Electrostatics: Charge, Field, Potential and Capacitors
What this domain tests
Electrostatics links charge, field and potential with Gauss's law and capacitors. Numericals usually combine Coulomb's law with superposition, or ask for capacitance, stored energy and the effect of a dielectric.
Core rules
- Coulomb's law: F = kq₁q₂/r², with k = 9×10⁹ N m² C⁻² and ε₀ = 8.85×10⁻¹² C² N⁻¹ m⁻²; valid for point charges at rest.
- Field E = F/q₀; potential V = W/q₀ is a scalar and adds algebraically; E = −dV/dr.
- Inside a conductor E = 0 and V is constant; just outside the surface E = σ/ε₀.
- Gauss's law: ∮E·dS = q_inside/ε₀ - choose a surface where E is constant or parallel to dS.
- Capacitors: series 1/C = Σ1/Cᵢ (same charge), parallel C = ΣCᵢ (same voltage); U = ½CV² = Q²/2C; a dielectric of constant K multiplies C by K.
Worked example
A parallel-plate capacitor has plate area 2×10⁻² m², separation 1 mm, air between the plates, and is charged to 100 V. Find C and the stored energy. C = ε₀A/d = (8.85×10⁻¹²)(2×10⁻²)/(1×10⁻³) = 1.77×10⁻¹⁰ F ≈ 177 pF. Then U = ½CV² = ½(1.77×10⁻¹⁰)(100)² = 8.85×10⁻⁷ J. Convert the separation to metres first; leaving 1 mm as 1 makes the answer 1000 times too large.
Traps that cost marks
- Adding potentials as vectors - potentials are scalars; only fields add by components.
- Forgetting that a dielectric reduces E inside but raises C by K at fixed charge.
- Assuming series capacitors share voltage equally; they share charge.
- Writing U = CV instead of ½CV².