Question

A conductor at potential V = 0 has the shape of an infinite plane except for a hemispherical bulge of radius a. A charge q is placed above the center of the bulge, a distance p from the plane. What is the force on the charge?

Answer 1

Let the z-axis be the symmetry axis.

Placing image charges -q at z = - p,

q' = -qR/p at z = R²/p and -q' = qR/p at z = -R²/p

on the z-axis makes the surface of the conductor an equipotential surface. The field everywhere outside the conductor is the same as that due to q and the image charges (uniqueness theorem). The force on q therefore is
F = k(-q²/(4p²) + qq'/(p - R²/p)² - qq'/(p + R²/p)²)k

= -kq²(1/(4p²) + 4R^3p3/(p⁴ - R⁴)²)k.
The charge is pulled towards the conductor.