Electrostatic Potential and Capacitance

1. Introduction

Electric potential is the scalar counterpart of the electric field. Capacitance measures the ability of a system to store charge and energy — essential for understanding circuits and electronic devices.

2. Electric Potential

2.1 Potential Due to a Point Charge

V = (1/4πε₀) × q/r

2.2 Potential Difference

V_B - V_A = -∫_A^B E · dl

The potential at a point is the work done per unit charge to bring a test charge from infinity to that point.

2.3 Potential Due to a Dipole

V = (1/4πε₀) × p cos θ/r²

At axial point (θ = 0): V = p/(4πε₀r²). At equatorial point (θ = 90°): V = 0.

2.4 Potential Due to a Charge Distribution

V = (1/4πε₀) × Σ qᵢ/rᵢ for discrete charges. V = (1/4πε₀) × ∫ dq/r for continuous distributions.

3. Equipotential Surfaces

Surfaces where potential is constant. The electric field is perpendicular to equipotential surfaces. No work is done moving a charge along an equipotential.

4. Capacitance

4.1 Capacitor

A device that stores charge and energy. C = Q/V.

4.2 Parallel Plate Capacitor

C = ε₀A/d (with vacuum between plates).

4.3 Effect of Dielectric

C = εᵣC₀ = KC₀, where K is the dielectric constant.

Inside a dielectric, E = E₀/K and induced charge appears on the dielectric surfaces.

4.4 Combinations

Series: 1/C_eq = 1/C₁ + 1/C₂ + 1/C₃ + ... Parallel: C_eq = C₁ + C₂ + C₃ + ...

4.5 Energy Stored in a Capacitor

U = (1/2)QV = (1/2)CV² = (1/2)Q²/C

Energy density (between plates): u = (1/2)ε₀E²

5. Van de Graaff Generator

A device that produces very high voltages (up to several million volts) using electrostatic principles.

Principle: Charge is transferred to a large hollow conducting sphere by a moving belt. The charge accumulates on the outer surface, producing a high potential.

Construction: A motor-driven insulating belt passes over pulleys. Sharp spray points (combs) spray charge onto the belt at the bottom, and a collector comb transfers charge to the hollow metallic sphere at the top.

Applications: Accelerating charged particles for nuclear physics experiments.

6. Dielectrics and Polarisation

6.1 Polar and Non-Polar Molecules

Non-polar molecules: Centre of positive and negative charges coincide. In an external field, induced dipole moment develops.

Polar molecules: Permanent dipole moment (e.g., H₂O). In an external field, dipoles align partially, producing polarisation.

6.2 Polarisation Vector

P = total dipole moment per unit volume.

For a linear dielectric: P = ε₀χₑE, where χₑ is electric susceptibility.

6.3 Dielectric Constant

K = ε/ε₀ = 1 + χₑ. Relation between dielectric constant and susceptibility.

7. Energy Stored in a Capacitor

U = (1/2)QV = (1/2)CV² = (1/2)Q²/C

Energy density (between plates): u = (1/2)ε₀E²

6. Worked Problems

Problem 1: Three capacitors 2 μF, 3 μF, and 6 μF are connected in series across 100V. Find total capacitance and charge on each. Solution: 1/C = 1/2 + 1/3 + 1/6 = (3+2+1)/6 = 1. So C = 1 μF. Q = CV = 1 × 10^{-6} × 100 = 10^{-4} C. In series, charge is same on each = 100 μC.

Problem 2: A parallel plate capacitor has plates of area 100 cm² separated by 1 mm with air. Find capacitance. If a dielectric of K = 4 fills the gap, find new capacitance. Solution: C₀ = ε₀A/d = (8.85×10^{-12} × 100×10^{-4})/10^{-3} = 8.85×10^{-11} F = 88.5 pF. With dielectric: C = KC₀ = 354 pF.

Problem 3: A 10 μF capacitor is charged to 100V and then disconnected. It is then connected to an uncharged 20 μF capacitor. Find final voltage and energy loss. Solution: Initial Q = CV = 10×10^{-6}×100 = 10^{-3} C. C_eq = 30 μF. V_final = 10^{-3}/(30×10^{-6}) = 33.3V. Initial U = ½×10^{-5}×10⁴ = 0.05 J. Final U = ½×30×10^{-6}×(33.3)² = 0.0167 J. Loss = 0.0333 J.

7. Common Mistakes

'Students often confuse formulas for series and parallel combinations. In series, charge is same. In parallel, voltage is same.'

'When a dielectric is inserted in a charged capacitor disconnected from battery, Q stays constant but V changes. When connected, V stays constant but Q changes.'

8. ISC Exam Focus

9. Self-Test Questions

1. Find the potential at the centre of a square of side 1 m with charges 2, -2, 3, -3 μC at vertices.
2. Two capacitors of 4 μF and 8 μF are connected in series across 200V. Find charge and voltage across each.
3. Derive the expression for energy stored in a capacitor.
4. A parallel plate capacitor is charged and then disconnected from the battery. If the plates are pulled apart, what happens to charge, capacitance, voltage, and energy?
5. Find the equivalent capacitance between points A and B with capacitors 2, 3, 6 μF in delta configuration.