Electric field and electrostatic potential
This domain tests Coulomb's law, the superposition of electric fields and potentials, and the use of Gauss's law for symmetric charge distributions. PAU questions frequently ask for a field vector at a point, a potential difference, or the motion of a charged particle.
Core rules
- Coulomb: F = k|q₁q₂|/r² with k = 1/(4πε₀) ≈ 9.0×10⁹ N·m²/C². The force is along the line of centres: repulsive for like charges, attractive for unlike.
- Field of a point charge: E = kq/r², radially outward for q > 0 and inward for q < 0. Superposition is vector addition.
- Potential of a point charge: V = kq/r, a scalar. Superposition is algebraic, so positive and negative contributions can cancel.
- Uniform field: E = -ΔV/d for parallel plates; work by the field is qΔV, and a positive charge moves toward lower potential.
- Gauss: Φ = ∮E·dA = Q_enclosed/ε₀. Outside a spherical shell the field is that of a point charge at the centre; inside an ideal conducting shell the field is zero. A line charge gives E = λ/(2πε₀r); an infinite sheet gives E = σ/(2ε₀).
Worked example
Charges q₁ = +2.0 μC at x = 0 and q₂ = -2.0 μC at x = 0.30 m. At the midpoint x = 0.15 m, the field from q₁ points +x and has magnitude (9.0×10⁹)(2.0×10⁻⁶)/(0.15²) = 8.0×10⁵ N/C. The field from q₂ points toward the negative charge, also +x, with the same magnitude. Total E = 1.6×10⁶ N/C in +x. The potentials are equal and opposite, so V = 0 at that point even though E is not zero.
Traps
Do not add fields algebraically unless they are collinear; resolve components. Do not confuse potential with potential energy: U = qV. A zero field does not imply zero potential, and zero potential does not imply zero field. Gauss's law is useful only when symmetry makes E constant over the surface; the enclosed charge, not the total charge, controls the flux. Inside a conductor in electrostatic equilibrium E = 0 and the whole conductor is equipotential.