91Ó°ÊÓ

The electric field, $E_{\text{out}}$and electric potential, $V_{\text{out}}$outside the sphere $(râ‰\yen R)$is that of a point charge Q. The electric field inside the sphere is given by,

$E_{in}=\frac{1}{4\mathrm{π}{∈}_0}\frac{Q}{R^3}r$

$V_{\text{in}}=V_{\text{out}}$at the surface of the sphere

Formula Used:

If $\stackrel{→}{E}$ represents electric field and V represents potential associated with it then they are related by $V=-∫Edz$

We have to find the potential at a point inside the sphere.

$E_{\text{out}}(râ‰\yen R)=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{Q}{r^2}=\frac{kQ}{r^2}$

$E_{\text{in}}(r≤R)=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{Q}{R^3}r=\frac{kQ}{R^3}r$

Also,

$V_{\text{out}}(râ‰\yen R)=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{Q}{r}=\frac{kQ}{r}$

- Since $V=-∫Edz$the potential inside the sphere is given by,

$$\begin{gathered} V_{\text{in}}=-{∫}_{=}^R{E}_{ow}dr-{∫}_{R^{\text{'}}}^r{E}_{in}dr=-{∫}_{=}^R\frac{kQ}{r^2}dr-{∫}_R^r\frac{kQ}{R^3}rdr \\ {\left.V_{\text{in}}=kQâˆ\pounds \frac{1}{r}\right]}_{-}^R-\frac{kQ}{R^3}{\left[\frac{r^2}{2}\right]}_R^r=\frac{kQ}{R}-\frac{kQ}{2R^3}\left(r^2-R^2\right) \end{gathered}$$

Therefore, $V_{in}=\frac{kQ}{2R}\left(3-\frac{r^2}{R^2}\right)$

From the expression for potential inside the sphere,

$V_{\text{center}}(r=0)=\frac{3kQ}{2R}$

$V_{\text{surface}}(r=R)=\frac{kQ}{R}$

Therefore, $\frac{V_{\text{center}}}{V_{\text{suffure}}}=\frac{3}{2}$

Graph of $V\text{versus}r\text{for}0≤r≤3RV\left(r\right)$