For the configuration of Prob. 2.16, find the potential difference between a point on the axis and a point on the outer cylinder. Note that it is not necessary to commit yourself to a particular reference point, if you use Eq. 2.22.

The line integral of electric field is known electric potential. The electric potential is described as,

$\mathbf{V}\left(r\right)\mathbf{=}\mathbf{-}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{r}}\mathbf{E}\mathbf{\times }\mathbf{dl}$

Here, V(r)is the potential at position r, Eis the electric field, and is the length of small element.

Consider the two points are aand b.

The electric potential difference between two points determined as,

$$\begin{gathered} V\left(b\right)-V\left(b\right)=-{\int }_0^{b}E-dl+{\int }_0^{a}E.dl \\ =-{\int }_0^{b}E-dl-{\int }_a^{0}E.dl \\ =-{\int }_a^{b}E-dl \end{gathered}$$

Here, V(a)is the potential at position a.

V(b) is the potential at position b.

Write the configuration of electric field for cylinder.

$E\left(s\right)=\left\{\begin{array}{cc}\frac{pS}{2{\epsilon }_0}& s<a\\ \frac{p{a}^2}{2s{\epsilon }_0}& a<s<b\\ 0,& s>b\end{array}\right.$

Here, ais the inner radius of the cylinder, bis the outer radius of the radius of sphere, sis the radius of the Gaussian surface, pis the volume charge density and ${\epsilon }_0$is the permittivity of the free space.

The electric potential is,

$$\begin{gathered} \nabla V=V\left(b\right)-V\left(a\right) \\ =-{\int }_a^{b}E\left(s\right)ds \end{gathered}$$

Now, substituting the value of electric field in above equation,

$V\left(b\right)-V\left(a\right)=-{\int }_{s-b}^{s-a}\frac{p{a}^2}{2s{\epsilon }_0}ds-{\int }_{s-a}^{s-0}\frac{ps}{2{\epsilon }_0}ds-0$

Solve for the voltage difference as,

$$\begin{gathered} V\left(b\right)-V\left(a\right)=-{\int }_{s-b}^{s-a}\frac{p{a}^2}{2s{\epsilon }_0}ds--{\int }_{s-a}^{s-0}\frac{ps}{2{\epsilon }_0}ds \\ =-\frac{p{a}^2}{2{\epsilon }_0}{\int }_{s-b}^{s-a}\frac{1}{s}ds-\frac{p}{2{\epsilon }_0}{\int }_{s-a}^{s-0}ds \\ =-\frac{p{a}^2}{2{\epsilon }_0}InS{|}_b^a-\frac{p}{2{\epsilon }_0}{\left(\frac{s^2}{2}\right)}_a^0 \end{gathered}$$

Further solving above equation,

$$\begin{gathered} V\left(b\right)-V\left(a\right)=-\frac{p{a}^2}{2{\epsilon }_0}(Ina-Inb)-\frac{p}{2{\epsilon }_0}\left(0-\frac{a^2}{2}\right) \\ =\left(\frac{2}{2}\right)\frac{p{a}^2}{2{\epsilon }_0}In\left(\frac{a}{b}\right)+\frac{p}{2{\epsilon }_0}\left(\frac{a^2}{2}\right) \\ =\frac{2p{a}^2}{4{\epsilon }_0}In\left(\frac{a}{b}\right)+\frac{p}{4{\epsilon }_0}\left(a^2\right) \\ =\frac{p{a}^2}{4{\epsilon }_0}\left(1-2In\left(\frac{a}{b}\right)\right) \\ \mathrm{Thus},\mathrm{the}\mathrm{potential}\mathrm{difference}\mathrm{is}\frac{p{a}^2}{4{\epsilon }_0}\left(1-2In\left(\frac{a}{b}\right)\right). \end{gathered}$$