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 = R^{2}/p and -q' = qR/p at z =
-R^{2}/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(-q^{2}/(4p^{2}) + qq'/(p -
R^{2}/p)^{2} - qq'/(p +
R^{2}/p)^{2})**k**

= -kq^{2}(1/(4p^{2}) + 4R^{3p}^{3}/(p^{4} -
R^{4})^{2})**k**.

The charge is pulled towards the conductor.

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