Potential Due to a Continuous Charge Distribution — Question 7

Question 7

A shell integral explains constant interior potential directly

A thin spherical shell of radius R carries uniform charge Q. Let the observation distance from its center be r>0 and use infinity reference. A surface ring at polar angle $\theta$ has $dq=(Q/2)\sin\theta\,d\theta$.

1. Integrate the potential using source-observer distance $s=\sqrt{R^2+r^2-2Rr\cos\theta}$.

2. Reduce the result for r<R and r>R, and check the center and surface limits.

Question 7 - Solution

A shell integral explains constant interior potential directly

(a) Method and calculation

$$V(r)=\frac{kQ}{2}\int_0^\pi\frac{\sin\theta\,d\theta}{\sqrt{R^2+r^2-2Rr\cos\theta}}.$$$$s\,ds=Rr\sin\theta\,d\theta\Rightarrow V=\frac{kQ}{2Rr}[(R+r)-|R-r|].$$

(b) Interpretation and further calculation

$$V(r)=\begin{cases}kQ/R&r<R,\\kQ/r&r>R.\end{cases}$$$$V(0)=\lim_{r\to0}V=kQ/R,\quad V(R^-)=V(R^+)=kQ/R.$$

Physical check

The absolute value in the lower distance limit is essential; omitting it incorrectly extends the exterior formula into the shell.