{ cm}$. What is the force on a charge of $1\,\mu\text{C}$ placed at the centre of the square?

Net force on $1\,\mu\text{C}$ at the center = $0\text{ N}$.

Geometry:
Diagonal of square $= 10\sqrt{2}\text{ cm}$. Distance from each vertex to center $O$ is $r = \frac{10\sqrt{2}}{2} = 5\sqrt{2}\text{ cm} = 5\sqrt{2} \times 10^{-2}\text{ m}$.

Diagonal AC:
$q_A = +2\,\mu\text{C}$ repels $q_0 = +1\,\mu\text{C}$ along $OA \to OC$.
$q_C = +2\,\mu\text{C}$ repels $q_0 = +1\,\mu\text{C}$ along $OC \to OA$.
Since $q_A = q_C$ and distances are identical, $\mathbf{F}_A + \mathbf{F}_C = \mathbf{0}$.

Diagonal BD:
$q_B = -5\,\mu\text{C}$ attracts $q_0$ along $OB$.
$q_D = -5\,\mu\text{C}$ attracts $q_0$ along $OD$.
Since $q_B = q_D$ and distances are identical, $\mathbf{F}_B + \mathbf{F}_D = \mathbf{0}$.

Resultant: $\mathbf{F}_{\text{net}} = (\mathbf{F}_A + \mathbf{F}_C) + (\mathbf{F}_B + \mathbf{F}_D) = \mathbf{0}$.

Ex 1.7 Field Line Properties

1.7 (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? (b) Explain why two field lines never cross each other at any point?

(a) Why no sudden breaks:
An electric field line represents the trajectory along which a positive test charge experiences a continuous electrostatic force. Since the electric field exists continuously in space and changes smoothly from point to point, a charge cannot jump discontinuously from one position to another. Hence field lines have no sudden breaks.

(b) Why field lines never cross:
The tangent to an electric field line at any point indicates the unique direction of the resultant electric field $\mathbf{E}$ at that point. If two lines intersected, two tangents could be drawn at the point of intersection, implying two distinct directions of net electric force at that single point, which is physically impossible.

Ex 1.8 Field and Force at Midpoint

1.8 Two point charges $q_A = 3\,\mu\text{C}$ and $q_B = -3\,\mu\text{C}$ are located $20\text{ cm}$ apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude $1.5 \times 10^{-9}\text{ C}$ is placed at this point, what is the force experienced by the test charge?

(a) $E = 5.4 \times 10^6\text{ N C}^{-1}$ along $OA \to OB$ • (b) $F = 8.1 \times 10^{-3}\text{ N}$ along $OB \to OA$.

(a) Electric field at midpoint O ($r = 10\text{ cm} = 0.1\text{ m}$):
$\mathbf{E}_A$ due to $+3\,\mu\text{C}$: directed along $OB$ (away from A).
$E_A = \frac{9 \times 10^9 \times 3 \times 10^{-6}}{(0.1)^2} = 2.7 \times 10^6\text{ N C}^{-1}$.
$\mathbf{E}_B$ due to $-3\,\mu\text{C}$: directed along $OB$ (towards B).
$E_B = \frac{9 \times 10^9 \times 3 \times 10^{-6}}{(0.1)^2} = 2.7 \times 10^6\text{ N C}^{-1}$.
$E_{\text{net}} = E_A + E_B = 5.4 \times 10^6\text{ N C}^{-1}$ directed along $AB$ (towards B).

(b) Force on negative test charge ($q = -1.5 \times 10^{-9}\text{ C}$):
$\mathbf{F} = q\mathbf{E} = (-1.5 \times 10^{-9}\text{ C}) \times (5.4 \times 10^6\text{ N C}^{-1}\text{ along }AB) = 8.1 \times 10^{-3}\text{ N}$ directed along $BA$ (towards A).

Ex 1.9 Dipole Moment Coordinates

1.9 A system has two charges $q_A = 2.5 \times 10^{-7}\text{ C}$ and $q_B = -2.5 \times 10^{-7}\text{ C}$ located at points $A(0, 0, -15\text{ cm})$ and $B(0, 0, +15\text{ cm})$, respectively. What are the total charge and electric dipole moment of the system?

Total Charge $Q = 0\text{ C}$ • Dipole Moment $\mathbf{p} = -7.5 \times 10^{-8}\,\hat{\mathbf{k}}\text{ C m}$ (along $-z$).

1. Total Charge:
$Q_{\text{total}} = q_A + q_B = 2.5 \times 10^{-7} + (-2.5 \times 10^{-7}) = 0\text{ C}$.

2. Dipole Separation:
Charges are on the z-axis: $z_A = -15\text{ cm}$, $z_B = +15\text{ cm}$. Separation $2a = 30\text{ cm} = 0.3\text{ m}$.

3. Dipole Moment Vector:
Direction of dipole moment is from negative charge ($q_B$ at $+15\text{ cm}$) to positive charge ($q_A$ at $-15\text{ cm}$), i.e., along the negative z-axis ($-\hat{\mathbf{k}}$).
$p = q(2a) = 2.5 \times 10^{-7} \times 0.3 = 7.5 \times 10^{-8}\text{ C m}$ along $-z$ direction ($\mathbf{p} = -7.5 \times 10^{-8}\,\hat{\mathbf{k}}\text{ C m}$).

Ex 1.10 Torque on Dipole

1.10 An electric dipole with dipole moment $4 \times 10^{-9}\text{ C m}$ is aligned at $30^\circ$ with the direction of a uniform electric field of magnitude $5 \times 10^4\text{ N C}^{-1}$. Calculate the magnitude of the torque acting on the dipole.

Torque $\tau = 10^{-4}\text{ N m}$.

Given:
$p = 4 \times 10^{-9}\text{ C m}$, $E = 5 \times 10^4\text{ N C}^{-1}$, $\theta = 30^\circ$.

Formula:
$\tau = p E \sin\theta = (4 \times 10^{-9}) \times (5 \times 10^4) \times \sin 30^\circ$.

Calculation:
$\tau = 20 \times 10^{-5} \times 0.5 = 10 \times 10^{-5} = 10^{-4}\text{ N m}$.

Ex 1.11 Electron & Mass Transfer

1.11 A polythene piece rubbed with wool is found to have a negative charge of $3 \times 10^{-7}\text{ C}$. (a) Estimate the number of electrons transferred (from which to which?). (b) Is there a transfer of mass from wool to polythene?

(a) $n = 1.875 \times 10^{12}$ electrons from wool to polythene • (b) Yes, $\Delta m = 1.71 \times 10^{-18}\text{ kg}$.

(a) Electrons transferred:
$n = \frac{q}{e} = \frac{3 \times 10^{-7}\text{ C}}{1.6 \times 10^{-19}\text{ C}} = 1.875 \times 10^{12}$ electrons transferred from wool to polythene.

(b) Mass transfer:
Yes, electrons possess mass ($m_e = 9.11 \times 10^{-31}\text{ kg}$).
$\Delta m = n \times m_e = 1.875 \times 10^{12} \times 9.11 \times 10^{-31}\text{ kg} \approx 1.71 \times 10^{-18}\text{ kg}$ (extremely small, but physically real).

Ex 1.12 Coulomb Scaling

1.12 (a) Two insulated charged copper spheres A and B have their centres separated by a distance of $50\text{ cm}$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{-7}\text{ C}$? (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

(a) $F = 1.52 \times 10^{-2}\text{ N}$ • (b) $F' = 0.243\text{ N}$ (16 times initial force).

(a) Initial Force:
$F = \frac{9 \times 10^9 \times (6.5 \times 10^{-7})^2}{(0.5)^2} = \frac{9 \times 10^9 \times 42.25 \times 10^{-14}}{0.25} = 1.521 \times 10^{-2}\text{ N}$.

(b) Modified Force:
$q' = 2q, r' = r/2 \implies F' = \frac{k (2q)(2q)}{(r/2)^2} = 16 F$.
$F' = 16 \times 1.521 \times 10^{-2}\text{ N} = 0.24336\text{ N} \approx 0.24\text{ N}$.

Ex 1.13 Charge-to-Mass Ratio Tracks

1.13 Figure 1.30 shows tracks of three charged particles 1, 2, 3 in a uniform electrostatic field between parallel plates (top plate positive, bottom plate negative). Give the signs of the three charges. Which particle has the highest charge to mass ratio ($q/m$)?

Particles 1 & 2: Negative (−) • Particle 3: Positive (+) • Particle 3 has highest $q/m$ ratio.

1. Signs of charges:
Particles 1 and 2 deflect towards the positive upper plate → Negatively charged (−).
Particle 3 deflects towards the negative lower plate → Positively charged (+).

2. Charge-to-mass ratio ($q/m$):
Vertical displacement $y = \frac{1}{2} a t^2 = \frac{1}{2}\left(\frac{qE}{m}\right)\left(\frac{x}{v_x}\right)^2 \propto \frac{q}{m}$.
From the textbook figure, Particle 3 exhibits the largest transverse deflection $y$, hence Particle 3 has the highest charge-to-mass ratio ($q/m$).

Ex 1.14 Flux Through Square

1.14 Consider a uniform electric field $\mathbf{E} = 3 \times 10^3\,\hat{\mathbf{i}}\text{ N/C}$. (a) What is the flux of this field through a square of $10\text{ cm}$ on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a $60^\circ$ angle with the x-axis?

(a) $\Phi = 30\text{ N m}^2\text{ C}^{-1}$ • (b) $\Phi = 15\text{ N m}^2\text{ C}^{-1}$.

Area: $S = (0.1\text{ m})^2 = 10^{-2}\text{ m}^2$.

(a) Plane parallel to yz plane:
Normal vector $\hat{\mathbf{n}}$ is along the x-axis ($\theta = 0^\circ$).
$\Phi = E S \cos 0^\circ = (3 \times 10^3) \times (10^{-2}) \times 1 = 30\text{ N m}^2\text{ C}^{-1}$.

(b) Normal at $\theta = 60^\circ$:
$\Phi = E S \cos 60^\circ = 30 \times 0.5 = 15\text{ N m}^2\text{ C}^{-1}$.

Ex 1.15 Uniform Field Through Cube

1.15 What is the net flux of the uniform electric field of Exercise 1.14 ($\mathbf{E} = 3 \times 10^3\,\hat{\mathbf{i}}\text{ N/C}$) through a cube of side $20\text{ cm}$ oriented so that its faces are parallel to the coordinate planes?

Net flux $\Phi_{\text{net}} = 0\text{ N m}^2\text{ C}^{-1}$.

Explanation:
The electric field is completely uniform throughout space. The number of field lines entering the left face is identically equal to the number of field lines exiting the right face ($\Phi_{\text{in}} = \Phi_{\text{out}} = E A$). All other four faces are parallel to $\mathbf{E}$ ($\Phi = 0$).
Therefore, net outward flux $\Phi = -EA + EA = 0$. (Also by Gauss's law, no charge is enclosed within the cube, so $q_{\text{enc}} = 0 \implies \Phi = 0$).

Ex 1.16 Black Box Flux & Net Charge

1.16 Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0 \times 10^3\text{ N m}^2\text{ C}^{-1}$. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?

(a) $q = 7.08 \times 10^{-8}\text{ C} \approx 0.07\,\mu\text{C}$ • (b) No, it only implies NET charge is zero.

(a) Charge inside box:
By Gauss's law: $q = \varepsilon_0 \Phi = (8.854 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}) \times (8.0 \times 10^3\text{ N m}^2\text{ C}^{-1}) = 7.08 \times 10^{-8}\text{ C} = 0.0708\,\mu\text{C}$.

(b) Zero flux interpretation:
No. Zero net flux only guarantees that the algebraic sum of all charges inside is zero ($\sum q = 0$). The box could enclose equal amounts of positive and negative charges (e.g. electric dipoles).

Ex 1.17 Flux Through One Face of Cube

1.17 A point charge $+10\,\mu\text{C}$ is a distance $5\text{ cm}$ directly above the centre of a square of side $10\text{ cm}$, as shown in Fig. 1.31. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

$\Phi_{\text{face}} = 1.88 \times 10^5\text{ N m}^2\text{ C}^{-1}$.

Symmetry Construction:
Construct a cube of side $10\text{ cm}$ centered around the charge. The given square forms one of the 6 symmetric faces of this cube, with the charge exactly at the center ($5\text{ cm}$ from each face).

Total Flux Through Cube:
$\Phi_{\text{total}} = \frac{q}{\varepsilon_0}$.

Flux Through Single Face:
By symmetry, flux divides equally among the 6 identical faces:
$\Phi_{\text{face}} = \frac{1}{6}\frac{q}{\varepsilon_0} = \frac{10 \times 10^{-6}\text{ C}}{6 \times 8.854 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}} = \frac{10^{-5}}{5.3124 \times 10^{-11}} \approx 1.88 \times 10^5\text{ N m}^2\text{ C}^{-1}$.

Ex 1.18 Net Flux from Cubic Surface

1.18 A point charge of $2.0\,\mu\text{C}$ is at the centre of a cubic Gaussian surface $9.0\text{ cm}$ on edge. What is the net electric flux through the surface?

$\Phi_{\text{net}} = 2.26 \times 10^5\text{ N m}^2\text{ C}^{-1}$.

Gauss's Law:
$\Phi_{\text{net}} = \frac{q}{\varepsilon_0} = \frac{2.0 \times 10^{-6}\text{ C}}{8.854 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}} \approx 2.258 \times 10^5\text{ N m}^2\text{ C}^{-1} \approx 2.26 \times 10^5\text{ N m}^2\text{ C}^{-1}$.

Note: The side length ($9.0\text{ cm}$) does not affect the total flux through the closed surface.

Ex 1.19 Spherical Gaussian Flux & Charge Value

1.19 A point charge causes an electric flux of $-1.0 \times 10^3\text{ N m}^2\text{ C}^{-1}$ to pass through a spherical Gaussian surface of $10.0\text{ cm}$ radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

(a) $\Phi = -1.0 \times 10^3\text{ N m}^2\text{ C}^{-1}$ (Unchanged) • (b) $q = -8.85\text{ nC}$.

(a) Doubling radius:
Electric flux $\Phi = q/\varepsilon_0$ depends strictly on enclosed charge, not on radius $r$. Hence flux remains $-1.0 \times 10^3\text{ N m}^2\text{ C}^{-1}$.

(b) Value of point charge $q$:
$q = \varepsilon_0 \Phi = (8.854 \times 10^{-12}) \times (-1.0 \times 10^3) = -8.854 \times 10^{-9}\text{ C} = -8.85\text{ nC}$.

Ex 1.20 Conducting Sphere Unknown Charge

1.20 A conducting sphere of radius $10\text{ cm}$ has an unknown charge. If the electric field $20\text{ cm}$ from the centre of the sphere is $1.5 \times 10^3\text{ N/C}$ and points radially inward, what is the net charge on the sphere?

$q = -6.67\text{ nC} = -6.67 \times 10^{-9}\text{ C}$.

Formula:
Outside a conducting sphere ($r = 0.2\text{ m} > R = 0.1\text{ m}$), $E = \frac{1}{4\pi\varepsilon_0}\frac{|q|}{r^2}$.
$|q| = \frac{E r^2}{k} = \frac{1.5 \times 10^3 \times (0.2)^2}{9 \times 10^9} = \frac{60}{9 \times 10^9} = 6.67 \times 10^{-9}\text{ C} = 6.67\text{ nC}$.

Sign: Since the electric field vector points radially inward, the charge must be negative ($q = -6.67\text{ nC}$).

Ex 1.21 Charged Conducting Sphere

1.21 A uniformly charged conducting sphere of $2.4\text{ m}$ diameter has a surface charge density of $80.0\,\mu\text{C/m}^2$. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

(a) $q = 1.45 \times 10^{-3}\text{ C} = 1.45\text{ mC}$ • (b) $\Phi = 1.63 \times 10^8\text{ N m}^2\text{ C}^{-1}$.

Given: Radius $R = 2.4/2 = 1.2\text{ m}$, $\sigma = 80.0 \times 10^{-6}\text{ C m}^{-2}$.

(a) Charge on Sphere:
$q = \sigma(4\pi R^2) = 80 \times 10^{-6} \times 4 \times \pi \times (1.2)^2 = 80 \times 10^{-6} \times 18.096 = 1.448 \times 10^{-3}\text{ C} \approx 1.45\text{ mC}$.

(b) Total Flux Leaving Surface:
$\Phi = \frac{q}{\varepsilon_0} = \frac{1.448 \times 10^{-3}}{8.854 \times 10^{-12}} \approx 1.63 \times 10^8\text{ N m}^2\text{ C}^{-1}$.

Ex 1.22 Infinite Line Charge Density

1.22 An infinite line charge produces a field of $9 \times 10^4\text{ N/C}$ at a distance of $2\text{ cm}$. Calculate the linear charge density.

Linear charge density $\lambda = 10^{-7}\text{ C m}^{-1} = 0.1\,\mu\text{C m}^{-1}$.

Given: $E = 9 \times 10^4\text{ N/C}$, $r = 2\text{ cm} = 0.02\text{ m}$.

Formula:
$E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{2k\lambda}{r} \implies \lambda = \frac{E r}{2k}$.

Calculation:
$\lambda = \frac{(9 \times 10^4) \times 0.02}{2 \times (9 \times 10^9)} = \frac{1800}{18 \times 10^9} = 10^{-7}\text{ C m}^{-1} = 0.1\,\mu\text{C m}^{-1}$.

Ex 1.23 Oppositely Charged Parallel Plates

1.23 Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17.0 \times 10^{-22}\text{ C/m}^2$. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

(a) $E = 0\text{ N/C}$ • (b) $E = 0\text{ N/C}$ • (c) $E = 1.92 \times 10^{-10}\text{ N/C}$.

Plate Fields: Each plate produces field $E_0 = \frac{\sigma}{2\varepsilon_0}$. Plate 1 ($+\sigma$) produces outward field; Plate 2 ($-\sigma$) produces inward field.

(a) Outer region of Plate 1 (Region I):
$E_I = E_+ - E_- = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0$.

(b) Outer region of Plate 2 (Region III):
$E_{III} = E_+ - E_- = 0$.

(c) Between the plates (Region II):
Both fields point in the same direction (from positive plate to negative plate):
$E_{II} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{17.0 \times 10^{-22}\text{ C m}^{-2}}{8.854 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}} \approx 1.92 \times 10^{-10}\text{ N C}^{-1}$.

04 / Rapid Reference

Chapter Summary & Points to Ponder

15 core summary principles, formula quick-reference sheet, physical quantities table, and all 11 official NCERT Points to Ponder.

100% SYLLABUS

Section 1.4 • Properties of Charge

Q = ne • Additive • Conserved

Quantised in steps of $e = 1.6 \times 10^{-19}\text{ C}$. Charge is scalar and conserved in all isolated reactions.

Section 1.5 • Coulomb's Law

F = (1/4πε₀) (q₁q₂/r²)

$k \approx 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$. Inverse square law along the line connecting charges.

Section 1.7 • Electric Field

E = F/q = (1/4πε₀) (Q/r²)

Force per unit positive test charge. Outward for $+Q$, inward for $-Q$.

Section 1.8 • Field Lines

Continuous • Never Cross • No Loops

Density represents field strength. Tangent gives direction of $\mathbf{E}$.

Section 1.10 • Dipole Fields

E_axial = 2kp/r³ • E_eq = -kp/r³

Dipole moment $\mathbf{p} = q(2\mathbf{a})$ (from $-q$ to $+q$). Field falls off as $1/r^3$.

Section 1.11 • Torque on Dipole

τ = p × E = pE sinθ

Net force is zero in uniform field. Maximum torque at $\theta = 90^\circ$.

Section 1.13 • Gauss's Law

Φ = ∮ E·dS = q_enc / ε₀

Total electric flux depends only on enclosed charge, independent of surface shape/size.

Section 1.14 • Gauss Applications

Wire: λ/(2πε₀r) • Sheet: σ/(2ε₀)

Spherical shell: $E_{\text{in}} = 0$, $E_{\text{out}} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$.

NCERT Official Table

Physical Quantities & Dimensions

Physical Quantity	Symbol	Dimensions	SI Unit	Remarks
Vector area element	ΔS	$[L^2]$	m²	$\Delta\mathbf{S} = \Delta S\,\hat{\mathbf{n}}$
Electric field	E	$[M L T^{-3} A^{-1}]$	V m⁻¹ (or N C⁻¹)	$\mathbf{F} = q\mathbf{E}$
Electric flux	Φ	$[M L^3 T^{-3} A^{-1}]$	V m (or N m² C⁻¹)	$\Delta\Phi = \mathbf{E}\cdot\Delta\mathbf{S}$
Dipole moment	p	$[L T A]$	C m	Vector from $-q$ to $+q$
Linear charge density	λ	$[L^{-1} T A]$	C m⁻¹	Charge / length
Surface charge density	σ	$[L^{-2} T A]$	C m⁻²	Charge / area
Volume charge density	ρ	$[L^{-3} T A]$	C m⁻³	Charge / volume

NCERT Official

Points to Ponder

Nuclear Binding: Protons inside a nucleus stay bound due to the strong nuclear force (effective range $\sim 10^{-14}\text{ m}$), overcoming mutual Coulomb repulsion. Electrons cannot sit inside the nucleus due to quantum mechanics.
Gravity vs Electrostatics: Coulomb force can be attractive or repulsive (enabling cancellation in bulk matter), whereas gravity is always attractive. Hence gravity dominates cosmology despite being much weaker.
Value of Constant $k$: In SI, current (A) is defined via Ampere's Law and $1\text{ C} = 1\text{ A s}$. Thus $k \approx 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$ is fixed experimentally.
Large Unit Size: $1\text{ C}$ is an extremely large unit for electrostatics because magnetic forces used to define the ampere are much weaker than electric forces.
Scalar Additivity: Electric charge has no direction; it adds algebraically with proper signs.
Relativistic Invariance: Charge is invariant under rotation and reference frames in relative motion (unlike kinetic energy).
Conservation vs Invariance: Charge conservation refers to invariance in time in an isolated system.
Quantisation of Mass: Charge is quantised ($q = ne$), but no analogous universal mass quantisation law exists.
Superposition Principle: Pairwise Coulomb force between two charges is unaffected by third-party charges, and no extra 3-body forces exist.
Discontinuity of Field: Electric field due to discrete charges is undefined at the charge location. For surface charges, $\mathbf{E}$ is discontinuous across the boundary.
Zero Net Charge Fields: A dipole with zero total charge still produces a non-zero field decaying as $1/r^3$.

05 / CBSE 10-Year PYQ Bank

Chapter 1: Previous Year Questions

Complete 66 official CBSE Board & Sample Paper questions (2012–2026). Continuous scrolling across all topics with smooth jump navigation between sub-topics.

66 QUESTIONS

SUBTOPIC 1

Topic 1: Electric Charge, Conservation & Coulomb's Law

18 Questions (PYQ 1–18)

PYQ 1 MCQ — 1M | CBSE Term-1 SQP 2021

Topic 1

1. Three microscopic latex spheres are sprayed into a chamber and become charged with $+3e$, $+5e$ and $-3e$ respectively. All three come in contact simultaneously and get separated. Which of the following are possible values for the final charges?

(a) $+5e, -1e, +5e$

(b) $+6e, +6e, -7e$

(d) $+5e, -8e, +7e$

Correct Answer: (b) $+6e, +6e, -7e$

Step 1: Total Initial Charge (Conservation Law)

According to the Law of Conservation of Electric Charge, for an isolated system, the algebraic sum of electric charges remains constant before and after interaction.
$$Q_{\text{initial}} = q_1 + q_2 + q_3 = (+3e) + (+5e) + (-3e) = +5e$$

Step 2: Checking Options for Total Final Charge

Evaluating the algebraic sum for each given option:
• Option (a): $(+5e) + (-1e) + (+5e) = +9e \ne +5e$ (Violates conservation)
• Option (b): $(+6e) + (+6e) + (-7e) = +5e$ (Satisfies conservation)
• Option (c): $(-4e) + (+3e) + (+5e) = +4e \ne +5e$ (Violates conservation)
• Option (d): $(+5e) + (-8e) + (+7e) = +4e \ne +5e$ (Violates conservation)

Step 3: Quantisation Condition

Furthermore, the charges on each sphere in (b) are all integral multiples of the elementary charge $e$, which strictly obeys the principle of quantisation of charge ($q = \pm ne$). Thus, (b) is the correct and only possible option.

PYQ 2 MCQ — 1M | CBSE 2017

Topic 1

2. One of two hydrogen atoms has a slightly different charge. If the net of electrostatic force and gravitational force between two hydrogen atoms placed at distance $d$ apart is zero, then $\Delta e$ is of the order of: [Given $m_H = 1.67 \times 10^{-27}\text{ kg}$]

(a) $10^{-23}\text{ C}$

(b) $10^{-37}\text{ C}$

(d) $10^{-20}\text{ C}$

Correct Answer: (b) $10^{-37}\text{ C}$

Step 1: Electrostatic Repulsion vs Gravitational Attraction

Let the net residual charge on each hydrogen atom be $\Delta e$. For equilibrium between the two atoms, the electrostatic repulsive force must exactly balance the universal gravitational attractive force:
$$F_{\text{net}} = 0 \implies F_e = F_g$$
$$\frac{1}{4\pi\varepsilon_0} \frac{(\Delta e)^2}{d^2} = G \frac{m_H^2}{d^2}$$

Step 2: Eliminating Distance $d$ and Substituting Constants

$$(\Delta e)^2 = 4\pi\varepsilon_0 G m_H^2$$
Given:
• $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2\text{ C}^{-2} \implies 4\pi\varepsilon_0 = \frac{1}{9 \times 10^9}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}$
• $G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}$
• $m_H = 1.67 \times 10^{-27}\text{ kg}$

Step 3: Calculating Value & Order of Magnitude

$$(\Delta e)^2 = \frac{6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2}{9 \times 10^9} = \frac{6.67 \times 2.7889 \times 10^{-65}}{9 \times 10^9} = \frac{18.602 \times 10^{-65}}{9 \times 10^9} \approx 2.067 \times 10^{-74}\text{ C}^2$$
$$\Delta e = \sqrt{2.067 \times 10^{-74}} \approx 1.44 \times 10^{-37}\text{ C}$$
Therefore, $\Delta e$ is of the order of $10^{-37}\text{ C}$.

PYQ 3 VSA — 1M | CBSE Recurring

Topic 1

3. What does "electric charge is quantised" mean? Write the relation.

Step 1: Physical Meaning of Quantisation

Quantisation of electric charge means that any physically existing and observable charge on a body is not continuous, but always occurs as an integral multiple of a basic elementary quantum of charge $e$ (magnitude of charge on an electron/proton, $e = 1.602 \times 10^{-19}\text{ C}$). Charge cannot exist in fractional values of $e$ in free states.

Step 2: Mathematical Formula & Symbols

$$q = \pm n e$$
where:
• $q =$ total electric charge on the body,
• $n = 1, 2, 3, \dots$ (an integer),
• $e = 1.602 \times 10^{-19}\text{ C}$ (elementary fundamental charge).

PYQ 4 VSA — 1M | CBSE Recurring

Topic 1

4. Does the force between two point charges change if the dielectric constant of the medium in which they are placed increases? Justify.

Step 1: Direct Statement

Yes, the electrostatic force between the two charges decreases.

Step 2: Justification from Coulomb's Law in Medium

The electrostatic force between two stationary point charges $q_1$ and $q_2$ separated by distance $r$ in a medium of dielectric constant $K$ (or relative permittivity $\varepsilon_r$) is given by:
$$F_m = \frac{1}{4\pi\varepsilon}\frac{q_1 q_2}{r^2} = \frac{1}{4\pi K\varepsilon_0}\frac{q_1 q_2}{r^2} = \frac{F_0}{K}$$
where $F_0$ is the electrostatic force in vacuum/air.
Since $F_m \propto \frac{1}{K}$, as the dielectric constant $K$ increases, the electrostatic force $F_m$ decreases by a factor of $K$.

PYQ 5 SA — 2M | CBSE Delhi 2017

Topic 1

5. Does the charge given to a metallic sphere depend on whether it is hollow or solid? Give reason for your answer.

Step 1: Direct Answer

No, the charge given to a metallic sphere does not depend on whether the sphere is hollow or solid.

Step 2: Electrostatic Principle in Conductors

In a metallic conductor, free conduction electrons are mobile. When excess electric charge is supplied to a conductor, the mutual repulsive Coulomb forces drive all the charges as far apart as possible. Consequently, all excess charge redistributes itself and resides exclusively on the outer surface of the sphere. The electrostatic field in the interior of a conductor is identically zero ($E_{\text{inside}} = 0$).

Step 3: Comparison between Solid and Hollow Spheres

Since a solid metallic sphere and a hollow metallic sphere of the identical radius $R$ have the exact same outer surface area ($A = 4\pi R^2$), both spheres can hold the exact same maximum amount of charge up to the dielectric breakdown limit of surrounding air.

PYQ 6 SA — 2M | CBSE Recurring

Topic 1

6. When a glass rod is rubbed with a silk cloth, charges appear on both. Explain how this observation is consistent with the law of conservation of charge.

Step 1: Initial Neutral State

Before rubbing, both the glass rod and silk cloth are electrically neutral. Hence, the initial net charge of the system is:
$$Q_{\text{initial}} = Q_{\text{glass}} + Q_{\text{silk}} = 0 + 0 = 0$$

Step 2: Mechanism of Charging by Friction

When the glass rod is rubbed vigorously with the silk cloth, work done against friction provides the energy needed to transfer loosely bound valence electrons from the glass atoms to the silk atoms.
• The glass rod loses $N$ electrons and acquires a positive charge: $Q_{\text{glass}} = +N e$
• The silk cloth gains the exact same $N$ electrons and acquires an equal negative charge: $Q_{\text{silk}} = -N e$

Step 3: Total Final Charge & Conservation Verification

The total final charge of the isolated combined system is:
$$Q_{\text{final}} = Q_{\text{glass}} + Q_{\text{silk}} = (+N e) + (-N e) = 0$$
Since $Q_{\text{initial}} = Q_{\text{final}} = 0$, no new charge is created or destroyed; charges are merely transferred from one body to another. Hence, the observation is completely consistent with the Law of Conservation of Charge.

PYQ 7 MCQ — 1M | CBSE 2026 Set 55/5/1

Topic 1

7. Two small identical metallic balls having charges $q$ and $-2q$ are kept at separation $r$. They are brought in contact and then separated to a distance $r/2$. Compared to the initial force $F$, they will now:

(A) attract with force $F/2$

(B) repel with force $F/2$

(D) attract with force $F$

Correct Answer: (B) repel with force $F/2$

Step 1: Initial Force Expression

Before contact, the charges are $q_1 = q$ and $q_2 = -2q$ at distance $r$. By Coulomb's law:
$$F = \frac{1}{4\pi\varepsilon_0} \frac{|q(-2q)|}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{2q^2}{r^2} \quad \text{(Attractive in nature)}$$

Step 2: Charge Redistribution on Contact

Because the two metallic balls are identical, when brought into physical contact, the total charge divides equally between them:
$$q_{\text{total}} = q + (-2q) = -q$$
$$q'_1 = q'_2 = \frac{q_{\text{total}}}{2} = -\frac{q}{2}$$

Step 3: New Force at Distance $r' = r/2$

When separated to a new distance $r' = \frac{r}{2}$:
$$F' = \frac{1}{4\pi\varepsilon_0} \frac{(-q/2)(-q/2)}{(r/2)^2} = \frac{1}{4\pi\varepsilon_0} \frac{q^2/4}{r^2/4} = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{r^2} = \frac{1}{2} \left( \frac{1}{4\pi\varepsilon_0} \frac{2q^2}{r^2} \right) = \frac{F}{2}$$
Since both spheres carry like charges (both negative, $-q/2$), they will now repel with force $F/2$.

PYQ 8 MCQ — 1M | CBSE Term-1 SQP 2021

Topic 1

8. Two point charges placed in a medium of dielectric constant $5$ are at distance $r$ between them and experience an electrostatic force $F$. The electrostatic force between them in vacuum at the same distance $r$ will be:

(a) $5F$

(b) $F$

(d) $F/5$

Correct Answer: (a) $5F$

Step 1: Relation Between Force in Vacuum and Medium

The electrostatic force between two point charges in a medium of dielectric constant $K$ is related to force in vacuum $F_{\text{vac}}$ by:
$$F_{\text{med}} = \frac{F_{\text{vac}}}{K}$$

Step 2: Substituting Values

Given $F_{\text{med}} = F$ and $K = 5$:
$$F = \frac{F_{\text{vac}}}{5} \implies F_{\text{vac}} = 5F$$
Hence, option (a) $5F$ is the correct answer.

PYQ 9 MCQ — 1M | CBSE 2016

Topic 1

9. Two identical charged spheres suspended from a common point by massless strings of length $l$ are initially at distance $d$ ($d \ll l$) apart because of their mutual repulsion. The charges begin to leak from both spheres at a constant rate. As a result, the spheres approach each other with velocity $v$. Then $v$ varies as a function of the distance $x$ between the spheres as:

(a) $v \propto x^{1/2}$

(b) $v \propto x$

(d) $v \propto x^{-1}$

Correct Answer: (c) $v \propto x^{-1/2}$

Step 1: Equilibrium of Suspended Sphere

Let $x$ be the separation between the spheres at any instant. The forces acting on each sphere are: Tension $T$, weight $mg$, and Coulomb repulsive force $F_e$.
In equilibrium:
$$T \cos\theta = mg \quad \text{and} \quad T \sin\theta = F_e = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{x^2}$$
$$\tan\theta = \frac{F_e}{mg} = \frac{k q^2}{x^2 mg}$$

Step 2: Small-Angle Approximation ($x \ll l$)

For small angle $\theta$, $\tan\theta \approx \sin\theta = \frac{x/2}{l}$:
$$\frac{x}{2l} = \frac{k q^2}{x^2 mg} \implies q^2 = \left( \frac{mg}{2kl} \right) x^3 \implies q \propto x^{3/2}$$

Step 3: Differentiating with Respect to Time

Differentiating $q$ with respect to time $t$:
$$\frac{dq}{dt} \propto \frac{3}{2} x^{1/2} \frac{dx}{dt} = x^{1/2} v$$
Since charge leaks at a constant rate, $\frac{dq}{dt} = \text{constant}$:
$$\text{constant} = x^{1/2} v \implies v \propto \frac{1}{x^{1/2}} = x^{-1/2}$$
Hence, option (c) $v \propto x^{-1/2}$ is correct.

PYQ 10 Numerical — 2M | CBSE/NCERT Recurring

Topic 1

10. 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 $30\text{ cm}$ apart in air?

Step 1: Given Data & Unit Conversions

• Charge on first sphere, $q_1 = 2 \times 10^{-7}\text{ C}$
• Charge on second sphere, $q_2 = 3 \times 10^{-7}\text{ C}$
• Distance of separation, $r = 30\text{ cm} = 0.3\text{ m} = 3 \times 10^{-1}\text{ m}$
• Electrostatic constant in air, $k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$

Step 2: Applying Coulomb's Law

$$F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$$
$$F = \frac{(9 \times 10^9) \times (2 \times 10^{-7}) \times (3 \times 10^{-7})}{(0.3)^2}$$

Step 3: Stepwise Calculation & Nature of Force

$$F = \frac{54 \times 10^{-5}}{0.09} = \frac{54 \times 10^{-5}}{9 \times 10^{-2}} = 6 \times 10^{-3}\text{ N}$$
Since both charges are positive ($q_1 > 0, q_2 > 0$), like charges repel each other.
Final Answer: Electrostatic force is $6 \times 10^{-3}\text{ N}$ (Repulsive).

PYQ 11 SA — 2M | CBSE 2015

Topic 1

11. Two equal balls having equal positive charge $q$ coulombs are suspended by insulating strings of equal length. What would be the effect on the force between the balls if a polythene sheet is introduced between the two balls without touching them?

Step 1: Property of Polythene Sheet

Polythene is an electrical insulator and behaves as a dielectric medium with a dielectric constant $K > 1$ (typically $K \approx 2.3$ for polyethylene).

Step 2: Effect on Mutual Coulomb Force

In air/vacuum ($K = 1$), electrostatic repulsive force is $F_0 = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2}$.
When a polythene sheet is introduced between the two charged spheres, the permittivity of the medium increases to $\varepsilon = K\varepsilon_0$.
The new electrostatic force becomes:
$$F = \frac{F_0}{K}$$
Since $K > 1$, the electrostatic repulsive force between the balls decreases.

Step 3: Physical Effect on Suspended Strings

Due to the reduction in mutual repulsive force, the balls move towards each other, resulting in a decrease in the angle of separation between the suspended strings.

PYQ 12 Numerical — 2M | CBSE Foreign 2013

Topic 1

12. The sum of two point charges is $7\;\mu\text{C}$. They repel each other with a force of $1\text{ N}$ when kept $30\text{ cm}$ apart in free space. Calculate the value of each charge.

Step 1: Given Equations & Parameters

• Sum of charges: $q_1 + q_2 = 7\;\mu\text{C} = 7 \times 10^{-6}\text{ C} \quad \text{--- [Equation 1]}$
• Separation: $r = 30\text{ cm} = 0.3\text{ m}$
• Electrostatic force: $F = 1\text{ N}$
• Electrostatic constant: $k = 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$

Step 2: Applying Coulomb's Law to Find Product $q_1 q_2$

$$F = \frac{k q_1 q_2}{r^2} \implies 1 = \frac{(9 \times 10^9) \times q_1 q_2}{(0.3)^2} = \frac{9 \times 10^9 \times q_1 q_2}{0.09} = 10^{11} q_1 q_2$$
$$q_1 q_2 = \frac{1}{10^{11}} = 10^{-11}\text{ C}^2 = 10 \times 10^{-12}\text{ C}^2 = 10\;(\mu\text{C})^2 \quad \text{--- [Equation 2]}$$

Step 3: Using Algebraic Identity to Solve for $q_1$ and $q_2$

$$(q_1 - q_2)^2 = (q_1 + q_2)^2 - 4 q_1 q_2 = (7)^2 - 4(10) = 49 - 40 = 9\;(\mu\text{C})^2$$
$$q_1 - q_2 = \sqrt{9} = 3\;\mu\text{C} \quad \text{--- [Equation 3]}$$
Adding Equation (1) and Equation (3):
$$2q_1 = 10\;\mu\text{C} \implies q_1 = 5\;\mu\text{C}$$
$$q_2 = 7 - 5 = 2\;\mu\text{C}$$
Final Answer: The values of the two charges are $5\;\mu\text{C}$ and $2\;\mu\text{C}$.

PYQ 13 SA — 3M | CBSE 2023

Topic 1

13. State Coulomb's law. Write it in vector form. Two point charges $q_1 = 7\;\mu\text{C}$ and $q_2 = -2\;\mu\text{C}$ are placed $20\text{ cm}$ apart in vacuum. Find the position of the neutral point on the line joining the two charges.

Step 1: Statement of Coulomb's Law & Vector Notation

Statement: The magnitude of the electrostatic force of interaction between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them, acting along the line joining the centers of the two charges.
Vector Form:
$$\mathbf{F}_{12} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}^2} \hat{\mathbf{r}}_{21} \quad \text{and} \quad \mathbf{F}_{21} = -\mathbf{F}_{12}$$

Step 2: Identifying Position of Neutral Point ($E_{\text{net}} = 0$)

Since the charges $q_1 = +7\;\mu\text{C}$ and $q_2 = -2\;\mu\text{C}$ have opposite polarities, the neutral point cannot lie between them (where both electric fields add in the same direction towards $q_2$).
It must lie on the extended line joining the charges on the side of the charge having smaller numerical magnitude (i.e. to the right of $q_2 = -2\;\mu\text{C}$).
Let the neutral point $P$ be at a distance $x\text{ cm}$ from $q_2$. Then its distance from $q_1$ is $(20 + x)\text{ cm}$.

Step 3: Equating Field Magnitudes and Solving for $x$

At neutral point $P$: $|\mathbf{E}_1| = |\mathbf{E}_2|$
$$\frac{1}{4\pi\varepsilon_0} \frac{7 \times 10^{-6}}{(20 + x)^2} = \frac{1}{4\pi\varepsilon_0} \frac{2 \times 10^{-6}}{x^2}$$
$$\frac{7}{(20 + x)^2} = \frac{2}{x^2} \implies \frac{\sqrt{7}}{20 + x} = \frac{\sqrt{2}}{x}$$
$$\sqrt{7} x = 20\sqrt{2} + \sqrt{2} x \implies x(\sqrt{7} - \sqrt{2}) = 20\sqrt{2}$$
$$x = \frac{20(1.414)}{2.646 - 1.414} = \frac{28.28}{1.232} \approx 22.95\text{ cm}$$
Final Answer: The neutral point lies on the line joining charges at a distance of $22.95\text{ cm}$ from the $-2\;\mu\text{C}$ charge (on the side away from $7\;\mu\text{C}$).

PYQ 14 SA — 3M | CBSE Foreign 2015

Topic 1

14. Three point electric charges $+q$ each are kept at the vertices of an equilateral triangle of side $a$. Determine the magnitude and sign of the charge to be kept at the centroid of the triangle so that the charges at the vertices remain in equilibrium.

Step 1: Geometry of Equilateral Triangle and Centroid Distance

Let vertices be $A, B, C$ each with charge $+q$. Side $= a$.
Height of triangle $h = a\sin 60^\circ = \frac{\sqrt{3}}{2}a$.
Distance of centroid $O$ from each vertex is $r = \frac{2}{3} h = \frac{2}{3} \left(\frac{\sqrt{3}}{2}a\right) = \frac{a}{\sqrt{3}}$.

Step 2: Net Force on Vertex Charge due to Other Two Vertices

Consider charge $+q$ at vertex $A$.
• Force due to $+q$ at $B$: $F_B = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{a^2}$ along $BA$ extended.
• Force due to $+q$ at $C$: $F_C = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{a^2}$ along $CA$ extended.
• Angle between $\mathbf{F}_B$ and $\mathbf{F}_C$ is $60^\circ$.
Resultant repulsive force on $A$:
$$F_R = \sqrt{F_B^2 + F_C^2 + 2F_B F_C \cos 60^\circ} = \sqrt{F^2 + F^2 + 2F^2(1/2)} = \sqrt{3} F = \sqrt{3} \left( \frac{1}{4\pi\varepsilon_0} \frac{q^2}{a^2} \right)$$
This resultant $\mathbf{F}_R$ acts radially outward along the line joining centroid $O$ to vertex $A$.

Step 3: Equilibrium by Centroid Charge $Q$

For the charge at vertex $A$ to be in equilibrium, the charge $Q$ placed at centroid $O$ must exert an equal and opposite attractive force directed along $AO$:
$$F_Q = \frac{1}{4\pi\varepsilon_0} \frac{q Q}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{q Q}{(a/\sqrt{3})^2} = \frac{1}{4\pi\varepsilon_0} \frac{3 q Q}{a^2}$$
For net force $\mathbf{F}_{\text{net}} = \mathbf{F}_R + \mathbf{F}_Q = 0$:
$$\sqrt{3}\left(\frac{1}{4\pi\varepsilon_0}\frac{q^2}{a^2}\right) + \frac{1}{4\pi\varepsilon_0}\frac{3 q Q}{a^2} = 0 \implies 3 Q = -\sqrt{3} q \implies Q = -\frac{q}{\sqrt{3}}$$
Final Answer: The charge at centroid must have negative sign and magnitude $Q = -\frac{q}{\sqrt{3}}$.

PYQ 15 SA — 3M | CBSE 2009; 2020 (55/5/1)

Topic 1

15. Two identical point charges $q$ each are kept $2\text{ m}$ apart in air. A third point charge $Q$ of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of $Q$.

Step 1: Equilibrium of the Third Charge $Q$ (Finding Position)

Let charge $q_1 = q$ be at $x = 0$ and $q_2 = q$ at $x = 2\text{ m}$. Let charge $Q$ be placed at distance $x$ from $x = 0$.
For charge $Q$ to be in equilibrium, the forces exerted on it by $q_1$ and $q_2$ must be equal in magnitude and opposite in direction:
$$\frac{1}{4\pi\varepsilon_0} \frac{q Q}{x^2} = \frac{1}{4\pi\varepsilon_0} \frac{q Q}{(2 - x)^2}$$
$$x^2 = (2 - x)^2 \implies x = 2 - x \implies 2x = 2 \implies x = 1\text{ m}$$
Hence, $Q$ must be placed at the exact midpoint ($1\text{ m}$ from either charge).

Step 2: Equilibrium of the Outer Charge $q$ (Finding Magnitude and Sign)

For the entire three-charge system to be in equilibrium, each outer charge $q$ at $x=0$ must also experience zero net force:
$$F_{\text{net on } q_1} = F_{\text{due to } q_2} + F_{\text{due to } Q} = 0$$
$$\frac{1}{4\pi\varepsilon_0} \frac{q^2}{(2)^2} + \frac{1}{4\pi\varepsilon_0} \frac{q Q}{(1)^2} = 0$$
$$\frac{q^2}{4} + q Q = 0 \implies q\left( \frac{q}{4} + Q \right) = 0 \implies Q = -\frac{q}{4}$$

Step 3: Conclusion & Mark Scheme Summary

Summary:
• Position: Midpoint of the line joining the two charges ($1\text{ m}$ from each charge).
• Nature/Sign: Negative.
• Magnitude: $|Q| = \frac{q}{4}$.

PYQ 16 SA — 3M | CBSE 2018

Topic 1

16. Three point charges $q, -4q$ and $2q$ are placed at the vertices of an equilateral triangle $ABC$ of side $l$ as shown in figure. Obtain the expression for the magnitude of the resultant electric force acting on charge $q$.

Step 1: Identifying Forces Acting on Charge $q$ at Vertex $A$

Let vertex $A$ hold charge $q$, vertex $B$ hold $-4q$, and vertex $C$ hold $+2q$. Side of equilateral triangle $= l$.
• Attractive force on $q$ due to $-4q$ at $B$ (directed along $AB$ towards $B$):
$$F_B = \frac{1}{4\pi\varepsilon_0} \frac{q(4q)}{l^2} = 4 \left( \frac{1}{4\pi\varepsilon_0} \frac{q^2}{l^2} \right) = 4 F_0$$
• Repulsive force on $q$ due to $+2q$ at $C$ (directed along $CA$ extended away from $C$):
$$F_C = \frac{1}{4\pi\varepsilon_0} \frac{q(2q)}{l^2} = 2 \left( \frac{1}{4\pi\varepsilon_0} \frac{q^2}{l^2} \right) = 2 F_0$$
where $F_0 = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{l^2}$.

Step 2: Angle Between Force Vectors

The interior angle of equilateral triangle is $\angle BAC = 60^\circ$.
Since $\mathbf{F}_B$ is along $AB$ and $\mathbf{F}_C$ is along $CA$ extended, the angle between the vectors $\mathbf{F}_B$ and $\mathbf{F}_C$ is:
$$\theta = 180^\circ - 60^\circ = 120^\circ$$

Step 3: Vector Addition for Resultant Force

$$F_{\text{res}} = \sqrt{F_B^2 + F_C^2 + 2 F_B F_C \cos 120^\circ}$$
$$F_{\text{res}} = \sqrt{(4 F_0)^2 + (2 F_0)^2 + 2(4 F_0)(2 F_0)\left(-\frac{1}{2}\right)} = \sqrt{16 F_0^2 + 4 F_0^2 - 8 F_0^2} = \sqrt{12 F_0^2} = 2\sqrt{3} F_0$$
Substituting $F_0 = \frac{q^2}{4\pi\varepsilon_0 l^2}$:
$$F_{\text{res}} = 2\sqrt{3} \left( \frac{q^2}{4\pi\varepsilon_0 l^2} \right) = \frac{\sqrt{3} q^2}{2\pi\varepsilon_0 l^2}$$

PYQ 17 SA — 3M | CBSE 2018

Topic 1

17. Four point charges $Q, q, Q$ and $q$ are placed at the corners of a square of side 'a'. Find the resultant electric force on a charge $Q$ placed at one of the corners.

Step 1: Configuration & Geometry

Let square corners in order be $A(Q), B(q), C(Q), D(q)$. Side length $= a$.
Consider the resultant force acting on charge $Q$ at corner $A$.
• Distance to adjacent charge $q$ at $B = a$
• Distance to adjacent charge $q$ at $D = a$
• Distance to diagonal charge $Q$ at $C = a\sqrt{2}$

Step 2: Individual Force Contributions on $Q$ at $A$

• Force due to $q$ at $B$: $F_1 = \frac{1}{4\pi\varepsilon_0} \frac{Q q}{a^2}$ along $BA$ extended (along side $x$).
• Force due to $q$ at $D$: $F_2 = \frac{1}{4\pi\varepsilon_0} \frac{Q q}{a^2}$ along $DA$ extended (along side $y$, perpendicular to $F_1$).
• Force due to $Q$ at $C$: $F_3 = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{(a\sqrt{2})^2} = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{2a^2}$ along diagonal $CA$ extended.

Step 3: Vector Sum along the Diagonal

The resultant of the two mutually perpendicular forces $F_1$ and $F_2$ acts symmetrically along the diagonal $CA$ extended:
$$F_{12} = \sqrt{F_1^2 + F_2^2} = \sqrt{2} F_1 = \sqrt{2} \left( \frac{1}{4\pi\varepsilon_0} \frac{Q q}{a^2} \right)$$
Since $F_{12}$ and $F_3$ point in the exact same direction along the diagonal:
$$F_{\text{net}} = F_{12} + F_3 = \frac{1}{4\pi\varepsilon_0} \frac{\sqrt{2} Q q}{a^2} + \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{2a^2} = \frac{Q}{4\pi\varepsilon_0 a^2} \left( \sqrt{2}q + \frac{Q}{2} \right)$$
Final Answer: Resultant force is $\frac{Q}{4\pi\varepsilon_0 a^2} \left( \sqrt{2}q + \frac{Q}{2} \right)$ along the diagonal.

PYQ 18 SA — 3M | CBSE 2019 (53/3/1)

Topic 1

18. Five point charges, each of charge $+q$, are placed on five vertices of a regular hexagon of side $l$. Find the magnitude of the resultant force on a charge $-q$ placed at the centre of the hexagon.

Step 1: Symmetry in a Complete Regular Hexagon

In a regular hexagon of side $l$, the distance from the centre $O$ to any of the 6 vertices is exactly equal to side length $l$.
If all 6 vertices $A_1, A_2, A_3, A_4, A_5, A_6$ had equal charges $+q$, then by symmetry, the attractive forces exerted on the central charge $-q$ by opposite pairs of vertices would cancel pairwise ($A_1 \leftrightarrow A_4$, $A_2 \leftrightarrow A_5$, $A_3 \leftrightarrow A_6$), yielding a net force of zero:
$$\sum_{i=1}^6 \mathbf{F}_i = 0$$

Step 2: Effect of One Missing Charge

Let vertex $A_6$ be vacant. The resultant force of the remaining 5 charges is:
$$\mathbf{F}_{\text{net}} = \sum_{i=1}^5 \mathbf{F}_i = -\mathbf{F}_6$$
This means the net force is equal in magnitude and opposite in direction to the force that would have been exerted by a $+q$ charge at the vacant vertex $A_6$.

Step 3: Force Calculation & Direction

The charge $+q$ at $A_3$ (diametrically opposite to vacant vertex $A_6$) attracts the central charge $-q$ towards $A_3$ with magnitude:
$$F_{\text{net}} = \frac{1}{4\pi\varepsilon_0} \frac{|q(-q)|}{l^2} = \frac{q^2}{4\pi\varepsilon_0 l^2}$$
Final Answer: Force magnitude is $\frac{q^2}{4\pi\varepsilon_0 l^2}$ directed towards the vacant vertex $A_6$.

End of Topic 1

Ready for Topic 2: Electric Field & Electric Field Lines (Q19–Q30)?

SUBTOPIC 2

Topic 2: Electric Field & Electric Field Lines

12 Questions (PYQ 19–30)

PYQ 19 VSA — 1M | CBSE Recurring

Topic 2

19. Name the physical quantity whose SI unit is N/C. Is it a scalar or a vector?

Step 1: Identification of Physical Quantity

The physical quantity is Electric Field Intensity (or simply Electric Field $\mathbf{E}$).

Step 2: Nature & Formula Definition

From the definition $\mathbf{E} = \frac{\mathbf{F}}{q_0}$ (Force per unit positive test charge):
• Unit: $\frac{\text{Newton}}{\text{Coulomb}} = \text{N C}^{-1}$
• Nature: It is a vector quantity, having both magnitude and direction (the direction of force on a positive test charge).

PYQ 20 MCQ — 1M | CBSE 2025 Set 55/2/1

Topic 2

20. Two charges $-q$ each are placed at the vertices A and B of an equilateral triangle ABC. If M is the mid-point of AB, the net electric field at C will point along:

(a) CM

(b) MC

(d) CB

Correct Answer: (a) CM

Step 1: Determining Individual Electric Field Vectors at Point C

Place a positive test charge $+q_0$ at vertex $C$.
• The negative charge $-q$ at $A$ attracts $+q_0$, so the electric field $\mathbf{E}_A$ is directed along vector $CA$ (towards $A$).
• The negative charge $-q$ at $B$ attracts $+q_0$, so the electric field $\mathbf{E}_B$ is directed along vector $CB$ (towards $B$).

Step 2: Symmetry and Vector Resultant

Since triangle $ABC$ is equilateral ($AC = BC$) and charges have equal magnitudes ($|-q| = |-q|$), the magnitudes of the electric fields are equal: $|\mathbf{E}_A| = |\mathbf{E}_B| = E_0$.
The resultant of two equal vectors bisects the angle between them ($\angle ACB = 60^\circ$).

Step 3: Direction along Median CM

The line bisecting $\angle ACB$ in an equilateral triangle is the median $CM$ drawn to the midpoint $M$ of side $AB$.
Therefore, the resultant electric field points straight downwards along CM (towards $M$).

PYQ 21 MCQ — 1M | CBSE SQP Term-1 2021

Topic 2

21. Two point charges $+8q$ and $-2q$ are located at $x = 0$ and $x = L$ respectively. The point on the x-axis at which the net electric field is zero due to these charges is:

(a) $8L$

(b) $4L$

(d) $L$

Correct Answer: (c) $2L$

Step 1: Locating the Region for $E_{\text{net}} = 0$

Because the charges have opposite signs ($+8q$ and $-2q$):
• Between $x=0$ and $x=L$: Both fields point in the $+x$ direction (away from $+8q$, towards $-2q$), so $E_{\text{net}} \ne 0$.
• To the left of $x=0$ ($x < 0$): Charge $+8q$ is larger in magnitude and closer, so its field always dominates.
• To the right of $x=L$ ($x > L$): The field of smaller magnitude charge $-2q$ can be matched by the field of the larger $+8q$ charge which is farther away.
Let the neutral point be at distance $d$ to the right of $x=L$ (i.e., coordinate $x = L + d$).

Step 2: Equating Field Magnitudes

$$|\mathbf{E}_{+8q}| = |\mathbf{E}_{-2q}|$$
$$\frac{1}{4\pi\varepsilon_0} \frac{8q}{(L + d)^2} = \frac{1}{4\pi\varepsilon_0} \frac{2q}{d^2}$$
$$\frac{8}{(L + d)^2} = \frac{2}{d^2} \implies \frac{4}{(L + d)^2} = \frac{1}{d^2}$$

Step 3: Solving for Position Coordinate $x$

Taking square root on both sides:
$$\frac{2}{L + d} = \frac{1}{d} \implies 2d = L + d \implies d = L$$
Coordinate on the x-axis:
$$x = L + d = L + L = 2L$$
Hence, option (c) $2L$ is the correct answer.

PYQ 22 MCQ — 1M | CBSE 2024 Delhi Set 1

Topic 2

22. The magnitude of the electric field due to a point charge object at a distance of $4.0\text{ m}$ is $9\text{ N/C}$. From the same charged object, the electric field of magnitude $16\text{ N/C}$ will be at a distance of:

(A) $1.5\text{ m}$

(B) $2.0\text{ m}$

(D) $6.0\text{ m}$

Correct Answer: (C) $3.0\text{ m}$

Step 1: Inverse-Square Relationship

The electric field intensity due to a point charge $Q$ in air is given by:
$$E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} \implies E \propto \frac{1}{r^2} \implies E_1 r_1^2 = E_2 r_2^2$$

Step 2: Substituting Given Values

Given:
• $E_1 = 9\text{ N/C}$ at $r_1 = 4.0\text{ m}$
• $E_2 = 16\text{ N/C}$ at distance $r_2$

Step 3: Calculating $r_2$

$$9 \times (4.0)^2 = 16 \times r_2^2$$
$$9 \times 16 = 16 \times r_2^2 \implies r_2^2 = 9 \implies r_2 = 3.0\text{ m}$$
Hence, option (C) $3.0\text{ m}$ is the correct answer.

PYQ 23 MCQ — 1M | CBSE 2026 Set 55/5/2

Topic 2

23. Two point charges $-Q$ and $+Q$ are located at points $(d, 0)$ and $(0, d)$ respectively in the x-y plane. The electric field $\mathbf{E}$ at the origin will be:

(A) $\frac{Q\sqrt{2}}{4\pi\varepsilon_0 d^2} (\hat{\mathbf{i}} - \hat{\mathbf{j}})$

(B) $\frac{Q\sqrt{2}}{4\pi\varepsilon_0 d^2} (-\hat{\mathbf{i}} - \hat{\mathbf{j}})$

(D) $\frac{Q}{4\pi\varepsilon_0 d^2} (-\hat{\mathbf{i}} + \hat{\mathbf{j}})$

Correct Answer: (C) $\frac{Q}{4\pi\varepsilon_0 d^2} (\hat{\mathbf{i}} - \hat{\mathbf{j}})$

Step 1: Field at Origin due to $-Q$ at $(d, 0)$

Distance from origin is $d$. Since the charge is negative, it attracts a positive test charge towards itself (along $+x$ direction):
$$\mathbf{E}_1 = +\frac{1}{4\pi\varepsilon_0} \frac{Q}{d^2} \hat{\mathbf{i}}$$

Step 2: Field at Origin due to $+Q$ at $(0, d)$

Distance from origin is $d$. Since the charge is positive, it repels a positive test charge away from itself (along $-y$ direction):
$$\mathbf{E}_2 = -\frac{1}{4\pi\varepsilon_0} \frac{Q}{d^2} \hat{\mathbf{j}}$$

Step 3: Vector Addition of Fields at Origin

$$\mathbf{E}_{\text{net}} = \mathbf{E}_1 + \mathbf{E}_2 = \frac{Q}{4\pi\varepsilon_0 d^2} (\hat{\mathbf{i}} - \hat{\mathbf{j}})$$
Hence, option (C) $\frac{Q}{4\pi\varepsilon_0 d^2} (\hat{\mathbf{i}} - \hat{\mathbf{j}})$ is correct.

PYQ 24 SA — 3M | CBSE 2016

Topic 2

24. A charge is distributed uniformly over a ring of radius $a$. Obtain an expression for the electric intensity $E$ at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge.

Step 1: Setup & Element Charge $dq$

Consider a circular ring of radius $a$ carrying total charge $Q$ uniformly distributed. Linear charge density $\lambda = \frac{Q}{2\pi a}$.
Let point $P$ be on the axis at distance $x$ from the centre $O$.
Distance of any infinitesimal element $dl$ (carrying charge $dq = \lambda dl$) from point $P$ is $r = \sqrt{a^2 + x^2}$.

Step 2: Resolving Components & Integrating along Axis

Magnitude of field due to element $dq$:
$$dE = \frac{1}{4\pi\varepsilon_0} \frac{dq}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{dq}{a^2 + x^2}$$
Resolving $dE$:
• Perpendicular components $dE_\perp = dE\sin\theta$ cancel out pairwise for diametrically opposite elements.
• Axial components $dE_\parallel = dE\cos\theta$ add up in the same direction along the axis.
From geometry, $\cos\theta = \frac{x}{r} = \frac{x}{\sqrt{a^2 + x^2}}$.
$$E = \int dE_\parallel = \int \frac{1}{4\pi\varepsilon_0} \frac{dq}{a^2 + x^2} \left( \frac{x}{\sqrt{a^2 + x^2}} \right) = \frac{x}{4\pi\varepsilon_0 (a^2 + x^2)^{3/2}} \int dq$$
Since $\int dq = Q$:
$$E = \frac{1}{4\pi\varepsilon_0} \frac{Q x}{(a^2 + x^2)^{3/2}}$$

Step 3: Limiting Case for Large Distance ($x \gg a$)

For points at large distances ($x \gg a$), the radius $a$ can be neglected in comparison with $x$ ($a^2 + x^2 \approx x^2$):
$$E \approx \frac{1}{4\pi\varepsilon_0} \frac{Q x}{(x^2)^{3/2}} = \frac{1}{4\pi\varepsilon_0} \frac{Q x}{x^3} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{x^2}$$
This is identical to the electrostatic field produced by a point charge $Q$. Hence, at large distances, the charged ring behaves exactly like a point charge.

PYQ 25 SA — 2M | CBSE Recurring

Topic 2

25. Why do two electric field lines never intersect each other?

Step 1: Tangent Definition of Field Direction

By definition, the tangent drawn to an electric field line at any point gives the unique direction of the resultant electric field vector $\mathbf{E}$ at that point.

Step 2: Contradiction at Intersection Point

If two electric field lines were to intersect at a common point, two distinct tangents could be drawn at the intersection point. This would imply two different directions for the resultant electric field at that single point in space, which is physically impossible.

PYQ 26 SA — 2M | CBSE Recurring

Topic 2

26. State three important properties of electric field lines.

Property 1 (Continuity & Polarity)

Electric field lines are continuous curves originating from positive charges and terminating at negative charges. They do not form closed loops because electrostatic fields are conservative.

Property 2 (Direction of Field)

The tangent drawn at any point on an electric field line gives the direction of the electric field intensity $\mathbf{E}$ at that point.

Property 3 (Non-Intersection & Conductor Boundary)

Two electric field lines never intersect each other, and they always enter or leave the surface of a charged conductor normally ($90^\circ$).

PYQ 27 SA — 2M | CBSE 2019 (55/2/3)

Topic 2

27. Sketch the electric field lines for two point charges $q_1$ and $q_2$ for (i) $q_1 = q_2$ and (ii) $q_1 > q_2$, separated by distance $d$.

Step 1: Case (i) $q_1 = q_2$ (Equal Positive Charges)

Field lines diverge symmetrically outward from both charges. Due to lateral repulsion, lines bend away from each other. The neutral point $N$ ($E = 0$) is located exactly at the geometric midpoint ($d/2$) on the line joining the two charges.

Step 2: Case (ii) $q_1 > q_2$ (Unequal Positive Charges)

The density of electric field lines emerging from $q_1$ is higher than from $q_2$ (since number of lines $\propto |q|$). The neutral point $N$ ($E = 0$) shifts closer to the smaller charge $q_2$.

PYQ 28 SA — 2M | CBSE 2019

Topic 2

28. Draw the pattern of electric field lines when a point charge $-Q$ is kept near an uncharged conducting plate.

Step 1: Electrostatic Induction on Plate

The negative point charge $-Q$ attracts free electrons away from the near face towards the far face of the metal plate, inducing a positive charge density on the near surface facing $-Q$.

Step 2: Field Line Pattern & Boundary Conditions

Electric field lines originate perpendicularly ($90^\circ$) from the positive induced charges on the flat conducting plate and converge radially inwards onto the negative point charge $-Q$.

PYQ 29 MCQ A-R — 1M | CBSE 2025 Delhi Set 1

Topic 2

29. Assertion (A): Equal amounts of positive and negative charges are distributed uniformly on two halves of a thin circular ring. The resultant electric field at the centre O of the ring is along OC.
Reason (R): It is so because the net potential at O is not zero.

(A) Both A and R are true and R is the correct explanation of A.

(B) Both A and R are true but R is not the correct explanation of A.

(D) Both A and R are false.

Correct Answer: (C) A is true but R is false.

Step 1: Evaluating Assertion (A)

The upper half has charge $+q$ (repels away from it) and the lower half has charge $-q$ (attracts towards it). Both halves produce electric fields pointing in the same downward direction along the perpendicular bisector $OC$. Hence, Assertion (A) is TRUE.

Step 2: Evaluating Reason (R)

Electric potential $V$ is a scalar quantity. The total potential at the centre $O$ is:
$$V = V_+ + V_- = \frac{1}{4\pi\varepsilon_0}\frac{+q}{R} + \frac{1}{4\pi\varepsilon_0}\frac{-q}{R} = 0$$
The net potential at $O$ is strictly ZERO. Hence, Reason (R) is FALSE.

Step 3: Conclusion

Assertion is true but Reason is false. Correct option is (C).

PYQ 30 Case Study — 4M | CBSE 2023

Topic 2

30. Case Study — Electron Beam Between Charged Plates: An electron beam is directed horizontally between two parallel charged plates of length $L$ (upper plate positive, lower plate negative) creating a uniform vertical electric field $E$.
(i) What is the direction of electric field $E$ between the plates?
(ii) Write the expression for the electric force acting on the electron.
(iii) What is the acceleration of the electron?
(iv) If $E = 2.0 \times 10^4\text{ N/C}$, calculate the magnitude and direction of the electrostatic force on the electron.

Step 1: (i) Direction of Electric Field

The electric field always points from the region of higher potential (positive plate) to lower potential (negative plate). Thus, $\mathbf{E}$ is directed vertically downwards.

Step 2: (ii) Expression for Force

$$\mathbf{F} = q \mathbf{E} = -e \mathbf{E} \implies F = e E$$

Step 3: (iii) Acceleration of Electron

By Newton's second law ($F = m_e a$):
$$a = \frac{e E}{m_e}$$
Since the electron carries a negative charge, the force and acceleration are directed vertically upwards towards the positive plate (opposite to the direction of $\mathbf{E}$).

Step 4: (iv) Numerical Calculation

Given $e = 1.6 \times 10^{-19}\text{ C}$ and $E = 2.0 \times 10^4\text{ N/C}$:
$$F = e E = (1.6 \times 10^{-19}\text{ C}) \times (2.0 \times 10^4\text{ N/C}) = 3.2 \times 10^{-15}\text{ N}$$
Direction: Vertically upwards towards the positive plate.

End of Topic 2

Ready for Topic 3: Electric Dipole, Electric Field & Torque (Q31–Q39)?

SUBTOPIC 3

Topic 3: Electric Dipole, Electric Field & Torque

9 Questions (PYQ 31–39)

PYQ 31 VSA — 1M | CBSE Recurring

Topic 3

31. Name the physical quantity whose SI unit is C·m.

Step 1: Identification

The physical quantity is Electric Dipole Moment ($\mathbf{p}$).

Step 2: Dimensional Check

$$\mathbf{p} = q(2\mathbf{a}) \implies \text{Unit} = [\text{Coulomb}] \times [\text{metre}] = \text{C}\cdot\text{m}$$

PYQ 32 MCQ — 1M | CBSE 2020 Covid

Topic 3

32. The electric field at a point on the equatorial plane at a distance $r$ from the centre of a dipole having dipole moment $p$ is given by ($r \gg 2a$):

(A) $p/4\pi\varepsilon_0 r^3$

(B) $2p/4\pi\varepsilon_0 r^3$

(D) $p/4\pi\varepsilon_0 r^2$

Correct Answer: (C) $-p/4\pi\varepsilon_0 r^3$

Step 1: Vector Formula for Equatorial Field

For a short electric dipole of dipole moment vector $\mathbf{p}$ (directed from $-q$ to $+q$), the electric field on its equatorial line at distance $r$ is directed antiparallel to $\mathbf{p}$:
$$\mathbf{E}_{\text{eq}} = -\frac{1}{4\pi\varepsilon_0} \frac{\mathbf{p}}{r^3}$$

Step 2: Conclusion

The negative sign indicates that the electric field direction is opposite to the dipole moment vector. Hence, (C) $-p/4\pi\varepsilon_0 r^3$ is the correct answer.

PYQ 33 VSA — 1M | CBSE Recurring

Topic 3

33. Define electric dipole moment. Is it a scalar or vector quantity? Write its SI unit.

Step 1: Definition of Dipole Moment

The electric dipole moment of an electric dipole is defined as the product of the magnitude of either charge $q$ and the vector distance $2\mathbf{a}$ separating them:
$$\mathbf{p} = q(2\mathbf{a})$$

Step 2: Vector Nature & Direction

It is a vector quantity. By convention in physics, its direction points from the negative charge ($-q$) to the positive charge ($+q$) along the dipole axis.

Step 3: SI Unit

The SI unit of electric dipole moment is $\text{Coulomb}\cdot\text{metre} (\text{C}\cdot\text{m})$.

PYQ 34 SA — 3M | CBSE 2013

Topic 3

34. Two small identical electrical dipoles AB and CD, each of dipole moment $p$, are kept at an angle of $120^\circ$. What is the resultant dipole moment of this combination? If this system is subjected to a uniform external field $\mathbf{E}$, what is the net torque acting on it?

Step 1: Vector Addition of Dipole Moments

Let $\mathbf{p}_1$ and $\mathbf{p}_2$ be the dipole moments of $AB$ and $CD$ respectively. Given $|mathbf{p}_1| = |mathbf{p}_2| = p$ and angle between them $\theta = 120^\circ$.
Resultant dipole moment magnitude:
$$p_{\text{res}} = \sqrt{p_1^2 + p_2^2 + 2 p_1 p_2 \cos 120^\circ} = \sqrt{p^2 + p^2 + 2 p^2 \left(-\frac{1}{2}\right)} = \sqrt{2p^2 - p^2} = \sqrt{p^2} = p$$

Step 2: Direction of Resultant Dipole Moment

By symmetry, the resultant dipole moment vector $\mathbf{p}_{\text{res}}$ bisects the $120^\circ$ angle, making an angle of $60^\circ$ with either dipole.

Step 3: Torque on the Combination in Uniform Field $\mathbf{E}$

The net torque acting on the system in a uniform external field $\mathbf{E}$ inclined at angle $\alpha$ with $\mathbf{p}_{\text{res}}$ is:
$$\boldsymbol{\tau} = \mathbf{p}_{\text{res}} \times \mathbf{E} \implies |\boldsymbol{\tau}| = p E \sin\alpha$$
Maximum torque occurs when $\alpha = 90^\circ$ ($ heta_{\text{max}} = pE$).

PYQ 35 LA — 5M | CBSE 2017; Recurring

Topic 3

35. (a) Derive expression for electric field $E$ due to dipole on axial line.
(b) Draw graph of $E$ versus $r$ for $r \gg a$.
(c) Stable and unstable equilibrium positions and torque in uniform field.

Step 1: (a) Derivation of Axial Electric Field

Consider an electric dipole consisting of charges $-q$ and $+q$ separated by distance $2a$. Center is $O$.
Let $P$ be a point on the axial line at distance $r$ from center $O$.
• Distance of $P$ from $+q$: $r_1 = r - a$
• Distance of $P$ from $-q$: $r_2 = r + a$
Electric field due to $+q$ (repulsive, along $OP$): $E_+ = \frac{1}{4\pi\varepsilon_0}\frac{q}{(r - a)^2}$
Electric field due to $-q$ (attractive, along $PO$): $E_- = \frac{1}{4\pi\varepsilon_0}\frac{q}{(r + a)^2}$
Resultant axial field:
$$E_{\text{axial}} = E_+ - E_- = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{(r - a)^2} - \frac{1}{(r + a)^2} \right] = \frac{q}{4\pi\varepsilon_0} \left[ \frac{(r + a)^2 - (r - a)^2}{(r^2 - a^2)^2} \right]$$
$$E_{\text{axial}} = \frac{q}{4\pi\varepsilon_0} \left[ \frac{4ar}{(r^2 - a^2)^2} \right] = \frac{1}{4\pi\varepsilon_0} \frac{2(q \cdot 2a)r}{(r^2 - a^2)^2} = \frac{1}{4\pi\varepsilon_0} \frac{2pr}{(r^2 - a^2)^2}$$
For short dipole ($r \gg a$):
$$E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^3} \quad \text{(along }\mathbf{p}\text{)}$$

Step 2: (b) Graph of $E$ vs $r$ for $r \gg a$

Since $E_{\text{axial}} \propto \frac{1}{r^3}$, the graph is a smooth curve decaying asymptotically towards zero much steeper than the $1/r^2$ point-charge field curve.

Step 3: (c) Equilibrium States in Uniform External Field

Torque on dipole: $\tau = pE\sin\theta$, Potential Energy: $U = -pE\cos\theta$.
• Stable Equilibrium: $\theta = 0^\circ$ (Dipole moment $\mathbf{p}$ aligned parallel to $\mathbf{E}$). Here $\tau = 0$ and $U = -pE$ (Minimum potential energy).
• Unstable Equilibrium: $\theta = 180^\circ$ (Dipole moment $\mathbf{p}$ antiparallel to $\mathbf{E}$). Here $\tau = 0$ and $U = +pE$ (Maximum potential energy).

PYQ 36 LA — 5M | CBSE 2025; 2026 Set 55/1/1; Recurring

Topic 3

36. (a) Derive equatorial field $E$ of a dipole.
(b) Show that for $r \gg a$, equatorial field is half of axial field.

Step 1: (a) Derivation of Equatorial Electric Field

Consider an electric dipole of length $2a$ with charges $-q$ and $+q$. Let $P$ be a point on the equatorial line at distance $r$ from center $O$.
Distance of $P$ from either charge is $\sqrt{r^2 + a^2}$.
Magnitudes of fields due to $+q$ and $-q$ are equal:
$$E_1 = E_2 = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + a^2}$$
Resolving components:
• Vertical components $E_1\sin\theta$ and $E_2\sin\theta$ are equal and opposite, cancelling each other.
• Horizontal components $E_1\cos\theta$ and $E_2\cos\theta$ add up parallel to the dipole axis (opposite to $\mathbf{p}$):
$$E_{\text{eq}} = 2 E_1 \cos\theta = 2 \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + a^2} \right) \cos\theta$$
From triangle, $\cos\theta = \frac{a}{\sqrt{r^2 + a^2}}$:
$$E_{\text{eq}} = \frac{1}{4\pi\varepsilon_0} \frac{2qa}{(r^2 + a^2)^{3/2}} = \frac{1}{4\pi\varepsilon_0} \frac{p}{(r^2 + a^2)^{3/2}}$$
For short dipole ($r \gg a$):
$$E_{\text{eq}} = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3}$$

Step 2: (b) Comparing Axial and Equatorial Fields

For a short dipole at the same distance $r$ ($r \gg a$):
$$E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^3}$$
$$E_{\text{eq}} = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3}$$
Dividing the two expressions:
$$\frac{E_{\text{eq}}}{E_{\text{axial}}} = \frac{\frac{p}{4\pi\varepsilon_0 r^3}}{\frac{2p}{4\pi\varepsilon_0 r^3}} = \frac{1}{2} \implies E_{\text{eq}} = \frac{1}{2} E_{\text{axial}}$$ (Hence proved).

PYQ 37 MCQ — 1M | CBSE SQP 2023

Topic 3

37. An electric dipole placed in an electric field of intensity $2 \times 10^5\text{ N/C}$ at an angle of $30^\circ$ experiences a torque equal to $4\text{ Nm}$. The charge on the dipole of dipole length $2\text{ cm}$ is:

(a) $7\;\mu\text{C}$

(b) $8\text{ mC}$

(d) $5\text{ mC}$

Correct Answer: (c) $2\text{ mC}$

Step 1: Formula for Torque on Dipole

$$\tau = p E \sin\theta = (q \cdot 2a) E \sin\theta$$

Step 2: Substituting Given Values

• $\tau = 4\text{ Nm}$
• $E = 2 \times 10^5\text{ N/C}$
• $\theta = 30^\circ \implies \sin 30^\circ = 0.5$
• Dipole length $2a = 2\text{ cm} = 0.02\text{ m}$

Step 3: Calculating Charge $q$

$$4 = q \times (0.02) \times (2 \times 10^5) \times 0.5$$
$$4 = q \times 2000 \implies q = \frac{4}{2000} = 2 \times 10^{-3}\text{ C} = 2\text{ mC}$$
Hence, option (c) $2\text{ mC}$ is the correct answer.

PYQ 38 LA — 5M | CBSE 2023

Topic 3

38. (a) Define dipole moment.
(b) Derive torque on dipole in uniform field.
(c) When is torque maximum and zero?
(d) Dipole of length $1\text{ cm}$ at $60^\circ$ in $10^5\text{ N/C}$ experiences torque $6\sqrt{3}\times 10^{-2}\text{ Nm}$. Find charge.

Step 1: (a) Definition of Dipole Moment

The electric dipole moment $\mathbf{p}$ is the product of magnitude of either charge $q$ and distance vector $2\mathbf{a}$ separating them: $\mathbf{p} = q(2\mathbf{a})$ directed from $-q$ to $+q$.

Step 2: (b) Derivation of Torque

Forces on charges $+q$ and $-q$ in uniform field $\mathbf{E}$ are $+q\mathbf{E}$ and $-q\mathbf{E}$.
Net translatory force $= q\mathbf{E} + (-q\mathbf{E}) = 0$.
These two equal, opposite, and non-collinear forces form a couple. Perpendicular distance between lines of action is $d = 2a\sin\theta$.
$$\tau = \text{Force} \times \text{Perpendicular distance} = (qE)(2a\sin\theta) = (q\cdot 2a)E\sin\theta = pE\sin\theta$$
In vector form: $\boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}$.

Step 3: (c) Maximum and Zero Torque Conditions

• Maximum Torque: Occurs when $\theta = 90^\circ$ ($ heta = 270^\circ$) $\implies \tau_{\text{max}} = pE$.
• Zero Torque: Occurs when $\theta = 0^\circ$ (aligned) or $\theta = 180^\circ$ (opposed) $\implies \tau = 0$.

Step 4: (d) Numerical Calculation

$$\tau = q(2a)E\sin 60^\circ$$
$$6\sqrt{3}\times 10^{-2} = q(0.01)(10^5)\left(\frac{\sqrt{3}}{2}\right) = q(1000)\left(\frac{\sqrt{3}}{2}\right) = 500\sqrt{3} q$$
$$q = \frac{6\sqrt{3}\times 10^{-2}}{500\sqrt{3}} = \frac{6 \times 10^{-2}}{500} = 1.2 \times 10^{-4}\text{ C} = 120\;\mu\text{C}$$

PYQ 39 Case Study — 4M | CBSE 2025

Topic 3

39. Case Study — Electric Dipole in Uniform Field:
(i) SI unit of $p$?
(ii) Torque expression and when max/zero?
(iii) Stable vs unstable orientations?
(iv) Work done rotating dipole of $p=10^{-7}\text{ Cm}$ from $0^\circ$ to $180^\circ$ in $E=10^4\text{ N/C}$.

Step 1: (i) SI Unit of Dipole Moment

$\text{Coulomb}\cdot\text{metre} (\text{C}\cdot\text{m})$.

Step 2: (ii) Torque Expression & Conditions

$\tau = pE\sin\theta$. Maximum at $\theta = 90^\circ$ ($ heta_{\text{max}} = pE$), Zero at $\theta = 0^\circ$ and $\theta = 180^\circ$.

Step 3: (iii) Stable vs Unstable Equilibrium

• Stable: $\theta = 0^\circ$ (Dipole aligned with field; minimum energy $U = -pE$).
• Unstable: $\theta = 180^\circ$ (Dipole antiparallel to field; maximum energy $U = +pE$).

Step 4: (iv) Work Done in Rotating Dipole

$$W = -pE(\cos\theta_2 - \cos\theta_1) = -pE(\cos 180^\circ - \cos 0^\circ) = -pE(-1 - 1) = 2pE$$
$$W = 2 \times (10^{-7}\text{ C m}) \times (10^4\text{ N/C}) = 2 \times 10^{-3}\text{ Joules}$$

End of Topic 3

Ready for Topic 4: Electric Flux & Gauss's Law (Q40–Q56)?

SUBTOPIC 4

Topic 4: Electric Flux & Gauss's Law

17 Questions (PYQ 40–56)

PYQ 40 SA — 2M | CBSE Foreign 2016

Topic 4

40. A spherical rubber balloon carries a charge that is uniformly distributed over its surface. As the balloon is blown up and increases in size, how does the total electric flux coming out of the surface change? Give reason for your answer.

Step 1: Direct Statement

The total electric flux coming out of the balloon surface remains completely unchanged.

Step 2: Reason via Gauss's Law

According to Gauss's Law, the total electric flux through any closed Gaussian surface enclosing charge $q_{\text{enclosed}}$ is:
$$\Phi_E = \oint \mathbf{E}\cdot d\mathbf{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$
As the balloon is inflated, the total electric charge $q$ distributed on its surface remains conserved. Since the enclosed charge does not change and Gauss's law is independent of the radius or geometry of the surface, the total outward electric flux remains constant.

PYQ 41 MCQ — 1M | CBSE 2022

Topic 4

41. A cylinder of radius $R$ and length $L$ is placed in a uniform electric field $E$ parallel to the cylinder axis. The total flux for the surface of the cylinder is given by:

(a) zero

(b) $\pi R^2 E$

(d) can be zero or $+\pi R^2 E$

Correct Answer: (a) zero

Step 1: Flux through Curved Surface

The electric field $\mathbf{E}$ is parallel to the cylinder axis. For the curved surface, the area vector $d\mathbf{A}$ is radially perpendicular to $\mathbf{E}$ ($\theta = 90^\circ$):
$$\Phi_{\text{curved}} = \int E dA \cos 90^\circ = 0$$

Step 2: Flux through Circular Flat End Faces

• Left flat face: Normal area vector points outward to the left (opposite to $\mathbf{E}$, $\theta = 180^\circ$): $\Phi_1 = E(\pi R^2)\cos 180^\circ = -E\pi R^2$
• Right flat face: Normal area vector points outward to the right (parallel to $\mathbf{E}$, $\theta = 0^\circ$): $\Phi_2 = E(\pi R^2)\cos 0^\circ = +E\pi R^2$

Step 3: Total Closed Flux

$$\Phi_{\text{total}} = \Phi_1 + \Phi_2 + \Phi_{\text{curved}} = -E\pi R^2 + E\pi R^2 + 0 = 0$$
Alternatively, since no charge is enclosed inside the cylinder ($q_{\text{enc}} = 0$), by Gauss's Law $\Phi = \frac{q_{\text{enc}}}{\varepsilon_0} = 0$. Hence (a) zero is correct.

PYQ 42 MCQ — 1M | CBSE 2022

Topic 4

42. A square surface of side $L$ metres is in the plane of paper. Electric field $E$ is limited only to the lower half of the square surface. The electric flux (in SI units) associated with the surface is:

(a) $EL^2/2$

(b) $EL^2$

(d) $EL^2/4$

Correct Answer: (c) zero

Step 1: Area Vector Direction

For a flat planar surface in the plane of paper, its area vector $d\mathbf{A}$ is directed perpendicular (normal) to the plane of the paper (out of or into the page).

Step 2: Angle between Electric Field and Area Vector

Given that the electric field $\mathbf{E}$ is in the plane of the paper, the angle between $\mathbf{E}$ and the surface normal $d\mathbf{A}$ is $\theta = 90^\circ$.

Step 3: Electric Flux Calculation

$$\Phi = \int \mathbf{E}\cdot d\mathbf{A} = E A \cos 90^\circ = 0$$
Hence, option (c) zero is correct.

PYQ 43 MCQ — 1M | CBSE 2019

Topic 4

43. A sphere encloses an electric dipole with charges $\pm 3 \times 10^{-6}\text{ C}$. What is the total electric flux across the sphere?

(a) $-3 \times 10^{-6}\text{ Nm}^2/\text{C}$

(b) zero

(d) $6 \times 10^{-6}\text{ Nm}^2/\text{C}$

Correct Answer: (b) zero

Step 1: Enclosed Charge Calculation

An electric dipole consists of equal and opposite charges $+q$ and $-q$. The total net charge enclosed within the sphere is:
$$q_{\text{enclosed}} = (+3 \times 10^{-6}\text{ C}) + (-3 \times 10^{-6}\text{ C}) = 0$$

Step 2: Gauss's Law Application

$$\Phi_E = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{0}{\varepsilon_0} = 0$$
Hence, option (b) zero is correct.

PYQ 44 VSA — 1M | CBSE Recurring

Topic 4

44. What is the electric flux through a cube of side $a$ if a charge $q$ is placed (i) at its centre, and (ii) at one of its corners?

Step 1: (i) Charge $q$ at Centre of Cube

The entire charge $q$ is completely enclosed by the 6 faces of the cube. By Gauss's Law:
$$\Phi_{\text{total}} = \frac{q}{\varepsilon_0}$$
(Flux through each single face $= \frac{q}{6\varepsilon_0}$).

Step 2: (ii) Charge $q$ at One Corner of Cube

A charge placed at a corner of a cube is shared equally by 8 identical symmetrical cubes meeting at that corner to completely enclose it.
Hence, total flux through one such cube is:
$$\Phi_{\text{cube}} = \frac{1}{8} \left( \frac{q}{\varepsilon_0} \right) = \frac{q}{8\varepsilon_0}$$

PYQ 45 VSA — 1M | CBSE Recurring

Topic 4

45. Define electric flux. Write its SI unit. Is it a scalar or a vector?

Step 1: Definition of Electric Flux

Electric flux ($Phi_E$) through a given surface is defined as the total number of electric field lines passing normally through that surface area:
$$\Phi_E = \int_S \mathbf{E}\cdot d\mathbf{A}$$

Step 2: SI Unit & Scalar Nature

• SI Unit: $\text{N}\cdot\text{m}^2\text{C}^{-1}$ (or $\text{Volt}\cdot\text{metre}, \text{V}\cdot\text{m}$).
• Nature: It is a scalar quantity because it is the dot product of two vectors.

PYQ 46 VSA — 1M | CBSE 2016

Topic 4

46. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected if the radius of the Gaussian surface is doubled?

Step 1: Direct Statement

The electric flux remains completely unchanged.

Step 2: Reason from Gauss's Law

By Gauss's Law, $\Phi = \frac{q_{\text{enc}}}{\varepsilon_0}$. The total electric flux depends solely on the amount of charge enclosed within the surface, and is completely independent of the size, radius, or shape of the Gaussian surface.

PYQ 47 Numerical — 2–3M | CBSE 2014

Topic 4

47. Given uniform field $\mathbf{E} = 5 \times 10^3\,\hat{\mathbf{i}}\text{ N/C}$, find flux through a $10\text{ cm}$ square: (a) plane parallel to y-z plane, (b) normal makes $30^\circ$ with x-axis. What if plane makes $60^\circ$ with x-axis?

Step 1: Surface Area of Square

$$A = (10\text{ cm})^2 = (0.1\text{ m})^2 = 0.01\text{ m}^2 = 10^{-2}\text{ m}^2$$

Step 2: (a) Square Plane Parallel to y-z Plane

Area normal vector $\hat{\mathbf{n}}$ points along $+x$ axis ($\hat{\mathbf{i}}$), so angle $\theta = 0^\circ$:
$$\Phi_a = E A \cos 0^\circ = (5 \times 10^3\text{ N/C}) \times (0.01\text{ m}^2) \times 1 = 50\text{ N m}^2\text{C}^{-1}$$

Step 3: (b) Normal Makes $30^\circ$ with x-axis & Plane at $60^\circ$

• When normal makes $30^\circ$ with x-axis ($\theta = 30^\circ$):
$$\Phi_b = E A \cos 30^\circ = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \approx 43.3\text{ N m}^2\text{C}^{-1}$$
• When plane of square makes $60^\circ$ with x-axis, the normal to the plane makes angle $\theta = 90^\circ - 60^\circ = 30^\circ$ with x-axis. The flux remains identical: $43.3\text{ N m}^2\text{C}^{-1}$.

PYQ 48 SA — 2M | CBSE 2016

Topic 4

48. Two charges of magnitudes $-2Q$ and $+Q$ are located at points $(a, 0)$ and $(4a, 0)$ respectively. What is the electric flux due to these charges through a sphere of radius $3a$ with its centre at the origin?

Step 1: Identifying Charges Enclosed by Gaussian Sphere

The Gaussian sphere is centered at origin $(0,0)$ and has radius $R = 3a$.
• Position of charge $-2Q$ is $(a, 0)$. Distance from origin $= a < 3a$ (Inside the sphere).
• Position of charge $+Q$ is $(4a, 0)$. Distance from origin $= 4a > 3a$ (Outside the sphere).

Step 2: Applying Gauss's Law

By Gauss's Law, only charges inside the closed surface contribute to total electric flux:
$$q_{\text{enclosed}} = -2Q$$
$$\Phi_E = \frac{q_{\text{enclosed}}}{\varepsilon_0} = -\frac{2Q}{\varepsilon_0}$$
Final Answer: Total electric flux is $-\frac{2Q}{\varepsilon_0}$.

PYQ 49 SA — 3M | CBSE 2016

Topic 4

49. S₁ and S₂ are two hollow concentric thin spherical shells enclosing charges Q and 2Q respectively.
(A) What is the ratio of the electric flux through S₁ and S₂?
(B) How will the electric flux through shell S₁ change if a medium of dielectric constant K is introduced in the space inside S₁ in place of air? Give reason.

Step 1: (A) Ratio of Flux through S₁ and S₂

• Charge enclosed by inner shell $S_1$: $q_1 = Q \implies \Phi_1 = \frac{Q}{\varepsilon_0}$
• Charge enclosed by outer shell $S_2$: $q_2 = Q + 2Q = 3Q \implies \Phi_2 = \frac{3Q}{\varepsilon_0}$
$$\text{Ratio } \frac{\Phi_1}{\Phi_2} = \frac{Q/\varepsilon_0}{3Q/\varepsilon_0} = \frac{1}{3} \implies \Phi_1 : \Phi_2 = 1 : 3$$

Step 2: (B) Effect of Dielectric Medium in S₁

When a medium of dielectric constant $K$ is filled inside shell $S_1$, the absolute permittivity of the medium becomes $\varepsilon = K\varepsilon_0$.
The new electric flux through $S_1$ becomes:
$$\Phi_1' = \frac{Q}{\varepsilon} = \frac{Q}{K\varepsilon_0} = \frac{\Phi_1}{K}$$
Reason: Due to dielectric polarization, bound charges reduce the net electric field at the surface, reducing the outward electric flux by a factor of $K$.

PYQ 50 Numerical — 3M | CBSE 2022

Topic 4

50. A hollow cylindrical box of length $1\text{ m}$ and area of cross-section $25\text{ cm}^2$ is placed in a coordinate system from $x=1\text{ m}$ to $x=2\text{ m}$. $E = 50x\,\hat{\mathbf{i}}$. Find net flux and enclosed charge.

Step 1: Electric Field and Area Vectors on Faces

Cross-sectional area $A = 25\text{ cm}^2 = 25 \times 10^{-4}\text{ m}^2$. Field $\mathbf{E} = 50x\,\hat{\mathbf{i}}$ points along $+x$.
• Flux through curved side $= 0$ ($\mathbf{E} \perp d\mathbf{A}$).

Step 2: Flux through Left Face ($x = 1\text{ m}$) and Right Face ($x = 2\text{ m}$)

• At left face ($x = 1\text{ m}$): $E_L = 50(1) = 50\text{ N/C}$, Area vector $\mathbf{A}_L = -A\hat{\mathbf{i}}$:
$$\Phi_L = E_L A \cos 180^\circ = -50 \times (25 \times 10^{-4}) = -0.125\text{ N m}^2\text{C}^{-1}$$
• At right face ($x = 2\text{ m}$): $E_R = 50(2) = 100\text{ N/C}$, Area vector $\mathbf{A}_R = +A\hat{\mathbf{i}}$:
$$\Phi_R = E_R A \cos 0^\circ = +100 \times (25 \times 10^{-4}) = +0.250\text{ N m}^2\text{C}^{-1}$$

Step 3: Net Flux & Enclosed Charge Calculation

$$\Phi_{\text{net}} = \Phi_R + \Phi_L = +0.250 - 0.125 = 0.125\text{ N m}^2\text{C}^{-1}$$
By Gauss's Law ($q = \varepsilon_0 \Phi_{\text{net}}$):
$$q = (8.854 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2}) \times 0.125 = 1.107 \times 10^{-12}\text{ C} = 1.11\text{ pC}$$
Final Answer: Net flux $= 0.125\text{ N m}^2\text{C}^{-1}$, Enclosed charge $= 1.11\text{ pC}$.

PYQ 51 SA — 3M | CBSE 2024

Topic 4

51. A point charge causes an electric flux of $-1.0 \times 10^3\text{ Nm}^2/\text{C}$ to pass through a spherical Gaussian surface of $10.0\text{ cm}$ radius centred on the charge.
(i) If radius doubled, how much flux passes?
(ii) Value of point charge?

Step 1: (i) Effect of Doubling Radius

By Gauss's Law, $\Phi_E = \frac{q}{\varepsilon_0}$. The flux through a closed surface depends only on the enclosed charge, not on the radius of the sphere.
Therefore, if the radius is doubled, the electric flux remains $-1.0 \times 10^3\text{ N m}^2\text{C}^{-1}$ (Unchanged).

Step 2: (ii) Calculating Value of Point Charge $q$

$$q = \varepsilon_0 \Phi_E$$
$$q = (8.854 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2}) \times (-1.0 \times 10^3\text{ N m}^2\text{C}^{-1})$$
$$q = -8.854 \times 10^{-9}\text{ C} = -8.85\text{ nC}$$
Final Answer: The point charge is $-8.85\text{ nC}$.

PYQ 52 SA — 2M | CBSE 2019

Topic 4

52. Measurement of field at surface of black box gives net outward flux $8.0 \times 10^3\text{ Nm}^2/\text{C}$.
(a) Net charge inside?
(b) If net flux were zero, could you conclude there were no charges inside?

Step 1: (a) Net Charge Inside the Box

By Gauss's Law:
$$q_{\text{net}} = \varepsilon_0 \Phi = (8.854 \times 10^{-12}) \times (8.0 \times 10^3) = 7.08 \times 10^{-8}\text{ C} = 0.071\;\mu\text{C}$$

Step 2: (b) Conclusion for Zero Net Flux

No, we cannot conclude that there are no charges inside.
Zero net flux only implies that the algebraic net sum of charges enclosed inside is zero (i.e. the box could contain equal amounts of positive and negative charges, or one/more electric dipoles).

PYQ 53 MCQ — 1M | CBSE Recurring

Topic 4

53. Which of the following statements is false regarding Gauss's law?

(a) Gauss's law is valid for any closed surface.

(b) In Gauss's law, the term q on the right side represents the total charge enclosed within the surface.

(d) Gauss's law is derived from Coulomb's law, which contains an inverse square dependence on distance.

Correct Answer: (c) When a system exhibits symmetry, Gauss's law may not be beneficial for calculating the electrostatic field.

Step 1: Analyzing Statement (c)

Statement (c) is false because Gauss's Law is specifically most useful and convenient for calculating electric fields when the charge distribution possesses high geometric symmetry (spherical, cylindrical, or planar).

PYQ 54 MCQ — 1M | CBSE 2025-26 SQP

Topic 4

54. Which Gaussian surface is most suitable for calculating the electric field of an infinitely long uniformly charged straight wire?

(a) Spherical

(b) Cylindrical

(d) Planar

Correct Answer: (b) Cylindrical

Step 1: Symmetry Matching

An infinitely long charged wire possesses radial cylindrical symmetry. A coaxial cylindrical Gaussian surface ensures that the electric field is everywhere normal and uniform over the curved surface, and parallel to flat end faces, making flux integration straightforward. Hence, (b) Cylindrical is the correct choice.

PYQ 55 VSA — 1M | CBSE Recurring

Topic 4

55. State Gauss's law in electrostatics. Write its mathematical form.

Step 1: Statement of Gauss's Law

Statement: The total electric flux through any closed Gaussian surface in free space is equal to $\frac{1}{\varepsilon_0}$ times the total net charge enclosed within that surface.

Step 2: Mathematical Form

$$\oint_S \mathbf{E}\cdot d\mathbf{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$

PYQ 56 Case Study — 4M | CBSE 2024

Topic 4

56. Case Study — Gauss's Law and Charge Distributions:
(i) State Gauss's law mathematically.
(ii) Point charge $Q$ at centre of sphere radius $R$, find $E$ on surface.
(iii) Flux if surface size doubled?
(iv) Can two field lines cross?

Step 1: (i) Mathematical Statement

$$\oint \mathbf{E}\cdot d\mathbf{A} = \frac{q_{\text{enc}}}{\varepsilon_0}$$

Step 2: (ii) Electric Field at Surface of Sphere

$$\oint E dA = E(4\pi R^2) = \frac{Q}{\varepsilon_0} \implies E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2}$$

Step 3: (iii) Flux if Surface Doubled

Flux remains $\frac{Q}{\varepsilon_0}$ (Unchanged), as it depends only on enclosed charge.

Step 4: (iv) Non-Crossing of Field Lines

No. If two field lines crossed, there would be two tangents and two directions of electric field at the intersection point, which is physically impossible.

End of Topic 4

Ready for Topic 5: Applications of Gauss's Law (Q57–Q66)?

SUBTOPIC 5

Topic 5: Applications of Gauss's Law

10 Questions (PYQ 57–66)

PYQ 57 SA — 3M | CBSE Foreign 2013

Topic 5

57. An infinitely long positively charged straight wire has linear charge density $\lambda\text{ C/m}$. An electron is revolving around the wire in a circular orbit.
(a) Deduce expression for kinetic energy.
(b) Plot graph of kinetic energy as a function of charge density $\lambda$.

Step 1: (a) Electric Field of Charged Wire

By Gauss's Law, the radial electric field produced by an infinitely long charged wire of linear charge density $\lambda$ at orbital radius $r$ is:
$$E = \frac{\lambda}{2\pi\varepsilon_0 r}$$

Step 2: Centripetal Force & Kinetic Energy Derivation

The electrostatic attractive force on the electron provides the required centripetal force for circular motion:
$$e E = \frac{m v^2}{r} \implies e \left( \frac{\lambda}{2\pi\varepsilon_0 r} \right) = \frac{m v^2}{r}$$
Multiplying both sides by $r$:
$$m v^2 = \frac{e \lambda}{2\pi\varepsilon_0}$$
Therefore, Kinetic Energy $K$ is:
$$K = \frac{1}{2} m v^2 = \frac{1}{2} \left( \frac{e \lambda}{2\pi\varepsilon_0} \right) = \frac{e \lambda}{4\pi\varepsilon_0}$$
Notice that Kinetic Energy is independent of orbital radius $r$.

Step 3: (b) Graph of $K$ versus $\lambda$

Since $K = \left( \frac{e}{4\pi\varepsilon_0} \right) \lambda \implies K \propto \lambda$, the graph of $K$ versus $\lambda$ is a straight line passing through the origin.

PYQ 58 LA — 5M | CBSE 2016; Recurring

Topic 5

58. (a) State Gauss's law.
(b) Using Gauss's law, derive expression for electric field due to infinitely long uniformly charged wire of linear charge density $\lambda$.

Step 1: (a) Statement of Gauss's Law

The total electric flux through any closed surface is equal to $\frac{1}{\varepsilon_0}$ times the total charge enclosed by that surface:
$$\oint_S \mathbf{E}\cdot d\mathbf{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$

Step 2: (b) Gaussian Surface Selection & Flux Evaluation

Consider an infinitely long thin straight wire of uniform linear charge density $\lambda$.
Choose a coaxial cylindrical Gaussian surface of radius $r$ and length $l$.
• Charge enclosed: $q_{\text{enc}} = \lambda l$.
• Total flux through Gaussian cylinder:
$$\Phi = \int_{\text{left face}} \mathbf{E}\cdot d\mathbf{A} + \int_{\text{right face}} \mathbf{E}\cdot d\mathbf{A} + \int_{\text{curved surface}} \mathbf{E}\cdot d\mathbf{A}$$
For flat end faces, $\mathbf{E} \perp d\mathbf{A}$ ($\theta = 90^\circ$), so $\Phi_{\text{ends}} = 0$.
For curved surface, $\mathbf{E}$ is everywhere parallel to $d\mathbf{A}$ ($\theta = 0^\circ$):
$$\Phi_{\text{curved}} = E \int dA = E(2\pi r l)$$

Step 3: Applying Gauss's Law

$$E(2\pi r l) = \frac{\lambda l}{\varepsilon_0} \implies E = \frac{\lambda}{2\pi\varepsilon_0 r}$$
In vector form: $\mathbf{E} = \frac{\lambda}{2\pi\varepsilon_0 r}\hat{\mathbf{n}}$ (where $\hat{\mathbf{n}}$ is radial unit vector normal to wire).

PYQ 59 Numerical — 3M | CBSE 2022

Topic 5

59. A long cylindrical wire has $\lambda = 2 \times 10^{-8}\text{ C/m}$. A particle of mass $m = 8 \times 10^{-25}\text{ kg}$ and charge $q = 4 \times 10^{-10}\text{ C}$ is revolving around the wire in a circular orbit. Find the radius $r$ of the orbit.

Step 1: Force Balance in Circular Orbit

Electrostatic force provides the necessary centripetal force:
$$q E = \frac{m v^2}{r}$$
Since $E = \frac{\lambda}{2\pi\varepsilon_0 r}$:
$$q \left( \frac{\lambda}{2\pi\varepsilon_0 r} \right) = \frac{m v^2}{r}$$

Step 2: Analyzing Radius Dependence

Cancelling $r$ from denominators on both sides:
$$m v^2 = \frac{q \lambda}{2\pi\varepsilon_0} = 2 \left( \frac{1}{4\pi\varepsilon_0} \right) q \lambda$$
$$v = \sqrt{\frac{q \lambda}{2\pi\varepsilon_0 m}}$$

Step 3: Physical Conclusion

Since orbital radius $r$ cancels out completely, the orbital speed is fixed by $\lambda$ and mass $m$, and circular orbit is dynamically stable for any arbitrary radius $r$.

PYQ 60 SA — 2–3M | CBSE 2026 Set 55/1/3

Topic 5

60. Using Gauss's law, deduce an expression for electric field at a point due to a uniformly charged infinite plane thin sheet. Draw the necessary diagram.

Step 1: Gaussian Pillbox Setup

Consider an infinite thin plane sheet of uniform surface charge density $\sigma\text{ C/m}^2$.
Choose a cylindrical Gaussian pillbox of cross-sectional area $A$ piercing the sheet normally, with its two flat faces at equal distance $r$ on either side.

Step 2: Flux Calculation Across Surfaces

• Charge enclosed inside pillbox: $q_{\text{enc}} = \sigma A$.
• Curved surface flux $= 0$ (since $\mathbf{E} \parallel$ surface, $\mathbf{E} \perp d\mathbf{A}$).
• Flat end caps flux $= E A + E A = 2 E A$.

Step 3: Gauss's Law Application

$$2 E A = \frac{\sigma A}{\varepsilon_0} \implies E = \frac{\sigma}{2\varepsilon_0}$$
In vector form: $\mathbf{E} = \frac{\sigma}{2\varepsilon_0}\hat{\mathbf{n}}$. Notice $E$ is completely independent of distance $r$.

PYQ 61 SA — 3M | CBSE 2026 Set 55/1/3

Topic 5

61. Two large thin plane sheets, each having surface charge density $\sigma$, are held close and parallel to each other in air. What is the net electric field at a point: (i) in between, and (ii) outside the sheets?

Step 1: Individual Field Magnitudes

Each infinite sheet produces a uniform electric field of magnitude $E_1 = E_2 = \frac{\sigma}{2\varepsilon_0}$ directed away from positive charge.

Step 2: (i) Field at a Point In Between the Sheets

The field due to left sheet points to the right ($+x$), and the field due to right sheet points to the left ($-x$).
$$E_{\text{in}} = E_1 - E_2 = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0$$

Step 3: (ii) Field at a Point Outside the Sheets

Both electric fields point in the same direction (directed away from both sheets):
$$E_{\text{out}} = E_1 + E_2 = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$$
Final Answer: Field in between is Zero; Field outside is $\frac{\sigma}{\varepsilon_0}$ directed outwards.

PYQ 62 LA — 5M | CBSE Delhi 2012; Central 2016; Recurring

Topic 5

62. (a) State Gauss's law.
(b) Prove field of infinite plane sheet is independent of distance.
(c) Direction of field for positive and negative sheets.

Step 1: (a) Statement of Gauss's Law

The surface integral of electric field $\mathbf{E}$ over any closed surface $S$ equals $\frac{q_{\text{enc}}}{\varepsilon_0}$.

Step 2: (b) Proof of Distance Independence

By applying cylindrical Gaussian pillbox of cross-section $A$:
$$\Phi = 2 E A = \frac{\sigma A}{\varepsilon_0} \implies E = \frac{\sigma}{2\varepsilon_0}$$
Since the expression contains no distance parameter $r$, the field is uniform at all distances from an infinite sheet.

Step 3: (c) Direction for Positive and Negative Sheets

• For positive sheet ($sigma > 0$): $\mathbf{E}$ is directed normally outwards away from the sheet.
• For negative sheet ($sigma < 0$): $\mathbf{E}$ is directed normally inwards towards the sheet.

PYQ 63 MCQ — 1M | CBSE 2025 Set 55/1/3

Topic 5

63. Three thin spherical shells A, B and C of radii R, 2R and 3R are given charges $-2q$, $+8q$ and $-10q$ respectively. The electric fields at a distance 3R from their centres are:

(A) $E_A > E_B > E_C$

(B) $E_A = E_B = E_C$

(D) $E_B > E_A > E_C$

Correct Answer: (C) $E_A < E_B < E_C$

Step 1: Electric Field Outside Spherical Shells ($r \ge R_{\text{shell}}$)

For any spherical shell of charge $Q$ and radius $R_{\text{shell}}$, at distance $r \ge R_{\text{shell}}$, the electric field magnitude is given by:
$$E = \frac{1}{4\pi\varepsilon_0} \frac{|Q|}{r^2}$$

Step 2: Calculating Field at $r = 3R$ for Each Shell

• Shell A (Radius $R$, charge $-2q$): $E_A = \frac{1}{4\pi\varepsilon_0} \frac{2q}{(3R)^2} = \frac{2kq}{9R^2}$
• Shell B (Radius $2R$, charge $+8q$): $E_B = \frac{1}{4\pi\varepsilon_0} \frac{8q}{(3R)^2} = \frac{8kq}{9R^2}$
• Shell C (Radius $3R$, charge $-10q$): $E_C = \frac{1}{4\pi\varepsilon_0} \frac{10q}{(3R)^2} = \frac{10kq}{9R^2}$

Step 3: Comparing Magnitudes

$$E_A < E_B < E_C$$
Hence, option (C) $E_A < E_B < E_C$ is correct.

PYQ 64 LA — 5M | CBSE 2018; 2019; 2024; Recurring

Topic 5

64. Using Gauss's law, obtain the expression for the electric field due to a uniformly charged thin spherical shell at a point: (i) outside ($r > R$), and (ii) inside ($r < R$). Plot graph of $E$ vs $r$.

Step 1: (i) Point Outside the Shell ($r > R$)

Consider a thin spherical shell of radius $R$ carrying uniform charge $q$.
Draw a concentric spherical Gaussian surface of radius $r > R$.
Total flux $\Phi = \oint E dA = E(4\pi r^2)$. Enclosed charge $= q$.
$$E(4\pi r^2) = \frac{q}{\varepsilon_0} \implies E_{\text{out}} = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \quad (r \ge R)$$

Step 2: (ii) Point Inside the Shell ($r < R$)

Draw a concentric spherical Gaussian surface of radius $r < R$.
Since all charge resides on the outer surface of the shell, the enclosed charge is zero ($q_{\text{enc}} = 0$).
$$E_{\text{in}}(4\pi r^2) = \frac{0}{\varepsilon_0} \implies E_{\text{in}} = 0 \quad (r < R)$$

Step 3: (iii) Graph of $E$ versus $r$

• $E = 0$ for $0 \le r < R$
• $E$ jumps discontinuously to maximum $E_{\text{max}} = \frac{1}{4\pi\varepsilon_0}\frac{q}{R^2}$ at surface $r = R$
• $E \propto \frac{1}{r^2}$ decays for $r > R$.

PYQ 65 LA — 5M | CBSE 2016; Recurring

Topic 5

65. (a) Find field of spherical shell for $r > R$ and $r < R$.
(b) Sphere $S_1$ of radius $r_1$ encloses $Q$. Concentric sphere $S_2$ of radius $r_2 > r_1$ has no extra charge. Ratio of flux through $S_1$ and $S_2$?

Step 1: (a) Expressions for Electric Field

$$E = \begin{cases} \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} & \text{for } r \ge R \\ 0 & \text{for } r < R \end{cases}$$

Step 2: (b) Flux Ratio for Concentric Spheres

• Flux through $S_1$: $\Phi_1 = \frac{Q}{\varepsilon_0}$
• Flux through $S_2$: $\Phi_2 = \frac{Q}{\varepsilon_0}$ (encloses the exact same charge $Q$)
$$\text{Ratio } \Phi_1 : \Phi_2 = 1 : 1$$

PYQ 66 LA — 5M | CBSE 2018; Recurring

Topic 5

66. (a) State and prove Gauss's law.
(b) Apply it to find electric field due to: (i) infinitely long wire, (ii) thin spherical shell outside and inside.

Step 1: (a) Statement and Proof of Gauss's Law

Statement: $\oint_S \mathbf{E}\cdot d\mathbf{A} = \frac{q}{\varepsilon_0}$.
Proof for Point Charge: Consider a point charge $q$ at center of sphere radius $r$. $\mathbf{E} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{\mathbf{r}}$ and $d\mathbf{A} = dA\hat{\mathbf{r}}$.
$$\Phi = \oint \mathbf{E}\cdot d\mathbf{A} = \oint \frac{kq}{r^2} dA = \frac{kq}{r^2} \oint dA = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} (4\pi r^2) = \frac{q}{\varepsilon_0}$$ (Proved).

Step 2: (b) (i) Infinitely Long Wire

$$E(2\pi r l) = \frac{\lambda l}{\varepsilon_0} \implies E = \frac{\lambda}{2\pi\varepsilon_0 r}$$

Step 3: (b) (ii) Thin Spherical Shell

• Outside ($r \ge R$): $E(4\pi r^2) = \frac{q}{\varepsilon_0} \implies E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$
• Inside ($r < R$): $E(4\pi r^2) = 0 \implies E = 0$.

05 / Practice Tests

3-Tier Practice Tests

Level 1 (Foundation), Level 2 (Application), and Level 3 (Challenge). Select options and submit to calculate score and review answers.

24 QUESTIONS

1. The SI unit of electric charge is:

2. The basic property representing $q = ne$ is called:

3. The value of permittivity of free space $\varepsilon_0$ in SI units is:

4. Electric field due to an isolated positive point charge is directed:

5. The direction of electric dipole moment vector $\mathbf{p}$ is:

6. Total electric flux through any closed surface enclosing charge $q$ is:

7. Electric field inside a uniformly charged conducting spherical shell is:

8. Number of electrons in $-1\text{ C}$ of charge is approximately:

1. If the distance between two point charges is halved, the electrostatic force becomes:

2. The ratio of axial to equatorial field of a short dipole at equal distance $r$ is:

3. An electric dipole in a uniform electric field experiences maximum torque when $\theta$ is:

4. Electric field due to an infinite line charge varies with distance $r$ as:

5. A cube encloses a dipole of moment $\mathbf{p}$. The total electric flux emerging from the cube is:

6. Force between two $+1\text{ C}$ charges separated by $1\text{ m}$ in vacuum is:

7. Electric field between two large oppositely charged plates with surface charge density $\sigma$ is:

8. What is the angle between electric dipole moment $\mathbf{p}$ and equatorial field $\mathbf{E}_{\text{eq}}$?

1. An electron and proton fall through distance $h$ in uniform field $E$. The ratio of fall times $t_e / t_p$ is:

2. If a charge $q$ is placed at one corner of a cube, the flux through one of the three non-adjacent faces is:

3. If $E = \alpha x^{1/2}$ along x-axis, flux depends on cube side $a$ as:

4. Stable equilibrium orientation of a dipole in uniform field $\mathbf{E}$ corresponds to:

5. Net electric force on a dipole placed in a NON-uniform electric field:

6. Total positive charge contained in $250\text{ g}$ of water ($\text{H}_2\text{O}$) is approximately:

7. For an early neutral atom model (nucleus $+Ze$ with uniform negative $\rho$ up to $R$), $E(r)$ for $r < R$ is:

8. Three charges $q, q, -q$ placed at vertices of equilateral triangle of side $l$. The vector sum of mutual forces on all 3 charges is:

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