91影视

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,${}^{{\mathbf{蔚}}_{\mathbf{0}}}$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

$q_{enclosed}={鈭�}_0^{r}pd蟿.$

Substitute krfor p, $4蟺r^{2}dr$for $d蟿$in the equation

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

Apply Gauss law on the Gaussian surface, by substituting $蟺k{r}^4$for $q_{enclosed}$, and $4{\mathrm{蟺谤}}^2$for dainto$鈭�E.da=\frac{q_{enclosed}}{{蔚}_0}$

localid="1654343673710" $$\begin{gathered} 鈭�E.da=\frac{q_{enclosed}}{{蔚}_0} \\ E\left(4{\mathrm{蟺谤}}^2\right)=\frac{{\mathrm{蟺碍谤}}^4}{{\mathrm{蔚}}_0} \\ =\frac{{\mathrm{蟺碍谤}}^4}{{\mathrm{蔚}}_0\left(4{\mathrm{蟺谤}}^2\right)} \\ =\frac{{\mathrm{Kr}}^2}{4{\mathrm{蔚}}_0}\hat{\mathrm{r}} \end{gathered}$$

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

$E=\frac{K{r}^2}{4{蔚}_0}\hat{r}$.