Electrostatics and Capacitors
What this domain tests
Electrostatics questions test Coulomb's law, electric field and potential, and capacitor behaviour in DC circuits. You should be able to calculate capacitance from plate geometry, combine capacitors, and find stored charge and energy.
Core rules
- Coulomb's law: F = kQ1Q2/r², with k ≈ 9 × 10⁹ N m² C⁻² in air; force acts along the line joining charges and reverses for opposite signs.
- Electric field E = F/Q, measured in V/m or N/C; potential V = W/Q.
- Capacitance C = Q/V. For parallel plates, C = ε0εrA/d, where ε0 = 8.854 × 10⁻¹² F/m.
- Capacitors in parallel add (C = C1 + C2); in series the reciprocals add, so the total is smaller than the smallest capacitor.
- Stored energy = ½CV² = ½QV = Q²/2C. A capacitor blocks steady DC but passes changing signals; voltage across a capacitor cannot change instantly.
Worked example
A parallel-plate capacitor has area 0.02 m², separation 1 mm and a dielectric of εr = 5. C = ε0εrA/d = (8.854 × 10⁻¹² × 5 × 0.02)/0.001 = 8.854 × 10⁻¹⁰ F, about 886 pF. At 100 V, Q = CV = 8.854 × 10⁻⁸ C and energy = ½CV² = 4.43 × 10⁻⁶ J. Note that reducing the spacing to 0.5 mm would double the capacitance because d is in the denominator.
Traps that cost marks
Using microfarads as farads without converting; adding series capacitors directly; thinking a capacitor dissipates energy as heat rather than storing it; and forgetting that the dielectric strength, not just εr, limits the safe voltage.