Q14P - Introduction to Electrodynamics

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.]

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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 kr for p, $4\pi r^{2}dr$ for $d\tau$ in the equation

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

 

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

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

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

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