Electrostatics: Fields, Potentials, Gauss's Theorem and Capacitors
Electrostatics opens the DHSE Plus Two course and carries heavy weight in the board paper: Coulomb's law, electric field and potential, dipoles, Gauss's theorem, conductors and capacitors. Questions usually pair a definition with a sign-sensitive calculation.
Core rules
- Coulomb's law in free space: F = (1/4πε₀)(q₁q₂/r²), with 1/4πε₀ = 9 × 10⁹ N m² C⁻². In a medium of dielectric constant K the force falls to F/K.
- The field inside a conductor in electrostatic equilibrium is zero; charge sits on the outer surface and E just outside is σ/ε₀, normal to the surface.
- Potential is a scalar: V = (1/4πε₀)(q/r). Field is the negative gradient, E = −dV/dr.
- Gauss's theorem: φ = ∮E·dS = q(enclosed)/ε₀. Flux depends only on the enclosed charge, not on its position or on outside charges.
- Capacitance C = Q/V; parallel plate C = ε₀A/d, becoming K times larger with a dielectric slab; series and parallel combine oppositely to resistors.
- Energy stored U = ½CV² = ½QV = Q²/2C.
Worked example
Charges +4 μC and −4 μC are 0.20 m apart. At the midpoint each is 0.10 m away, and both fields point from + to −, so they add: E = 2 × (9×10⁹ × 4×10⁻⁶)/(0.10)² = 2 × 3.6×10⁶ = 7.2×10⁶ N C⁻¹, directed toward the negative charge. The potential there, however, is zero because the two scalar contributions cancel.
Traps
Forgetting that potential adds with sign while field adds as a vector; assuming E = 0 implies V = 0 (false inside a charged shell); using r² instead of r for the axial dipole field; ignoring the dielectric when combining capacitors.