Class 12 — Electrostatic Potential and Capacitance

Chapter Overview & Weightage

Electrostatic Potential and Capacitance is one of the heaviest chapters in CBSE Class 12 Physics, fetching $7$–$8$ marks consistently across the last five years. It combines a conceptual layer (potential, equipotential surfaces) with a strong numerical layer (capacitor combinations, energy stored).

The chapter overlaps with the Electrostatics chapter from Class 11 NCERT (Chapter 1) — together they form a $13$–$15$ mark unit.

Key Concepts You Must Know

Electric Potential ($V$)

• Work done per unit positive test charge to bring it from infinity to a point.
• Scalar; SI unit volt = J/C.
• $V = \tfrac{kq}{r}$ for a point charge; signs matter.

Equipotential Surfaces

• All points at the same $V$.
• Always perpendicular to electric field lines.
• No work to move a charge along an equipotential.

Potential Energy of a System of Charges

• Sum over all pairs: $U = \tfrac{1}{4\pi\varepsilon_0}\sum_{i<j} \tfrac{q_iq_j}{r_{ij}}$.

Conductors in Electrostatic Equilibrium

• $E = 0$ inside the conductor.
• Whole conductor is an equipotential.
• Surface charge density highest at points (lightning rod).

Capacitance

• $C = Q/V$. SI unit farad.
• Parallel plate capacitor: $C = \tfrac{\varepsilon_0 A}{d}$.

Combinations

• Series: $\tfrac{1}{C_{\text{eq}}} = \sum \tfrac{1}{C_i}$.
• Parallel: $C_{\text{eq}} = \sum C_i$.

Energy Stored

• $U = \tfrac{1}{2}CV^2 = \tfrac{Q^2}{2C}$.

Effect of Dielectrics

• Capacitance increases by factor $K$ (dielectric constant).

Important Formulas

Solved Previous Year Questions

PYQ 1 — Capacitor Combination (CBSE 2024, $5$ marks)

Three capacitors of capacitances $C_1 = 2 \mu F$, $C_2 = 3 \mu F$ and $C_3 = 6 \mu F$ are connected such that $C_1$ and $C_2$ are in parallel and this combination is in series with $C_3$. Find equivalent capacitance and energy stored when connected to a $30$ V battery.

Solution:

Parallel: $C_{12} = C_1 + C_2 = 5 \mu F$.

Series of $C_{12}$ and $C_3$: $\tfrac{1}{C_{\text{eq}}} = \tfrac{1}{5} + \tfrac{1}{6} = \tfrac{11}{30}$. So $C_{\text{eq}} = 30/11 \mu F \approx 2.73 \mu F$.

Energy: $U = \tfrac{1}{2}C_{\text{eq}}V^2 = \tfrac{1}{2} \times \tfrac{30}{11} \times 10^{-6} \times 900 = \tfrac{27000}{22} \times 10^{-6} \approx 1.227 \text{ mJ}$.

PYQ 2 — Potential Due to Multiple Charges (CBSE 2023, $3$ marks)

Two point charges $+5 \mu C$ and $-5 \mu C$ are placed $20$ cm apart. Find the potential at the midpoint of the line joining them.

Solution:

At the midpoint, distance from each charge $= 10$ cm $= 0.1$ m.

The potentials from the two charges cancel exactly at the midpoint — even though the field there is non-zero.

PYQ 3 — Dielectric Insertion (CBSE 2022, $5$ marks)

A parallel plate capacitor of capacitance $C$ is charged by a battery of voltage $V$. The battery is disconnected, then a dielectric of constant $K$ is inserted. Find the new (a) capacitance, (b) charge, (c) potential difference, (d) energy.

Solution:

(a) New capacitance: $C' = KC$.

(b) Charge stays $Q = CV$ (battery disconnected, charge isolated).

(c) New potential: $V' = Q/C' = CV/(KC) = V/K$.

(d) New energy: $U' = Q^2/(2C') = (CV)^2/(2KC) = CV^2/(2K) = U/K$.

So energy decreases by factor $K$ — work was done by the system pulling the dielectric in.

Difficulty Distribution

Expert Strategy

Week 1 — Concept first, formulas second: Understand why equipotential surfaces are perpendicular to field lines, why a conductor’s interior has zero field. These conceptual points are worth $1$–$2$ marks each in the short-answer section.

Week 2 — Capacitor combinations: Practise $20$ combination problems mixing series, parallel, dielectrics. Watch for “battery connected” vs “battery disconnected” — they change the formula you use.

Week 3 — Numerical fluency: Solve every NCERT exercise problem and the past $5$ years of board PYQs. The pattern stabilises after about $30$ problems.

Common Traps

Quick Revision Card

• $V_A - V_B = -\int_A^B \vec{E} \cdot d\vec{l}$. Potential is the line integral of $-E$.
• Equipotential surfaces are perpendicular to $\vec{E}$.
• $C = \varepsilon_0 A/d$ for parallel plate. $K$ multiplies it when filled with dielectric.
• Battery connected $\Rightarrow$ $V$ constant. Battery disconnected $\Rightarrow$ $Q$ constant.
• $U = \tfrac{1}{2}CV^2 = Q^2/(2C)$.

This chapter rewards careful, methodical work. There are no shortcuts — but there are also no surprises.