Answered: Find the electric potential…

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter7: Electric Potential

Section: Chapter Questions

Problem 97AP: (a) Find x L limit of the potential of a finite uniformly charged rod and show that it coincides...

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Question

Find the electric potential distribution of charged spherical surface with the charge q and the radius R

Expert Solution

Step 1

Electric potential at a point in the field is defined as the work done in moving a unit positive test charge from infinity to that point against the electrostatic forces, along any path.
Gauss's law states that the surface integral of electrostatic field $\vec{E}$ produced by any sources over any closed surface S enclosing a volume V in vacuum is $\frac{1}{ε_{0}}$ times the total charge (Q) contained inside S, i.e.,
$ϕ_{E} = \oint \vec{E} . \vec{ds} = \frac{Q}{ε_{0}}$

Step 2

Let charge +q be distributed uniformly over the spherical surface.

To calculate electric field intensity at any point P, where OP=r, consider a sphere S₁ with center O and radius r. This is a Gaussian surface, at every point of which $\vec{E}$ is the same, directed radially outwards.

According to Gauss's theorem:

$\oint \vec{E} . \vec{ds} = \oint \vec{E} . \hat{n} ds = \frac{q}{ε_{0}} \Rightarrow E \oint ds = \frac{q}{ε_{0}}$

$\begin{array}{rcl} ∴ E . (4 {πr}^{2}) & = & \frac{q}{ε_{0}} \\ E & = & \frac{q}{4 {πε}_{0} r^{2}} \end{array}$

Clearly ,electric field intensity at any point outside the spherical surface is such, as if the entire charge was concentrated at the center of the spherical surface.

$∴$ Potential outside the spherical surface is given by-

$\begin{array}{rcl} V & = & - \int_{\infty}^{r} Edl \\ = & - \int_{\infty}^{r} \frac{q}{4 {πε}_{0} r^{2}} dl \\ = & \frac{q}{4 {πε}_{0}} | \frac{1}{r} |_{\infty}^{r} \\ = & \frac{q}{4 {πε}_{0}} [\frac{1}{r} - \frac{1}{\infty}] \\ = & \frac{q}{4 {πε}_{0} r} \end{array}$

Potential at a point on the surface of the spherical surface is given by replacing r for R in the expression for electric potential outside the charged surface and is equal to $\frac{q}{4 {πε}_{0} R}$

Now,
if the point P lies inside the spherical surface, then Gaussian surface is surface of a sphere S₂ of radius r'<R. Since the charge enclosed by spherical surface is zero, therefore the Gaussian surface in this case encloses no charge, i.e.,
$q = 0 ∴ E = 0$

So potential at a point r' from the center of spherical surface is given as-

$\begin{array}{rcl} V & = & - \int \vec{E} \vec{dl} \\ V & = & - \int_{\infty}^{R} Edl - \int_{R}^{r'} Edl \\ V & = & - \int_{\infty}^{R} (\frac{q}{4 {πε}_{0} r^{2}}) dl - \int_{R}^{r'} (0) dl \\ V & = & \frac{q}{4 {πε}_{0}} | \frac{1}{r} |_{\infty}^{R} \\ V & = & \frac{q}{4 {πε}_{0}} [\frac{1}{R} - \frac{1}{\infty}] \\ V & = & \frac{q}{4 {πε}_{0} R} \end{array}$

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