Method of Images: Sphere

Chapter 3 — Conductors

A point charge near a grounded conducting sphere also admits an image solution — a single well-chosen image charge, placed inside the sphere, reproduces the correct boundary condition. The answer isn't quite as clean as the plane (the image has a weird magnitude and location), but the method is the same.

Setup

A grounded conducting sphere of radius \(R\) is centered at the origin. A point charge \(q > 0\) is placed on the \(z\)-axis at \(z = a\), with \(a > R\). We want the field outside the sphere (and the force on \(q\)).

Where to put the image

By the uniqueness theorem, anything we put inside the sphere is fine as long as (i) the sources in the outside region are unchanged, and (ii) the sphere surface is still at \(V = 0\). Axial symmetry suggests a single image charge on the \(z\)-axis at some \(z = b\) with magnitude \(-Q\) (\(b < R\) so it's inside).

Enforce the boundary condition

Pick a point \(P\) on the surface at angle \(\theta\) from the \(z\)-axis. Distance from \(P\) to real charge: \(\sqrt{R^2 + a^2 - 2Ra\cos\theta}\). Distance from \(P\) to image: \(\sqrt{R^2 + b^2 - 2Rb\cos\theta}\). Setting \(V_P = 0\):

\[\frac{q}{\sqrt{R^2 + a^2 - 2Ra\cos\theta}} = \frac{Q}{\sqrt{R^2 + b^2 - 2Rb\cos\theta}}.\]

This has to hold for every \(\theta\). Squaring and matching constant terms versus \(\cos\theta\) terms gives two equations:

\[2Rbq^2 = 2RaQ^2,\qquad q^2(R^2+b^2) = Q^2(R^2+a^2).\]

Solving (discard the trivial \(b = a\) root) yields

\[\boxed{\ Q = \frac{R}{a}\,q,\qquad b = \frac{R^2}{a}\ }.\]

So the image is a charge \(-qR/a\) at distance \(R^2/a\) from the center. Since \(a > R\), we have \(b < R\): the image does live inside, as required.

Force on the real charge

The force \(q\) feels from the induced charge on the sphere equals the Coulomb force from the image (valid outside the sphere). The distance between \(q\) and its image is \(a - b = a - R^2/a\), so

\[F = \frac{1}{4\pi\epsilon_0}\frac{q\cdot(-Q)}{(a-b)^2} = -\frac{1}{4\pi\epsilon_0}\frac{qQ}{(a - R^2/a)^2} = -\frac{1}{4\pi\epsilon_0}\frac{q^2 Ra}{(a^2 - R^2)^2}.\]

Attractive toward the sphere. Sanity checks:

• As \(a\to R^+\): denominator vanishes, \(F\to -\infty\). Makes sense (charge approaches its mirror image).
• As \(a\to\infty\): \(F\to 0\). The sphere looks like a point and the induced charges spread out to negligible effect.
• In the limit \(R\to\infty\) (with \(a = R + d\)): reduces to the grounded plane result \(F = -kq^2/(2d)^2\). ✓

Variants

Ungrounded, neutral sphere: add a second image of charge \(+Q = +qR/a\) at the center. The first image takes care of the boundary condition; the second gives the sphere total charge zero without changing the spherical equipotential (a point charge at the center keeps the sphere an equipotential, just at a nonzero potential).

Sphere held at potential \(V_0\): add a charge \(4\pi\epsilon_0 R V_0\) at the center to shift the boundary from \(0\) to \(V_0\).

Practice Problems