Question 3.2: A point charge q is situated a distance a from the center of...

Examine the completely different configuration, consisting of the point charge q ogether with another point charge

\acute{q}=-\frac{R}{a}q,                 (3.15)

placed a distance

b=\frac{R^{2}}{a},                         (3.16)

to the right of the center of the sphere (Fig. 3.13). No conductor, now—just the two point charges. The potential of this configuration is

V(r)=\frac{1}{4\pi \epsilon _{0}}\left(\frac{q}{η }+\frac{\acute{q}}{\acute{η}} \right) ,                 (3.17)

where η and \acute{η} are the distances from q and \acute{q}, respectively. Now, it happens (see Prob. 3.8) that this potential vanishes at all points on the sphere, and therefore fits the boundary conditions for our original problem, in the exterior region.^{7} Conclusion: Eq. 3.17 is the potential of a point charge near a grounded conducting sphere. (Notice that b is less than R, so the “image” charge \acute{q} is safely inside the sphere—you cannot put image charges in the region where you are calculating V; that would change ρ, and you’d be solving Poisson’s equation with 7This solution is due to William Thomson (later Lord Kelvin), who published it in 1848, when he was just 24. It was apparently inspired by a theorem of Apollonius (200 BC) that says the locus of points with a fixed ratio of distances from two given points is a sphere. See J. C. Maxwell, “Treatise on Electricity and Magnetism, Vol. I,” Dover, New York, p. 245. I thank Gabriel Karl for this interesting history.

the wrong source.) In particular, the force of attraction between the charge and the sphere is

F=\frac{1}{4\pi \epsilon _{0}}\frac{q\acute{q}}{\left(a-b\right)^{2} } =\frac{1}{4\pi \epsilon _{0}}\frac{q^{2}Ra}{\left(a^{2}-b^{2}\right)^{2} }           (3.18)