91Ó°ÊÓ

It is given that a sphere carries a uniform volume charge density, which is proportional to the distance from the origin, as${}^{P=Kr}$, is a constant The electric field inside and outside the has to be evaluated.

If there is a surface area enclosing a volume, possessing a charge${}^{\mathbf{q}}$inside the volume then the electric field due to the surface or volume charge is given as

Here${}^{\mathbf{q}}$is the elemental surface area,${}^{{\mathbf{Î\mu }}_{\mathbf{0}}}$is the permittivity of free surface.

Consider a Gaussian surface of radius${}^r$such that${}^{r<R}$inside the sphere as shown below:

It is known that the spherical consist the charge density which varies as${}^{P=Kr}$.So, the charge enclosed by the Gaussian sphere of radius is obtained by integrating the charge density from 0 to${}^r$, as

$q_{enclosed}={∫}_0^{r}pd\mathrm{τ}.$

Substitute krfor p, $4\mathrm{Ï€}{r}^{2}dr$for $d\mathrm{Ï„}$in the equation

localid="1654599163389" $$\begin{gathered} q_{enclosed}={∫}_0^{r}kr\left(4\mathrm{π}{r}^{2}dr\right) \\ =4\mathrm{π}k{∫}_0^r{r}^{3}dr \\ =4\mathrm{π}k{\left(\begin{array}{c}\frac{r^4}{4}\end{array}\right)}_0^r \\ =4\mathrm{π}k{r}^4 \end{gathered}$$

Apply Gauss law on the Gaussian surface, by substituting $\mathrm{Ï€}k{r}^4$for $q_{enclosed}$, and $4{\mathrm{Ï€°ù}}^2$for dainto$âˆ\circledR E.da=\frac{q_{enclosed}}{{\mathrm{Î\mu }}_0}$

localid="1654343673710" $$\begin{gathered} âˆ\circledR E.da=\frac{q_{enclosed}}{{\mathrm{Î\mu }}_0} \\ E\left(4{\mathrm{Ï€°ù}}^2\right)=\frac{{\mathrm{Ï€°­°ù}}^4}{{\mathrm{Î\mu }}_0} \\ =\frac{{\mathrm{Ï€°­°ù}}^4}{{\mathrm{Î\mu }}_0\left(4{\mathrm{Ï€°ù}}^2\right)} \\ =\frac{{\mathrm{Kr}}^2}{4{\mathrm{Î\mu }}_0}\hat{\mathrm{r}} \end{gathered}$$

Thus, the electric field inside the non-uniformly charged solid sphere is

$E=\frac{K{r}^2}{4{\mathrm{Î\mu }}_0}\hat{r}$.