Physics – Class XII – Student Support Material (SSM 2025–26)

SECTION A – Objective Type Questions (MCQs & Assertion-Reason)

Section A mainly checks your conceptual clarity, formulas, and understanding of basic principles.

In this chapter, Section A includes:

Multiple Choice Questions (MCQs)

These questions test:

Coulomb’s Law

Inverse square relationship (F ∝ 1/r²)

Dielectric constant effect

Electric field due to dipole

Torque on dipole

Gauss’s law applications

Flux through surfaces

For example:

If distance between charges is halved → force becomes four times.
This is because Coulomb’s law states:

F=14πε0q1q2r2F = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r^2}F=4πε0​1​r2q1​q2​​

So force is inversely proportional to square of distance.

Similarly, MCQs test:

Field inside hollow conducting sphere = zero

Electric field of dipole on axial & equatorial line

Flux through one face of cube = q / 6ε₀

These questions require strong formula clarity.

Assertion–Reason Questions

These test deeper conceptual understanding.

For example:

Why dipole has translational motion in non-uniform field?

Why Gauss law works for any shape?

Why electric field inside conductor is zero?

Why electrostatic forces are conservative?

Here you must understand reasoning, not just formula.

Section A focuses on:

Theory clarity

Formula application

Conceptual correctness

SECTION B – Very Short & Short Answer Questions

Section B checks your explanation skills and derivation understanding.

Important Areas in Section B

1️⃣ Electric Dipole & Dipole Moment

Electric dipole = two equal and opposite charges separated by small distance.

Dipole moment:

p=q(2a)p = q (2a)p=q(2a)

It is a vector quantity.
Direction: from negative to positive charge.

You may be asked:

Define dipole moment

Stable and unstable equilibrium

Expression for torque

Torque on dipole:

τ=pEsin⁡θ\tau = pE \sin\thetaτ=pEsinθ

If θ = 0° → Stable equilibrium
If θ = 180° → Unstable equilibrium

2️⃣ Electric Flux

Electric flux = number of electric field lines passing through a surface.

ϕ=E⃗⋅S⃗\phi = \vec{E} \cdot \vec{S}ϕ=E⋅S

It is a scalar quantity.
SI unit = Nm²/C

Gauss Law:

ϕ=qinε0\phi = \frac{q_{in}}{\varepsilon_0}ϕ=ε0​qin​​

Section B may ask:

Define flux

State Gauss law

Calculate flux in cube

Conceptual explanation

3️⃣ Field Due to Line Charge

Using cylindrical Gaussian surface:

E=λ2πε0rE = \frac{\lambda}{2\pi \varepsilon_0 r}E=2πε0​rλ​

Important:
Field is radial
Depends only on distance r

4️⃣ Field Due to Plane Sheet

Using Gaussian pillbox:

E=σ2ε0E = \frac{\sigma}{2\varepsilon_0}E=2ε0​σ​

Field is constant
Independent of distance

5️⃣ Field Due to Spherical Shell

Using Gauss Law:

Outside (r > R):

E=14πε0qr2E = \frac{1}{4\pi \varepsilon_0} \frac{q}{r^2}E=4πε0​1​r2q​

Inside (r < R):

E=0E = 0E=0

Section B builds:

Derivation clarity

Diagram understanding

Conceptual explanation

SECTION C – Long Answer Questions (4–5 Marks)

This section checks derivations, graphical explanation, and deep conceptual clarity.

1️⃣ Derivation Based Questions

You must derive expressions like:

(a) Electric field due to spherical shell using Gauss Law

Important steps:

Choose Gaussian surface

Apply symmetry

Apply Gauss theorem

Solve for E

Conclusion:
Inside conductor → E = 0
Outside → behaves like point charge

(b) Electric Field due to Infinite Line Charge

Using cylindrical Gaussian surface:

E(2πrl)=λlε0E (2\pi r l) = \frac{\lambda l}{\varepsilon_0}E(2πrl)=ε0​λl​ E=λ2πε0rE = \frac{\lambda}{2\pi \varepsilon_0 r}E=2πε0​rλ​

You must clearly explain symmetry and reasoning.

(c) Torque on Electric Dipole

Show:

Net force = 0

But torque exists

Derive τ = pE sinθ

2️⃣ Graph Based Questions

For spherical shell:

Graph of E vs r:

For r < R → E = 0

For r > R → E decreases as 1/r²

 3️⃣ Source Based Questions

These test:

Charge interaction

Field line properties

Gauss theorem understanding

Conceptual applications

Example:
Two charges inside sphere → flux depends only on enclosed charge.