Electrostatic Potential and Capacitance — Previous Year Questions | eduPhysics
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Class XII · Chapter 02

Electrostatic Potential and Capacitance — Exam Practice

Practice questions in CBSE, NEET, and JEE Main style covering potential, equipotential surfaces, capacitance, and energy stored in capacitors.

📝 10 questions 🎯 CBSE · NEET · JEE Main
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.

1CBSE Board Exam Style

CBSE · 1 Mark
Q1. Why can two equipotential surfaces never intersect each other?
Answer If two equipotential surfaces intersected, the point of intersection would simultaneously have two different potential values — but potential at any point in space has a single, unique value, so this is impossible.
CBSE · 3 Marks
Q2. Derive an expression for the electric potential at a point due to a single point charge Q, at a distance r from it.
Answer Potential V at distance r is the work done per unit charge in bringing a test charge from infinity to that point: V = −∫∞r E · dr.

For a point charge, E = kQ/x² (radially). So V = −∫∞r (kQ/x²) dx = kQ[1/x]∞r = kQ/r.
CBSE · 5 Marks
Q3. (a) Derive the expression for energy stored in a parallel plate capacitor. (b) A parallel plate capacitor of capacitance 2 μF is charged to 100 V, then isolated. A dielectric of dielectric constant K = 3 is now inserted between the plates. Find the energy stored before and after inserting the dielectric.
Answer (a) Work done to add a small charge dq at potential V = q/C is dW = (q/C)dq. Total work to charge from 0 to Q: W = ∫₀Q (q/C)dq = Q²/2C. Using Q = CV, this can equivalently be written U = ½CV² = ½QV.

(b) Before: U₀ = ½CV² = ½ × 2×10⁻⁶ × (100)² = 0.01 J.
Since the capacitor is isolated, charge Q stays constant: Q = CV = 2×10⁻⁶ × 100 = 2×10⁻⁴ C. New capacitance: C′ = KC = 6×10⁻⁶ F.
New energy: U′ = Q²/(2C′) = (2×10⁻⁴)² / (2×6×10⁻⁶) ≈ 3.33×10⁻³ J.
Energy decreases — the field does positive work pulling the dielectric in, at the expense of stored energy.

2NEET Style (Single-Correct MCQ)

NEET Style
Q4. The capacitance of a parallel plate capacitor is independent of:
Plate area
Plate separation
✓ Charge on the plates
Dielectric constant of the medium
Solution C = ε₀A/d (or Kε₀A/d with a dielectric) depends only on geometry and the medium — not on how much charge or voltage is actually applied. C = Q/V stays constant because Q and V change proportionally.
NEET Style
Q5. The electric potential on the equatorial line of an electric dipole is:
Maximum
✓ Zero
Equal to the axial value
Negative infinity
Solution V = kp cosθ/r², and θ = 90° on the equatorial line, giving cos90° = 0 — so V = 0 there, even though the field itself is not zero.
NEET Style
Q6. Two identical capacitors, each of capacitance C, are connected in series. The equivalent capacitance is:
2C
✓ C/2
C
4C
Solution 1/Ceq = 1/C + 1/C = 2/C → Ceq = C/2.
NEET Style
Q7. The work done in moving a charge along an equipotential surface is:
Maximum
✓ Zero
Dependent on the path taken
Infinite
Solution W = qΔV. Since every point on an equipotential surface has the same potential, ΔV = 0, so no work is done moving a charge along it.

3JEE Main Style

JEE Main Style · Numerical
Q8. Three capacitors of 2 μF, 3 μF, and 6 μF are connected in series. Find the equivalent capacitance.
Solution 1/C = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 → C = 1 μF.
JEE Main Style · MCQ
Q9. A conducting sphere of radius R carries charge Q. The electric potential at a point inside the sphere, at distance r < R from the centre, is:
kQ/r
✓ kQ/R (constant)
Zero
kQr/R²
Solution Since E = 0 everywhere inside a conductor, potential doesn't change inside it — the entire conductor sits at the same potential as its surface, kQ/R. Field being zero does not mean potential is zero; it means potential is constant.
JEE Main Style · Numerical
Q10. A 4 μF capacitor is charged to 50 V, then disconnected from the source and connected to an uncharged 2 μF capacitor. Find the common potential and the loss in energy.
Solution Charge is conserved: Q = C₁V₁ = 4 × 50 = 200 μC.

Common potential: V = Q/(C₁+C₂) = 200/6 ≈ 33.3 V.

Initial energy: U₁ = ½C₁V₁² = ½ × 4 × 2500 = 5000 μJ.
Final energy: U₂ = ½(C₁+C₂)V² = ½ × 6 × (33.3)² ≈ 3333 μJ.
Loss in energy ≈ 5000 − 3333 = 1667 μJ (≈1.67×10⁻³ J) — dissipated as heat and a small amount of radiation during the brief redistribution current.
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