Class 12 Physics Chapter 1 Important Questions (2026-27)

Class 12 Physics Chapter 1 Electric Charges and Fields: important questions with answers on Gauss's law and the dipole for CBSE 2026-27.

Important Questions for Class 12 Physics Chapter 1 – Electric Charges and Fields

Looking for Class 12 Physics Chapter 1 important questions? This page covers Electric Charges and Fields with exam-focused questions and step-by-step answers, arranged by marks: 1 mark, 2–3 marks and 5 marks. Written for the CBSE 2026–27 syllabus, it focuses on Gauss's law, electric flux and the electric dipole. Attempt them after finishing the NCERT chapter.

Very Short Answer Questions (1 Mark)

Electric Dipole

Question 1

How should an electric dipole be oriented in a uniform electric field to be in stable equilibrium?

Answer: When the dipole moment vector \(\vec{p}\) is parallel to the electric field vector \(\vec{E}\).

Question 2

In which orientation is a dipole in a uniform electric field in (a) stable and (b) unstable equilibrium?

Answer:

Stable equilibrium: Dipole placed parallel to the electric field (\(\theta = 0^\circ\)).
Unstable equilibrium: Dipole placed antiparallel to the electric field (\(\theta = 180^\circ\)).

Question 3

Define electric dipole moment and state its SI unit.

Answer: The electric dipole moment of a dipole is the product of the magnitude of either charge and the dipole length. It is a vector directed from the negative to the positive charge. SI unit: coulomb metre (\(\text{C}\cdot\text{m}\)).

Electric Flux and Gauss's Law

Question 4

A Gaussian surface enclosing a charge has its radius reduced to half. What happens to the electric flux through it?

Answer: The flux remains unchanged. Electric flux depends only on the enclosed charge (\(\phi = Q/\varepsilon_0\)), not on the size of the Gaussian surface.

Question 5

Three point charges \(+2q\), \(-q\), and \(+3q\) are present; \(+2q\) and \(-q\) lie inside a surface \(S\). What is the flux through \(S\)?

Answer: Flux \(= \dfrac{\text{enclosed charge}}{\varepsilon_0} = \dfrac{+2q - q}{\varepsilon_0} = \dfrac{q}{\varepsilon_0}\).

Question 6

A charge \(q\) is placed at the centre of a cube of side \(l\). What is the electric flux through each face?

Answer: Total flux \(= q/\varepsilon_0\). Since the cube has 6 faces, flux through each face \(= \dfrac{q}{6\varepsilon_0}\).

Question 7

What is the electric flux through two opposite faces of the same cube?

Answer: Flux through two opposite faces \(= 2 \times \dfrac{q}{6\varepsilon_0} = \dfrac{q}{3\varepsilon_0}\).

Question 8

Can the electric field due to a charge configuration with total charge zero be necessarily zero? Justify.

Answer: No. For example, an electric dipole has net charge zero but produces a non-zero electric field around it.

Question 9

Charges \(-3Q\) and \(+2Q\) are at \((a, 0)\) and \((4a, 0)\). What is the flux through a sphere of radius \(5a\) centred at the origin?

Answer: The sphere encloses both charges. Total enclosed charge \(= -3Q + 2Q = -Q\). Flux \(= -\dfrac{Q}{\varepsilon_0}\).

Question 10

What is the electric flux through a cube that encloses an electric dipole?

Answer: Zero, because the net charge of a dipole (\(+q\) and \(-q\)) is zero.

Question 11

A point charge is enclosed by a spherical Gaussian surface. If the radius is increased, how does the flux change?

Answer: The flux remains unaffected — it depends only on the enclosed charge, not the radius.

Conductors and Electric Field Lines

Question 12

Why must the electrostatic field be zero inside a conductor?

Answer: In a static condition, all charge resides on the surface of a conductor, so there is no charge inside — hence no electric field inside.

Question 13

Why is the electrostatic field normal to the surface at every point of a charged conductor?

Answer: If the field had a tangential component, free charges on the surface would move. Since charges are at rest in electrostatic equilibrium, the field must be perpendicular to the surface.

Question 14

What is the direction of the electric field at the surface of a charged conductor with negative surface charge density (\(\sigma < 0\))?

Answer: The field is normal to the surface and directed inward.

Question 15

Why do electric field lines not form closed loops?

Answer: Electric field lines originate from positive charges and terminate on negative charges. They have a definite starting and ending point, so they cannot form closed loops.

Question 16

Why do electric field lines never intersect each other?

Answer: If two field lines intersected, there would be two directions of the electric field at the same point — which is impossible. Hence they never cross.

Question 17

Why are electric field lines perpendicular to an equipotential surface?

Answer: If the field had a component along the surface, work would be done in moving a test charge along it, contradicting the definition of an equipotential surface. Hence the field must be perpendicular.

Question 18

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

Answer: No. Charge resides only on the outer surface of a conductor, regardless of whether it is hollow or solid.

Electric Potential

Question 19

Name the physical quantity whose SI unit is \(\text{J/C}\). Is it scalar or vector?

Answer: Electric potential. It is a scalar quantity.

Short Answer Questions (2–3 Marks)

Electric Dipole

Question 20

Derive an expression for the torque experienced by an electric dipole in a uniform electric field.

Answer: Consider a dipole with charges \(+q\) and \(-q\) separated by \(2a\), placed in a uniform field \(\vec{E}\) at an angle \(\theta\). The dipole moment is \(p = q \times 2a\).

The forces on the two charges are equal and opposite, forming a couple:

Torque \(\tau = \text{Force} \times \text{perpendicular distance}\)
\(\tau = qE \times 2a \sin\theta\)
\(\tau = pE \sin\theta\)

In vector form: \(\vec{\tau} = \vec{p} \times \vec{E}\)

The direction of \(\vec{\tau}\) is perpendicular to both \(\vec{p}\) and \(\vec{E}\).

Question 21

An electric dipole is placed in a uniform field \(\vec{E}\) with \(\vec{p}\) parallel to \(\vec{E}\). Find (a) work done in turning the dipole until \(\vec{p}\) points opposite to \(\vec{E}\), (b) orientation for maximum torque.

Answer:

(a) Work done \(= pE(1 - \cos 180^\circ) = pE(1 + 1) =\) \(2pE\)

(b) Torque \(\tau = pE \sin\theta\) is maximum when \(\theta = 90^\circ\), i.e., \(\vec{p}\) perpendicular to \(\vec{E}\).

Question 22

Derive the expression for the electric field at a point on the equatorial line of an electric dipole.

Answer: Consider a dipole with charges \(+q\) and \(-q\) separated by \(2a\). Let \(P\) be a point on the equatorial line at distance \(r\) from the centre.

Field due to \(+q\): \(E_1 = \dfrac{kq}{r^2 + a^2}\), directed along \(AP\)
Field due to \(-q\): \(E_2 = \dfrac{kq}{r^2 + a^2}\), directed along \(PC\)

The components perpendicular to the equatorial line cancel. The components parallel to the dipole axis add up:

\(E = 2 \times \dfrac{kq}{r^2 + a^2} \times \cos\theta\), where \(\cos\theta = \dfrac{a}{\sqrt{r^2 + a^2}}\)

\(E = \dfrac{2kqa}{(r^2 + a^2)^{3/2}}\)

Since \(p = q \times 2a\):
\(E = \dfrac{kp}{(r^2 + a^2)^{3/2}}\)

For \(r \gg a\): \(E = \dfrac{kp}{r^3}\)

The direction is opposite to the dipole moment (from positive to negative).

Electric Flux

Question 23

Define electric flux and write its SI unit. A charge \(q\) is enclosed by a spherical surface of radius \(R\). If the radius is halved, how does the flux change?

Answer: Electric flux is the total number of electric field lines passing through a given area. It is a scalar quantity. SI unit: \(\text{N}\cdot\text{m}^2\cdot\text{C}^{-1}\) (or \(\text{V}\cdot\text{m}\)).

By Gauss's law, flux \(\phi = q/\varepsilon_0\). Since the enclosed charge doesn't change when the radius is halved, the flux remains the same.

Question 24

Define electric flux and write its SI unit. Calculate the flux of \(\vec{E} = 3 \times 10^3\,\hat{i}\ \text{N/C}\) through a square of side 10 cm held normal to the field.

Answer: Electric flux is the total number of field lines crossing an area normally. SI unit: \(\text{N}\cdot\text{m}^2/\text{C}\).

Flux \(= EA = 3 \times 10^3 \times (0.1)^2 =\) \(30\ \text{N}\cdot\text{m}^2/\text{C}\)

Question 25

A uniform electric field \(\vec{E} = 5 \times 10^3\,\hat{i}\ \text{N/C}\) passes through a square of side 10 cm. Find the flux when (a) the plane is parallel to the \(y\)-\(z\) plane, (b) the plane makes a \(30^\circ\) angle with the \(x\)-axis.

Answer: Area \(A = (0.1)^2 = 0.01\ \text{m}^2\)

(a) When plane is parallel to \(y\)-\(z\) plane, normal is along \(x\)-axis \(\Rightarrow \theta = 0^\circ\)
Flux \(= EA\cos 0^\circ = 5 \times 10^3 \times 0.01 =\) \(50\ \text{N}\cdot\text{m}^2/\text{C}\)

(b) When plane makes \(30^\circ\) with \(x\)-axis, normal makes \(60^\circ\) with \(x\)-axis \(\Rightarrow \theta = 60^\circ\)
Flux \(= EA\cos 60^\circ = 50 \times 0.5 =\) \(25\ \text{N}\cdot\text{m}^2/\text{C}\)

Question 26

A hollow cylindrical box of length 1 m and cross-sectional area \(25\ \text{cm}^2\) is placed in a region with \(E = 50x\,\hat{i}\ \text{N/C}\). Find (a) net flux through the cylinder, (b) charge enclosed.

Answer:

(a) Flux through left face (at \(x = 0\)): \(E = 0\), so flux \(= 0\)
Flux through right face (at \(x = 1\)): \(E = 50\ \text{N/C}\), flux \(= 50 \times 25 \times 10^{-4} = 0.125\ \text{N}\cdot\text{m}^2/\text{C}\)
Net flux \(=\) \(0.125\ \text{N}\cdot\text{m}^2/\text{C}\)

(b) Charge enclosed \(= \varepsilon_0 \times \text{flux} = 8.85 \times 10^{-12} \times 0.125 \approx\) \(1.1 \times 10^{-12}\ \text{C}\)

Applications of Gauss's Law

Question 27

State Gauss's law and use it to derive the electric field due to a uniformly charged infinite plane sheet.

Answer: Gauss's law: The total electric flux through any closed surface equals \(1/\varepsilon_0\) times the net charge enclosed by it.

For an infinite plane sheet with surface charge density \(\sigma\), choose a cylindrical Gaussian surface with axis perpendicular to the sheet, cross-sectional area \(A\), extending distance \(r\) on both sides.

Flux through curved surface \(= 0\) (\(\vec{E}\) parallel to surface).
Flux through two flat ends \(= 2EA\).

By Gauss's law: \(2EA = \dfrac{\sigma A}{\varepsilon_0}\)
\(E = \dfrac{\sigma}{2\varepsilon_0}\)

This is independent of distance \(r\) from the sheet, so the field is uniform.

Question 28

State Gauss's law and derive the electric field due to an infinitely long straight wire with linear charge density \(\lambda\).

Answer: Gauss's law: Total flux through a closed surface \(= \dfrac{1}{\varepsilon_0} \times\) enclosed charge.

Consider a cylindrical Gaussian surface of radius \(r\) and length \(l\) coaxial with the wire. By symmetry, \(E\) is radial and constant on the curved surface.

Flux through curved surface \(= E \times 2\pi rl\)
Enclosed charge \(= \lambda l\)

By Gauss's law: \(E \times 2\pi rl = \dfrac{\lambda l}{\varepsilon_0}\)
\(E = \dfrac{\lambda}{2\pi\varepsilon_0 r}\)

Question 29

Use Gauss's law to derive the electric field between two uniformly charged parallel sheets with surface charge densities \(+\sigma\) and \(-\sigma\).

Answer: Between the two sheets, the fields from both sheets point in the same direction (from positive to negative).

Field due to each sheet \(= \dfrac{\sigma}{2\varepsilon_0}\)
Total field between the sheets \(= \dfrac{\sigma}{2\varepsilon_0} + \dfrac{\sigma}{2\varepsilon_0} =\) \(\dfrac{\sigma}{\varepsilon_0}\)

Outside the sheets, the fields cancel, so \(E = 0\).

Question 30

A thin conducting spherical shell of radius \(R\) has charge \(Q\) uniformly spread on its surface. Using Gauss's law, derive \(E\) at a point outside the shell. Plot \(E\) versus \(r\) for \(0 \le r \le \infty\).

Answer: For a point at distance \(r > R\), choose a Gaussian sphere of radius \(r\).
By Gauss's law: \(E \times 4\pi r^2 = \dfrac{Q}{\varepsilon_0}\)
\(E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{r^2}\)

Graph of \(E\) vs \(r\):

For \(0 < r < R\): \(E = 0\) (inside the shell)
At \(r = R\): \(E = \dfrac{Q}{4\pi\varepsilon_0 R^2}\) (maximum)
For \(r > R\): \(E \propto \dfrac{1}{r^2}\) (decreases)

Question 31

A thin infinitely long conducting wire with linear charge density \(\lambda\) is enclosed by a cylindrical surface of radius \(r\) and length \(l\). Find the electric flux through the cylinder.

Answer: By Gauss's law, flux through the curved surface \(= E \times 2\pi rl\), where \(E = \dfrac{\lambda}{2\pi\varepsilon_0 r}\).

Flux \(= \dfrac{\lambda l}{\varepsilon_0}\)

(Flux through the two flat ends is zero since \(\vec{E}\) is parallel to those surfaces.)

Conductors

Question 32

Show that the electric field at the surface of a charged conductor is \(\vec{E} = \dfrac{\sigma}{\varepsilon_0}\hat{n}\).

Answer: Consider a charged spherical conductor of radius \(R\) with surface charge density \(\sigma\). Total charge \(Q = \sigma \times 4\pi R^2\).

Electric field at the surface: \(E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{R^2} = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{\sigma \times 4\pi R^2}{R^2} = \dfrac{\sigma}{\varepsilon_0}\)

In vector form: \(\vec{E} = \dfrac{\sigma}{\varepsilon_0}\hat{n}\), where \(\hat{n}\) is the unit vector normal to the surface in the outward direction.

Question 33

A spherical conducting shell has inner radius \(r_1\) and outer radius \(r_2\), with charge \(Q\). A charge \(q\) is placed at the centre. Find (a) surface charge density on inner and outer surfaces, (b) electric field at a point \(x > r_2\).

Answer:

(a) Charge on inner surface \(= -q\); surface charge density \(= -\dfrac{q}{4\pi r_1^2}\)
Charge on outer surface \(= Q + q\); surface charge density \(= \dfrac{Q + q}{4\pi r_2^2}\)

(b) For \(x > r_2\): \(E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q + Q}{x^2}\)

Question 34

A small metal sphere with charge \(+Q\) is at the centre of a spherical cavity in a large uncharged metallic shell. Write the charges on the inner and outer surfaces, and the field at a point \(P_1\) inside the cavity.

Answer:

Inner surface: \(-Q\)
Outer surface: \(+Q\)
Electric field at \(P_1\) (distance \(r_1\) from centre): \(E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{r_1^2}\)

Question 35

Two charged spherical conductors of radii \(R_1\) and \(R_2\) are connected by a conducting wire. Find the ratio of their surface charge densities.

Answer: When connected, both spheres have the same potential:
\(\dfrac{kQ_1}{R_1} = \dfrac{kQ_2}{R_2} \Rightarrow \dfrac{Q_1}{Q_2} = \dfrac{R_1}{R_2}\)

Surface charge densities: \(\sigma_1 = \dfrac{Q_1}{4\pi R_1^2}\), \(\sigma_2 = \dfrac{Q_2}{4\pi R_2^2}\)

\(\dfrac{\sigma_1}{\sigma_2} = \dfrac{Q_1}{Q_2} \times \dfrac{R_2^2}{R_1^2} = \dfrac{R_1}{R_2} \times \dfrac{R_2^2}{R_1^2} =\) \(\dfrac{R_2}{R_1}\)

So \(\sigma_1 : \sigma_2 = R_2 : R_1\)

Coulomb's Law and Field Calculations

Question 36

Plot a graph of Coulomb force \(F\) versus \(1/r^2\) for pairs \((1\,\mu\text{C}, 2\,\mu\text{C})\) and \((2\,\mu\text{C}, -3\,\mu\text{C})\). Interpret the graphs.

Answer:

For \((1\,\mu\text{C}, 2\,\mu\text{C})\): Both charges are positive \(\Rightarrow\) repulsive force \(\Rightarrow\) positive slope.
For \((2\,\mu\text{C}, -3\,\mu\text{C})\): Charges are opposite \(\Rightarrow\) attractive force \(\Rightarrow\) negative slope.

The magnitude of the slope is proportional to the product of the charges.

Question 37

Two point charges \(+3q\) and \(-4q\) are at vertices \(B\) and \(C\) of an equilateral triangle \(ABC\) of side \(a\). Find the magnitude and direction of the resultant electric field at vertex \(A\).

Answer:

Field due to \(+3q\) at \(A\): \(E_1 = \dfrac{k(3q)}{a^2}\), directed away from \(B\) along \(BA\).
Field due to \(-4q\) at \(A\): \(E_2 = \dfrac{k(4q)}{a^2}\), directed toward \(C\) along \(AC\).

Angle between \(E_1\) and \(E_2\) \(= 120^\circ\) (since it's an equilateral triangle).

Resultant: \(E = \sqrt{E_1^2 + E_2^2 + 2E_1E_2\cos 120^\circ}\)
\(= \dfrac{kq}{a^2}\sqrt{9 + 16 + 2 \times 3 \times 4 \times (-1/2)}\)
\(= \dfrac{kq}{a^2}\sqrt{25 - 12}\)
\(= \dfrac{kq}{a^2}\sqrt{13}\)

\(E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{\sqrt{13}\,q}{a^2}\)

Direction: Along the bisector of angle \(A\), pointing away from the triangle.

Question 38

A charge is uniformly distributed over a ring of radius \(a\). Derive an expression for the electric field at a point on its axis at distance \(x\) from the centre. Show that for large distances it behaves like a point charge.

Answer: Consider a small element of charge \(dq\) on the ring. Its field at point \(P\) on the axis is \(dE = \dfrac{k\,dq}{a^2 + x^2}\).

The component along the axis: \(dE\cos\theta = \dfrac{k\,dq\,x}{(a^2 + x^2)^{3/2}}\)

Integrating over the whole ring: \(E = \dfrac{kQx}{(a^2 + x^2)^{3/2}}\)

For \(x \gg a\): \(E \approx \dfrac{kQx}{x^3} = \dfrac{kQ}{x^2}\) — same as a point charge \(Q\) at the centre.

Question 39

Two thin concentric spherical shells of radii \(a\) and \(b\) (\(b > a\)) carry charges \(q\) and \(Q\). Find \(E\) at a point distance \(x\) from the centre for (i) \(0 < x < a\), (ii) \(a \le x < b\), (iii) \(b \le x < \infty\).

Answer:

(i) \(0 < x < a\): Point is inside both shells \(\Rightarrow\) \(E = 0\)

(ii) \(a \le x < b\): Point is outside the inner shell but inside the outer shell. Only \(q\) is enclosed.
\(E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q}{x^2}\)

(iii) \(x \ge b\): Point is outside both shells. Total enclosed charge \(= q + Q\).
\(E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q + Q}{x^2}\)

Question 40

A charge \(+Q\) is uniformly distributed within a sphere of radius \(R\). Find \(E\) at distance \(r\) from the centre for (i) \(r < R\), (ii) \(r > R\).

Answer:

(i) For \(r < R\): Charge enclosed \(= Q \times \dfrac{r^3}{R^3}\)
By Gauss's law: \(E \times 4\pi r^2 = \dfrac{Q}{\varepsilon_0} \cdot \dfrac{r^3}{R^3}\)
\(E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Qr}{R^3}\)

(ii) For \(r > R\): Enclosed charge \(= Q\)
\(E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{r^2}\)

Question 41

Sphere \(S_1\) of radius \(r_1\) encloses charge \(Q\). A concentric sphere \(S_2\) of radius \(r_2\) (\(r_2 > r_1\)) encloses charge \(2Q\). Find the ratio of flux through \(S_1\) and \(S_2\). How does flux through \(S_1\) change if a dielectric of constant \(K\) is introduced inside \(S_2\)?

Answer:

Flux through \(S_1 = \dfrac{Q}{\varepsilon_0}\)
Flux through \(S_2 = \dfrac{2Q}{\varepsilon_0}\)
Ratio \(=\) \(1:2\)

Introducing a dielectric inside \(S_2\) does not change the flux through \(S_1\) (the enclosed charge remains \(Q\)).

Long Answer Questions (5 Marks)

Applications of Gauss's Law

Question 42

(a) Using Gauss's law, derive \(E\) at a point outside a uniformly charged thin spherical shell of radius \(R\) and surface charge density \(\sigma\). Draw field lines for (i) positive and (ii) negative charge density. (b) A conducting sphere of diameter 2.5 m has surface charge density \(100\,\mu\text{C/m}^2\). Calculate (i) charge on the sphere, (ii) total flux through it.

Answer:

(a) For a point at distance \(r > R\), choose a Gaussian sphere of radius \(r\).
\(E \times 4\pi r^2 = \dfrac{Q}{\varepsilon_0}\)
\(E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{r^2} = \dfrac{\sigma R^2}{\varepsilon_0 r^2}\)

Field lines:

(i) Positive: radially outward
(ii) Negative: radially inward

(b) Radius \(R = 1.25\ \text{m}\), \(\sigma = 100 \times 10^{-6}\ \text{C/m}^2\)
(i) \(Q = \sigma \times 4\pi R^2 = 100 \times 10^{-6} \times 4\pi \times (1.25)^2 \approx\) \(1.96 \times 10^{-3}\ \text{C}\)
(ii) Flux \(= \dfrac{Q}{\varepsilon_0} = \dfrac{1.96 \times 10^{-3}}{8.85 \times 10^{-12}} \approx\) \(2.2 \times 10^8\ \text{N}\cdot\text{m}^2/\text{C}\)

Question 43

(a) Define electric flux and write its SI units. (b) The electric field components inside a cube of side 0.1 m are \(E_x = 500x\ (\text{N/C})\), \(E_y = 0\), \(E_z = 0\). Calculate (i) flux through the cube, (ii) charge inside the cube.

Answer:

(a) Electric flux is the total number of field lines crossing a surface normally. SI unit: \(\text{N}\cdot\text{m}^2/\text{C}\).

(b) Only the faces perpendicular to the \(x\)-axis contribute.
At \(x = 0\): \(E = 0\), flux \(= 0\)
At \(x = 0.1\): \(E = 500 \times 0.1 = 50\ \text{N/C}\), flux \(= 50 \times (0.1)^2 = 0.5\ \text{N}\cdot\text{m}^2/\text{C}\)

(i) Net flux \(=\) \(0.5\ \text{N}\cdot\text{m}^2/\text{C}\)
(ii) Charge \(= \varepsilon_0 \times \text{flux} = 8.85 \times 10^{-12} \times 0.5 \approx\) \(4.4 \times 10^{-12}\ \text{C}\)

Question 44

(a) Define electric flux and write its SI units. (b) Using Gauss's law, prove that the electric field due to a uniformly charged infinite plane sheet is independent of distance. (c) How is the field directed if the sheet is (i) positively charged, (ii) negatively charged?

Answer:

(a) Electric flux is the measure of field lines passing normally through an area. SI unit: \(\text{N}\cdot\text{m}^2/\text{C}\).

(b) For an infinite sheet with surface charge density \(\sigma\), choose a cylindrical Gaussian surface with axis perpendicular to the sheet, cross-sectional area \(A\), extending distance \(r\) on both sides.

Flux through curved surface \(= 0\)
Flux through two flat ends \(= 2EA\)
Enclosed charge \(= \sigma A\)

\(2EA = \dfrac{\sigma A}{\varepsilon_0} \Rightarrow\) \(E = \dfrac{\sigma}{2\varepsilon_0}\)

This expression has no \(r\), so \(E\) is independent of distance from the sheet.

Question 45

(a) Define electric flux and write its SI unit. (b) A small metal sphere with charge \(+Q\) is at the centre of a spherical cavity in a large uncharged metallic shell. Find expressions for \(E\) at points \(P_1\) (inside cavity) and \(P_2\) (inside the metal). (c) Draw the pattern of electric field lines.

Answer:

(a) Electric flux is the total number of field lines crossing an area normally. SI unit: \(\text{N}\cdot\text{m}^2/\text{C}\).

(b) At \(P_1\) (distance \(r_1\) from centre, inside cavity): The Gaussian surface encloses only \(+Q\).
\(E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{r_1^2}\) (radially outward)

At \(P_2\) (inside the metal): The Gaussian surface encloses \(+Q\) (on cavity surface) and \(-Q\) (on inner surface of shell), so net enclosed charge \(= 0\).
\(E = 0\)

(c) Field lines: Radially outward from \(+Q\), terminating on the inner surface of the shell. No field lines inside the metal. Field lines emerge from the outer surface of the shell (which has charge \(+Q\)).

Question 46

Using Gauss's law, derive \(E\) due to a uniformly charged spherical conducting shell of radius \(R\) at a point (i) outside and (ii) inside. Plot \(E\) versus \(r\).

Answer:

(i) Outside (\(r > R\)): Choose a Gaussian sphere of radius \(r\).
\(E \times 4\pi r^2 = \dfrac{Q}{\varepsilon_0}\)
\(E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{r^2}\)

(ii) Inside (\(r < R\)): The Gaussian surface encloses no charge.
\(E \times 4\pi r^2 = 0 \Rightarrow\) \(E = 0\)

Graph of \(E\) vs \(r\):

For \(r < R\): \(E = 0\)
At \(r = R\): \(E = \dfrac{Q}{4\pi\varepsilon_0 R^2}\) (maximum)
For \(r > R\): \(E \propto \dfrac{1}{r^2}\)

Question 47

(a) Define electric flux and write its SI unit. "Gauss's law is true for any closed surface, regardless of shape or size." Justify with an example. (b) Use Gauss's law to prove that \(E\) inside a uniformly charged spherical shell is zero.

Answer:

(a) Electric flux is the number of field lines crossing an area normally. SI unit: \(\text{N}\cdot\text{m}^2/\text{C}\).

Justification: For a closed surface containing an electric dipole, the net enclosed charge is zero, so total flux \(= 0\) — regardless of the surface's shape or size. This confirms Gauss's law's generality.

(b) For a point inside a uniformly charged shell (\(r < R\)), choose a Gaussian sphere of radius \(r\). It encloses no charge.
\(E \times 4\pi r^2 = 0 \Rightarrow\) \(E = 0\)

Question 48

(i) Use Gauss's law to find \(E\) due to a uniformly charged infinite plane sheet. What is the field direction for positive and negative charge densities? (ii) Find the ratio of potential differences that must be applied across parallel and series combinations of two capacitors \(C_1\) and \(C_2\) (ratio 1:2) so that the energy stored is the same.

Answer:

(i) \(E = \dfrac{\sigma}{2\varepsilon_0}\), directed away from the sheet if \(\sigma > 0\) and toward the sheet if \(\sigma < 0\).

(ii) Let \(C_1 = C\), \(C_2 = 2C\).

Parallel: \(C_p = C + 2C = 3C\)
Energy: \(U_p = \dfrac{1}{2}C_p V_p^2 = \dfrac{1}{2}(3C)V_p^2\)

Series: \(\dfrac{1}{C_s} = \dfrac{1}{C} + \dfrac{1}{2C} = \dfrac{3}{2C} \Rightarrow C_s = \dfrac{2C}{3}\)
Energy: \(U_s = \dfrac{1}{2}C_s V_s^2 = \dfrac{1}{2}\left(\dfrac{2C}{3}\right)V_s^2\)

For equal energy: \(\dfrac{1}{2}(3C)V_p^2 = \dfrac{1}{2}\left(\dfrac{2C}{3}\right)V_s^2\)
\(3V_p^2 = \dfrac{2}{3}V_s^2\)
\(\dfrac{V_s^2}{V_p^2} = \dfrac{9}{2}\)
\(\dfrac{V_s}{V_p} = \dfrac{3}{\sqrt{2}}\)

Question 49

(a) Use Gauss's theorem to find \(E\) due to a uniformly charged infinite plane sheet with surface charge density \(\sigma\). (b) Find the work done in bringing a point charge \(q\) from infinity to a point at distance \(r\) in front of the charged sheet.

Answer:

(a) \(E = \dfrac{\sigma}{2\varepsilon_0}\), directed away from a positively charged sheet.

(b) The field is uniform (constant magnitude \(\sigma/(2\varepsilon_0)\)) and directed away from the sheet.

Work done \(= -q\int \vec{E}\cdot d\vec{r}\) from \(\infty\) to \(r\). Since \(E\) is constant and points away from the sheet, the potential at distance \(r\) (taking \(V = 0\) at the sheet) is \(V = -\dfrac{\sigma r}{2\varepsilon_0}\).

Work done in bringing \(q\) from infinity to \(r\): \(W = qV =\) \(-\dfrac{q\sigma r}{2\varepsilon_0}\)

(Note: The field due to an infinite sheet extends to infinity, so the potential is defined relative to the sheet.)

Question 50

(a) State Gauss's law. Using it, derive \(E\) due to an infinitely long straight conductor with linear charge density \(\lambda\). (b) A wire \(AB\) of length \(L\) has linear charge density \(\lambda = kx\) (\(x\) measured from \(A\)). It is enclosed by a Gaussian hollow surface. Find the flux through the surface.

Answer:

(a) Gauss's law: Total flux through a closed surface \(= \dfrac{1}{\varepsilon_0} \times\) enclosed charge.

For an infinitely long wire with linear charge density \(\lambda\), using a cylindrical Gaussian surface of radius \(r\) and length \(l\):
\(E \times 2\pi rl = \dfrac{\lambda l}{\varepsilon_0}\)
\(E = \dfrac{\lambda}{2\pi\varepsilon_0 r}\)

(b) Total charge on the wire: \(Q = \int_0^L kx\,dx = \dfrac{kL^2}{2}\)

Flux through the Gaussian surface \(= \dfrac{Q}{\varepsilon_0} =\) \(\dfrac{kL^2}{2\varepsilon_0}\)

Electric Dipole

Question 51

Define electric dipole moment. Is it scalar or vector? Derive the expression for the electric field of a dipole at a point on the equatorial plane.

Answer: Electric dipole moment is the product of the magnitude of either charge and the distance between them. It is a vector directed from negative to positive charge. SI unit: \(\text{C}\cdot\text{m}\).

For a point \(P\) on the equatorial line at distance \(r\) from the centre:

Field due to \(+q\): \(E_1 = \dfrac{kq}{r^2 + a^2}\)
Field due to \(-q\): \(E_2 = \dfrac{kq}{r^2 + a^2}\)

The perpendicular components cancel; the parallel components add:

\(E = 2 \times \dfrac{kq}{r^2 + a^2} \times \dfrac{a}{\sqrt{r^2 + a^2}}\)
\(E = \dfrac{kp}{(r^2 + a^2)^{3/2}}\)

For \(r \gg a\): \(E = \dfrac{kp}{r^3}\), directed opposite to \(\vec{p}\).

Question 52

(a) Deduce the expression for torque on a dipole in a uniform electric field. (b) Two hollow concentric spheres \(S_1\) and \(S_2\) enclose charges \(2Q\) and \(4Q\). (i) Find the ratio of flux through them. (ii) How does flux through \(S_1\) change if a dielectric of constant \(\varepsilon_r\) is introduced inside \(S_1\)?

Answer:

(a) Torque \(\tau = pE\sin\theta\), vector form \(\vec{\tau} = \vec{p} \times \vec{E}\)

(b) (i) Flux through \(S_1 = \dfrac{2Q}{\varepsilon_0}\); flux through \(S_2 = \dfrac{2Q + 4Q}{\varepsilon_0} = \dfrac{6Q}{\varepsilon_0}\)
Ratio \(= \dfrac{2Q}{6Q} =\) \(1:3\)

(ii) Flux through \(S_1\) becomes \(\dfrac{2Q}{\varepsilon_r\varepsilon_0}\) — it decreases by a factor of \(\varepsilon_r\).

Question 53

(a) Derive the expression for \(E\) due to a dipole at distance \(x\) on its axial line. Show that for \(x \gg a\), \(E \to \dfrac{2p}{4\pi\varepsilon_0 x^3}\). (b) Only faces perpendicular to the \(x\)-axis contribute to electric flux through a cube. Explain.

Answer:

(a) On the axial line:

Field due to \(+q\): \(E_1 = \dfrac{kq}{(x - a)^2}\) (away from dipole)
Field due to \(-q\): \(E_2 = \dfrac{kq}{(x + a)^2}\) (toward dipole)

Net field: \(E = E_1 - E_2 = kq\left[\dfrac{1}{(x - a)^2} - \dfrac{1}{(x + a)^2}\right]\)
\(= kq \cdot \dfrac{(x + a)^2 - (x - a)^2}{(x^2 - a^2)^2}\)
\(= kq \cdot \dfrac{4ax}{(x^2 - a^2)^2}\)
\(= \dfrac{2k(2qa)x}{(x^2 - a^2)^2}\)
\(E = \dfrac{2kpx}{(x^2 - a^2)^2}\)

For \(x \gg a\): \(E \approx \dfrac{2kpx}{x^4} =\) \(\dfrac{2kp}{x^3} = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{2p}{x^3}\)

(b) For faces perpendicular to the \(x\)-axis, the area vector is parallel to \(E_x\), so flux \(= E_x A\). For faces parallel to the \(x\)-axis, \(E_x\) is parallel to the surface (perpendicular to area vector), so flux \(= 0\).

Question 54

(a) Derive \(E\) due to a dipole at distance \(r\) on its axial line. (b) Draw \(E\) versus \(r\) for \(r \gg a\). (c) For the dipole in a uniform external field, show diagrammatically the stable and unstable equilibrium positions and write torque expressions.

Answer:

(a) (Same as Question 53a) \(E = \dfrac{2kpr}{(r^2 - a^2)^2}\); for \(r \gg a\): \(E = \dfrac{2kp}{r^3}\)

(b) Graph: \(E\) decreases as \(1/r^3\) for large \(r\).

(c)

Stable equilibrium: \(\vec{p}\) parallel to \(\vec{E}\) (\(\theta = 0^\circ\)). Torque \(\tau = pE\sin 0^\circ =\) \(0\)
Unstable equilibrium: \(\vec{p}\) antiparallel to \(\vec{E}\) (\(\theta = 180^\circ\)). Torque \(\tau = pE\sin 180^\circ =\) \(0\)

In both cases torque is zero, but stable equilibrium is restored after small displacement, while unstable is not.

Capacitors and Energy

Question 55

(a) Derive the expression for energy stored in a parallel plate capacitor. Hence find the energy density. (b) A charged capacitor is connected across an identical uncharged capacitor. Show that the energy stored in the combination is less than the initial energy.

Answer:

(a) Work done in charging a capacitor to charge \(Q\):
\(dW = V\,dq = \dfrac{q}{C}\,dq\)
Total work \(W = \int_0^Q \dfrac{q}{C}\,dq = \dfrac{Q^2}{2C}\)

Energy stored: \(U = \dfrac{Q^2}{2C} = \dfrac{1}{2}CV^2 = \dfrac{1}{2}QV\)

Energy density: For a parallel plate capacitor, \(C = \dfrac{\varepsilon_0 A}{d}\), \(V = Ed\)
\(U = \dfrac{1}{2} \cdot \dfrac{\varepsilon_0 A}{d} \cdot (Ed)^2 = \dfrac{1}{2}\varepsilon_0 E^2 \times Ad\)
Energy density \(u = \dfrac{U}{Ad} =\) \(\dfrac{1}{2}\varepsilon_0 E^2\)

(b) Initially: \(U_1 = \dfrac{Q^2}{2C}\)
When connected to an identical capacitor, charge distributes equally: each gets \(Q/2\).
Final energy: \(U_2 = \dfrac{(Q/2)^2}{2C} + \dfrac{(Q/2)^2}{2C} = \dfrac{Q^2}{4C}\)

\(\dfrac{U_2}{U_1} = \dfrac{Q^2/4C}{Q^2/2C} = \dfrac{1}{2}\)

So \(U_2 = \dfrac{U_1}{2}\) — the energy in the combination is half the initial energy (the rest is dissipated as heat in the connecting wires).

Mixed Concepts

Question 56

Answer:

(a) Gauss's law: Total flux through a closed surface \(= \dfrac{1}{\varepsilon_0} \times\) enclosed charge.

(b) Total charge on the wire: \(Q = \int_0^L kx\,dx = \dfrac{kL^2}{2}\)

Flux through the Gaussian surface \(= \dfrac{Q}{\varepsilon_0} =\) \(\dfrac{kL^2}{2\varepsilon_0}\)

Question 57

Two point charges \(4\,\mu\text{C}\) and \(1\,\mu\text{C}\) are separated by 2 m in air. Find the point on the line joining them where the net electric field is zero.

Answer: Let the point be at distance \(x\) from the \(4\,\mu\text{C}\) charge (and \(2 - x\) from the \(1\,\mu\text{C}\) charge).

For net field to be zero (both charges positive, so the point must be between them):

\(\dfrac{k(4)}{x^2} = \dfrac{k(1)}{(2 - x)^2}\)

\(\dfrac{4}{x^2} = \dfrac{1}{(2 - x)^2}\)

Taking square root: \(\dfrac{2}{x} = \dfrac{1}{2 - x}\)
\(2(2 - x) = x\)
\(4 - 2x = x\)
\(3x = 4\)
\(x = \dfrac{4}{3}\ \text{m} \approx 1.33\ \text{m}\) from the \(4\,\mu\text{C}\) charge

Key Formulas to Remember

Concept	Formula
Coulomb's law	\(F = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q_1 q_2}{r^2}\)
Electric field	\(E = \dfrac{F}{q} = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{r^2}\)
Electric flux	\(\phi = E \cdot A \cdot \cos\theta\)
Gauss's law	\(\phi = \dfrac{Q_{\text{enclosed}}}{\varepsilon_0}\)
Dipole moment	\(p = q \times 2a\)
Torque on dipole	\(\tau = pE\sin\theta\)
Axial field of dipole	\(E = \dfrac{2kp}{r^3}\) (\(r \gg a\))
Equatorial field of dipole	\(E = \dfrac{kp}{r^3}\) (\(r \gg a\))
Infinite wire field	\(E = \dfrac{\lambda}{2\pi\varepsilon_0 r}\)
Infinite sheet field	\(E = \dfrac{\sigma}{2\varepsilon_0}\)
Conducting shell (outside)	\(E = \dfrac{kQ}{r^2}\)
Conducting shell (inside)	\(E = 0\)

FAQs

Are these questions enough for CBSE Class 12 Physics Chapter 1?

Yes. These questions cover every major topic — electric flux, Gauss's law, dipole behaviour, conductors, and field calculations — that CBSE has asked in previous board exams. Combine them with NCERT exercises and you'll be well prepared.

Which topics are most important for board exams?

Gauss's law applications (infinite wire, plane sheet, spherical shell), torque on a dipole, electric flux calculations, and field inside/outside conductors are consistently high-weightage topics.

How should I prepare for numericals in this chapter?

Practise flux calculations through cubes, cylinders, and spheres, and field problems involving point charges and dipoles. Focus on understanding the Gaussian surface selection for each case.