Q24P - Introduction to Electrodynamics

Question

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.

Step-by-Step Solution

Answer

The potential difference is$\frac{p{a}^2}{4{\epsilon }_0}\left(1-2In\left(\frac{a}{b}\right)\right)$.

Step 1: Define electric potential.

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 , E is the electric field, and is the length of small element.

Consider the two points are a and 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.

Step 2: Determine electric field configuration

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, a is the inner radius of the cylinder, b is the outer radius of the radius of sphere, s is the radius of the Gaussian surface, p is the volume charge density and ${\epsilon }_0$ is the permittivity of the free space.

Step 3: Determine electric potential.

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,

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