Electric field and electrostatic interactions
The electrostatics block tests Coulomb's law, the field and potential of point charges, superposition, and the behaviour of conductors and charged spheres. The PAU normally pairs a vector calculation (field at a point) with a scalar one (potential or energy), because that contrast is exactly where candidates slip.
Core rules
- Coulomb: F = kq₁q₂/r² with k = 9.0×10⁹ N m² C⁻²; the sign of the product tells you attraction or repulsion.
- Field of a point charge: E = kq/r², a vector pointing away from positive charge and towards negative charge.
- Potential: V = kq/r, a scalar that keeps the sign of q. Superposition means adding potentials algebraically but adding fields as vectors.
- Work done by the field on q moving between two points: W = q(V₁ − V₂); an external agent does the opposite amount.
- Uniform field between parallel plates: E = V/d, force qE, giving parabolic motion like a projectile.
- Gauss's law: flux through a closed surface equals Q_enclosed/ε₀. Hence E = 0 inside a conductor in equilibrium and E = kQ/r² outside a charged sphere.
Worked example
Charges q₁ = +3.0 µC at x = 0 and q₂ = −1.0 µC at x = 0.20 m. Where is the field zero? Between them the two fields point the same way, so the null point must lie outside, on the side of the smaller charge: x > 0.20 m. Equating magnitudes, 3/x² = 1/(x − 0.20)², so √3(x − 0.20) = x and x = 0.346/(√3 − 1) ≈ 0.47 m. The potential at that point is not zero, because potentials add as scalars and do not cancel there.
Marks lost here
Treating E and V as interchangeable: a point may have zero field and non-zero potential, or the reverse. Dropping the sign of the potential of a negative charge. Forgetting to add perpendicular fields as vectors rather than numerically. Using E = V/d outside a region where the field is uniform. Quoting the potential energy of a pair as kq₁q₂/r but then ignoring the sign when discussing binding.