Practice question

**Relation E = -∇V** shows field points down potential gradient. For system of opposite charges close together, equipotential near midpoint between them has V≈0, but shape distorted, not spherical, reflecting superposition of potentials V = k q₁/r₁ + k q₂/r₂. If the potential difference between two points is zero, they lie on the same equipotential surface. The path taken by the particle must lie entirely on this equipotential surface (where V is constant), as any deviation would imply a potential difference. Equipotential surfaces are perpendicular to the electric field, so the particle moves perpendicular to the field direction, doing no work against the field along