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Q24P

Expert-verifiedFound in: Page 83

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

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,

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

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

*V(b)* is the potential at position *b*.

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.

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,

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

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