91Ó°ÊÓ

1. A positive charge Ze is surrounded by a -Ze negative charge uniformly distributed in a sphere of radius R.
2. The electric potential at a distance r within the sphere,$\mathrm{V}=\frac{\mathrm{Ze}}{4{\mathrm{Ï€Î\mu }}_0}\left(\frac{1}{\mathrm{r}}-\frac{3}{2\mathrm{R}}+\frac{{\mathrm{r}}^3}{2{\mathrm{R}}^3}\right)$

Electric field is the negative space derivation of electric potential.

Using the given formula in the electric field equation, get the magnitude of the electric field by differentiating the given equation. Now, substituting the value of for , to get the value of both the electric field and potential at that point. Again, as there is no charge outside a uniformly charged sphere, its potential is zero outside the sphere surface.

Formula:

The relation of electric field due to charge and changing potential is,

$\mathrm{E}=-\frac{∂\mathrm{V}}{∂\mathrm{r}}$ ….. (1)

Using the given potential equation in equation (1), the magnitude of the electric field for the condition $0≤\mathrm{r}≤\mathrm{R}$can be given as follows:

$$\begin{gathered} \mathrm{E}=-\frac{∂}{∂\mathrm{r}}\left[\frac{\mathrm{Ze}}{4{\mathrm{Ï€Î\mu }}_0}\left(\frac{1}{\mathrm{r}}-\frac{3}{2\mathrm{R}}+\frac{{\mathrm{r}}^3}{2{\mathrm{R}}^3}\right)\right] \\ \mathrm{E}=\frac{\mathrm{Ze}}{4{\mathrm{Ï€Î\mu }}_0}\left(\frac{1}{{\mathrm{r}}^2}-\frac{\mathrm{r}}{{\mathrm{R}}^3}\right) \end{gathered}$$ ….. (2)

Hence, the value of the electric field is$\frac{\mathrm{Ze}}{4{\mathrm{Ï€Î\mu }}_0}\left(\frac{1}{{\mathrm{r}}^2}-\frac{\mathrm{r}}{{\mathrm{R}}^3}\right)$ .

Now, the electric field at r = R is given using equation (2) as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{r}=\mathrm{R}}=\frac{\mathrm{Ze}}{4{\mathrm{Ï€Î\mu }}_0}\left(\frac{1}{{\mathrm{R}}^2}-\frac{\mathrm{R}}{{\mathrm{R}}^3}\right) \\ =0 \end{gathered}$$

Thus, the field vanishes at r = R .

Since, the value of potential outside the sphere is , V = 0 this conclude that the electric field is also zero considering equation (1).

Hence, the value of the electric field for$\mathrm{r}â‰\yen \mathrm{R}$ is 0 .

Now, the potential at outside the sphere is V = 0 .

So, the value of potential at is given as follows:

${\mathrm{V}}_{\mathrm{r}=\mathrm{R}}=\frac{\mathrm{Ze}}{4{\mathrm{Ï€Î\mu }}_0}\left(\frac{1}{\mathrm{R}}-\frac{3}{2\mathrm{R}}+\frac{{\mathrm{R}}^3}{2{\mathrm{R}}^3}\right)$

Hence, the value of the potential for$\mathrm{r}â‰\yen \mathrm{R}$ is 0.