Short notes

Electrostatics Short Notes

Inside a parallel-plate capacitor with plate area A = 0.01 m² and separation d = 1 mm filled with a dielectric of εr = 4, the electric field is uniform at E = V/d = 100 V / 0.001 m = 100 kV/m. That field is the electrostatic quantity that determines breakdown voltage, capacitance (C = ε₀εr·A/d = 354 pF), and the energy stored in the dielectric. Everything in electrostatics — from charge distributions to boundary conditions — ultimately comes down to calculating this field.

EEE, ECE, EI

How it works

Coulomb's law gives the force between two point charges q₁ and q₂ separated by r as F = q₁q₂/(4πε₀r²) in free space. The electric field E = F/q = Q/(4πε₀r²) r̂. Gauss's law in integral form: ∮E·dS = Q_enclosed/ε₀. This is most powerful for symmetric geometries — a uniformly charged sphere of radius R with total charge Q gives E = Q/(4πε₀r²) outside (r > R) and E = Qr/(4πε₀R³) inside. Electric potential V = −∫E·dl; for a point charge, V = Q/(4πε₀r). Poisson's equation is ∇²V = −ρ_v/ε; Laplace's equation is ∇²V = 0 where ρ_v = 0.

Key points to remember

The boundary conditions at a dielectric interface: normal component of D is continuous if no surface charge (D₁ₙ = D₂ₙ); tangential component of E is always continuous (E₁ₜ = E₂ₜ). For a conductor surface, E is entirely normal and E_tangential = 0 inside. Energy density in the electric field is w_e = ½ε|E|² J/m³. The capacitance of a spherical capacitor with inner radius a and outer radius b is C = 4πε₀εr·ab/(b−a). Dipole moment p = Qd determines the torque in an external field as τ = p × E. Electric flux density D = ε₀εr·E.

Exam tip

The examiner always asks you to apply Gauss's law to a coaxial cable or spherical charge distribution and find E at various radii — draw the Gaussian surface, write the integral form, and state the enclosed charge explicitly before evaluating.

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