Question

Find the electric field inside a sphere that carries a charge density proportional to the distance from the origin,${}^{\mathbf{P}\mathbf{=}\mathbf{K}\mathbf{r}}$for some constant k. [Hint: This charge density is not uniform, and you must integrate to get the enclosed charge.]

Short answer

The electric field inside the non-uniformly charged solid sphere is$E=\frac{k{r}^2}{4{\epsilon }_0}\hat{r}.$

Step-by-step solution

Describe the given information

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.

Define the Gauss law

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,${}^{{\mathbf{\epsilon }}_{\mathbf{0}}}$is the permittivity of free surface.

Obtain the electric field inside the spherical shell

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

$q_{enclosed}={\int }_0^{r}pd\tau .$

Substitute krfor p, $4\pi r^{2}dr$for $d\tau$in the equation

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

Apply Gauss law on the Gaussian surface, by substituting $\pi k{r}^4$for $q_{enclosed}$, and $4{\mathrm{\pi r}}^2$for dainto$\oint E.da=\frac{q_{enclosed}}{{\epsilon }_0}$

localid="1654343673710" $$\begin{gathered} \oint E.da=\frac{q_{enclosed}}{{\epsilon }_0} \\ E\left(4{\mathrm{\pi r}}^2\right)=\frac{{\mathrm{\pi Kr}}^4}{{\mathrm{\epsilon }}_0} \\ =\frac{{\mathrm{\pi Kr}}^4}{{\mathrm{\epsilon }}_0\left(4{\mathrm{\pi r}}^2\right)} \\ =\frac{{\mathrm{Kr}}^2}{4{\mathrm{\epsilon }}_0}\hat{\mathrm{r}} \end{gathered}$$

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

$E=\frac{K{r}^2}{4{\epsilon }_0}\hat{r}$.