Chapter 23 Problem 35

Problem

A flat ring of inner radius $R_{1}$ and outer radius $R_{2}$ carries a uniform surface charge density $\sigma$. Determine the electric potential at points along the x axis

Solution

The disc from the chapter

This is just a disc removed from the full disc we already solved in the chapter. Let's recycle things from there.

$dq=\sigma dA=\sigma (2\pi RdR)$

$V={\frac {1}{4\pi \epsilon _{0}}}\int _{R_{1}}^{R_{2}}{\frac {dq}{r}}={\frac {1}{4\pi \epsilon _{0}}}\int _{R_{1}}^{R_{2}}{\frac {\sigma (2\pi RdR)}{\sqrt {x^{2}+R^{2}}}}$

$={\frac {\sigma }{2\epsilon _{0}}}\int _{R_{1}}^{R_{2}}{\frac {R}{\sqrt {x^{2}+R^{2}}}}dR$

$x^{2}+R^{2}=u\rightarrow 2RdR=du\rightarrow RdR={\frac {1}{2}}du$

$={\frac {\sigma }{2\epsilon _{0}}}\int {\frac {1}{2}}u^{-1/2}du$

$V=\left.{\frac {\sigma }{2\epsilon _{0}}}(x^{2}+R^{2})^{1/2}\right|_{R_{1}}^{R_{2}}$

$V={\frac {\sigma }{2\epsilon _{0}}}\left({\sqrt {x^{2}+{R_{2}}^{2}}}-{\sqrt {x^{2}+{R_{1}}^{2}}}\right)$

i.e. you just remove the inner hole's contribution from the total solution