91Ó°ÊÓ

The electric potential at infinity, $V=0$.

After reading the question, Charge on sphere of radiusris determined by using total charge q. By using Gauss theorem, field on this sphere is determined and by integrating in proper limits, potential atris determined.

Gauss Lawstates that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.

Formula:

The electric potential is defined by,

$V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}â‹\dots \frac{q}{R}$

Where, V is potential energy, R is the distance between the point charges, q is charge.

Charge of the sphere is q and radius is R . Potential at infinity is zero.

$q^{\text{'}}=\frac{q}{\left(\frac{4}{3}\mathrm{Ï€}{R}^3\right)}\left(\frac{4}{3}\mathrm{Ï€}{r}^3\right)$

Charge on the sphere of radiusis,

$r=q\left(\frac{r^3}{R^3}\right)$

Let E be the field strength.

From Gauss theorem, electric flux is $\frac{1}{{\mathrm{Î\mu }}_0}$ times charge enclosed.

$\begin{array}{c}4\mathrm{Ï€}{r}^2\left(E\right)=\frac{1}{{\mathrm{Î\mu }}_0}\left(q\left(\frac{r^3}{R^3}\right)\right)\\ 4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2\left(E\right)=q\left(\frac{r^3}{R^3}\right)\\ E=\frac{qr}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{R}^3}\end{array}$

Let $V_s$be the potential difference on the surface of the sphere,

$V_s=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}R}$

Potential at the point kept at a distance r from the center is,

role="math" localid="1662625140265" $\begin{array}{c}V=V_s−\underset{R}{\overset{r}{}}Edr\\ =\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}R}−\underset{R}{\overset{r}{}}\left(\frac{qr}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{R}^3}\right)dr\end{array}$

$\begin{array}{c}V=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}R}−\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{R}^3}\left(\frac{r^2−R^2}{2}\right)\\ =\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{1}{R}−\frac{r^2}{2R^3}+\frac{1}{2R}\right)\\ =\frac{q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0{R}^3}(3R^2−r^2)\end{array}$

Hence, the potential at the point kept at a distance r from the center is $V=\frac{q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0{R}^3}(3R^2−r^2)$.

Electric potential when $r=R$is,

$\begin{array}{c}{V}_R=\frac{q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0{R}^3}(3R^2−r^2)\\ =\frac{2q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_{0}R}\end{array}$

Electric potential when $r=0$is,

$$" width="9">V0=q8πε0R3(3R2−r2)=3q8πε0R

Potential difference is,

$\begin{array}{c}V=V_R−V_0\\ =\frac{2q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_{0}R}−\frac{3q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_{0}R}\\ =−\frac{q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_{0}R}\end{array}$

Hence,the requitedpotential difference is$V=−\frac{q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_{0}R}$.