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Found in: Page 83

Introduction to Electrodynamics

Book edition 4th edition
Author(s) David J. Griffiths
Pages 613 pages
ISBN 9780321856562

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 potential difference is$\frac{p{a}^{2}}{4{\epsilon }_{0}}\left(1-2In\left(\frac{a}{b}\right)\right)$.

See the step by step solution

Step 1: Define electric potential.

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

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

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,

$V\left(b\right)-V\left(b\right)=-{\int }_{0}^{b}E-dl+{\int }_{0}^{a}E.dl\phantom{\rule{0ex}{0ex}}=-{\int }_{0}^{b}E-dl-{\int }_{a}^{0}E.dl\phantom{\rule{0ex}{0ex}}=-{\int }_{a}^{b}E-dl$

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}}& sb\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,

$\nabla V=V\left(b\right)-V\left(a\right)\phantom{\rule{0ex}{0ex}}=-{\int }_{a}^{b}E\left(s\right)ds$

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,

$V\left(b\right)-V\left(a\right)=-{\int }_{s-b}^{s-a}\frac{p{a}^{2}}{2s{\epsilon }_{0}}ds--{\int }_{s-a}^{s-0}\frac{ps}{2{\epsilon }_{0}}ds\phantom{\rule{0ex}{0ex}}=-\frac{p{a}^{2}}{2{\epsilon }_{0}}{\int }_{s-b}^{s-a}\frac{1}{s}ds-\frac{p}{2{\epsilon }_{0}}{\int }_{s-a}^{s-0}ds\phantom{\rule{0ex}{0ex}}=-\frac{p{a}^{2}}{2{\epsilon }_{0}}InS{|}_{b}^{a}-\frac{p}{2{\epsilon }_{0}}{\left(\frac{{s}^{2}}{2}\right)}_{a}^{0}$

Further solving above equation,

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