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Electrostatic Potential and Capacitance
Physics · Class 12 · Part 1NCERT pp. 45–762 sets × 25 questionsFree
Key facts and formulas for Electrostatic Potential and Capacitance, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.
Formula sheet for this chapter
| Idea | Rule or fact |
|---|
| Potential of point charge | V = kQ/r, k = 9×10⁹ N m² C⁻²; scalar, so add with signs |
| Potential energy | U = qV; pair: U = kq₁q₂/r; system: sum over all pairs |
| Field from potential | E = −dV/dx; E is normal to equipotential surfaces |
| Dipole | Axis: V = kp/r²; equator: V = 0 but E ≠ 0; U = −pE cos θ |
| Shell and sphere | Outside: kQ/r; inside shell: kQ/R (constant); C = 4πε₀R |
| Parallel plates | C = ε₀A/d; with slab: C = ε₀A/(d − t + t/K); metal slab: ε₀A/(d − t) |
| Combinations | Series: 1/C = Σ1/Cᵢ; parallel: C = ΣCᵢ; series shares same Q |
| Energy | U = ½CV² = Q²/2C = ½QV; density u = ½ε₀E² |
| Battery vs isolated | Battery on: V fixed; battery off: Q fixed; sharing charge conserves Q, loses energy |
| Conductor | E = 0 inside, E = σ/ε₀ outside; force on plate F = Q²/(2ε₀A) |
Three sample questions
Q1. A positive test charge is moved slowly from a far point R to a nearer point P in the field of a fixed positive charge. The work done by the electric field is
- positive
- negative
- zero
- positive only if the path is straight
Hint and solution
Hint: Does the field push the test charge towards P or away from P? Compare that direction with the direction of motion.
- Force on +q from +Q is repulsive (outward) — The fixed positive charge pushes the test charge away from itself, so the field force points outward.
- Displacement is inward (R to P) — The charge is moved closer to the source, so the displacement is opposite to the force.
- W = F d cos 180° = −F d — When force and displacement are opposite, the work done by the field is negative. The outside agent does the equal positive work.
- Work by field is negative — The potential energy of the system rises, so the field loses energy. The answer is negative work. Option B.
Answer: B. The field repels the charge, but the charge moves inward. So the field does negative work, and the outside agent does the equal positive work.
Q2. The potential at 0.9 m from a point charge of 5 µC (1/4πε₀ = 9×10⁹ SI) is
- 4.5 × 10⁴ V
- 5 × 10⁴ N/C
- 5.6 × 10⁴ V
- 5 × 10⁴ V
Hint and solution
Hint: Use V = kQ/r with the charge in coulomb. Convert µC first.
- V = kQ / r — Potential of a point charge falls as 1/r. Convert 5 µC to 5×10⁻⁶ C.
- kQ = 9×10⁹ × 5×10⁻⁶ = 4.5×10⁴ — Multiply the constant and the charge first.
- V = 4.5×10⁴ / 0.9 = 5×10⁴ V — Divide by the distance 0.9 m.
- V = 5 × 10⁴ V — The potential is 5 × 10⁴ volt. Option D.
Answer: D. V = kQ/r = 9×10⁹ × 5×10⁻⁶ / 0.9 = 5×10⁴ V.
Q3. A charge of −3 nC is placed at a point where the potential is +60 V. Its potential energy is
- −180 nJ
- +180 nJ
- −20 nJ
- +20 nJ
Hint and solution
Hint: U = qV, and keep the sign of the charge.
- U = qV — Potential energy of a charge at a point equals charge times potential.
- U = (−3 nC)(+60 V) — Put the signed values. The charge is negative and the potential is positive.
- U = −180 nJ — 3 × 60 is 180, and a negative times a positive is negative.
- U = −180 nJ — So the potential energy is minus 180 nanojoule. Option A.
Answer: A. U = qV = (−3 nC)(60 V) = −180 nJ.
How to use this chapter
Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.