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Electrostatic Potential and Capacitance – Class 12 Physics notes, important questions and MCQs

NCERT Class 12 Physics, chapter 2. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Electric potential is work done per unit charge; capacitors store charge and energy, and dielectrics increase capacitance.

Potential

Potential of a point charge V = kq/r; potential difference V = W/q.
Equipotential surfaces are perpendicular to field lines; no work is done along them.
E = −dV/dr; potential energy of two charges U = kq₁q₂/r.

Capacitors

C = Q/V (farad). Parallel plate: C = ε₀A/d; with dielectric C = Kε₀A/d.
Series: 1/C = 1/C₁ + 1/C₂; parallel: C = C₁ + C₂.
Energy stored U = ½CV² = Q²/(2C).

Key terms

Equipotential surface: surface on which potential is the same everywhere
Dielectric constant: factor K by which capacitance increases with a dielectric
Capacitance: ratio of charge stored to potential difference, C = Q/V (unit farad)

Formulas

V = kq/r
C = ε₀A/d
U = ½CV²

Important questions with answers

Why is the electric field always perpendicular to an equipotential surface?2 marks

Moving a charge along an equipotential surface needs no work (ΔV = 0). Work is zero only if the force (and field) is perpendicular to the displacement, so E is normal to the surface.

Derive the capacitance of a parallel plate capacitor and show how it changes with a dielectric of constant K.3 marks

Plates of area A, separation d, charge ±Q. Field between plates E = σ/ε₀ = Q/(Aε₀). V = Ed = Qd/(Aε₀). C = Q/V = ε₀A/d. With a dielectric filling the gap, the field reduces to E/K, so V becomes V/K and C = Kε₀A/d, K times larger.

Derive the expression for energy stored in a capacitor and the energy density of the electric field. Two capacitors 4 µF and 6 µF are connected in series to a 200 V supply; find the charge and the energy stored in each.5 marks

Charging from 0 to Q: dW = V dq = (q/C) dq, so W = ∫₀^Q q/C dq = Q²/2C = ½CV² = ½QV. For a parallel plate capacitor, U = ½ε₀(A/d)(Ed)² = ½ε₀E² × (Ad), so energy density u = ½ε₀E². Series: C = (4 × 6)/(4 + 6) = 2.4 µF; Q = CV = 2.4 × 10⁻⁶ × 200 = 4.8 × 10⁻⁴ C (same on each). V₁ = Q/4 µF = 120 V, V₂ = Q/6 µF = 80 V. U₁ = ½QV₁ = 0.0288 J, U₂ = ½QV₂ = 0.0192 J (total 0.048 J).

Quick revision

Field is perpendicular to equipotentials.
Dielectric increases capacitance K times.
Energy ½CV².

Exam tips

Capacitors in series carry the same charge; in parallel they share the same voltage.
When a dielectric slab is inserted with the battery connected, V stays the same and Q increases; with the battery removed, Q stays the same and V decreases.

Practice MCQs

SI unit of capacitance is
1. farad
2. volt
3. coulomb
4. henry
Show answer
Answer: (a) farad. Farad.
Work done in moving a charge on an equipotential surface is
1. maximum
2. infinite
3. negative
4. zero
Show answer
Answer: (d) zero. No potential difference.
Two 4 µF capacitors in series give
1. 8 µF
2. 2 µF
3. 4 µF
4. 16 µF
Show answer
Answer: (b) 2 µF. 4 × 4/(4 + 4).
Energy stored in a 10 µF capacitor at 100 V is
1. 5 J
2. 0.5 J
3. 0.05 J
4. 0.1 J
Show answer
Answer: (c) 0.05 J. ½ × 10⁻⁵ × 10⁴.
Three capacitors of 6 µF each in series have equivalent capacitance
1. 2 µF
2. 18 µF
3. 6 µF
4. 3 µF
Show answer
Answer: (a) 2 µF. 1/C = 3/6.
A dielectric of K = 5 is inserted fully in a parallel plate capacitor (battery connected). Capacitance becomes
1. same
2. 1/5
3. 5 times
4. 25 times
Show answer
Answer: (c) 5 times. C' = KC.
Potential at the centre of a uniformly charged spherical shell of radius R and charge Q is
1. kQ/R
2. zero
3. kQ/R²
4. 2kQ/R
Show answer
Answer: (a) kQ/R. Same as on the surface.
A 10 µF capacitor charged to 100 V stores energy
1. 0.01 J
2. 0.5 J
3. 5 J
4. 0.05 J
Show answer
Answer: (d) 0.05 J. ½CV² = ½ × 10⁻⁵ × 10⁴.

All chapters of Class 12 Physics

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.