A non-conducting solid sphere has a uniform volume charge density P. Let$\overrightarrow{\mathbf{r}}$be the vector from the center of the sphere to a general point Pwithin the sphere.

(a) Show that the electric field at Pis given by$\mathbf{E}\mathbf{⃗}\mathbf{=}\mathbf{\rho r}\mathbf{⃗}\mathbf{/}\mathbf{3}{\mathbf{\epsilon }}_{\mathbf{0}}$(Note that the result is independent of the radius of the sphere.)

(b) A spherical cavity is hollowed out of the sphere, as shown in Fig. 23- 60. Using superposition concepts, show that the electric field at all points within the cavity is uniform and equal to $\mathbf{E}\mathbf{⃗}\mathbf{=}\mathbf{\rho r}\mathbf{⃗}\mathbf{/}\mathbf{3}{\mathbf{\epsilon }}_{\mathbf{0}}$where ais the position vector from the center of the sphere to the center of the cavity.

A non-conducting solid sphere has a uniform volume charge density p . Let,$\overrightarrow{r}$ be the vector from the center of the sphere to a general point P within the sphere.

Using the concept of the electric field at a point due to a charged particle, we can get the value of the field using the volume charge density. Similarly, using this value, we can get the electric field within the cavity points.

Formula:

The electric field at a point due to a charged particle,

$\overrightarrow{E}(r⃗)=\frac{q_{enc}}{4\pi {\epsilon }_0{r}^3}r⃗$ (i)

Where charge $q=4\pi \rho r^3/3$

Using the given data in equation (i), we can get the electric field expression as follows:

$$\begin{gathered} \overrightarrow{E}(r⃗)=\frac{1}{4\pi {\epsilon }_0}\frac{\left(4\pi {\epsilon }_0{r}^3/3\right)}{r^3}\overrightarrow{r} \\ =\rho r⃗/3{\epsilon }_0 \end{gathered}$$

Hence, the value of the electric field is$\rho r⃗/3{\epsilon }_0$ . It is proved.

The charge distribution, in this case, is equivalent to that of a whole sphere of charge density plus a smaller sphere of charge density -p that fills the void.

By using the superposition, the total electric field at all the points within the cavity using the above value can be given as:

$$\begin{gathered} \overrightarrow{E}\left(\overrightarrow{r}\right)=\frac{p\overrightarrow{r}}{3{\epsilon }_0}+\frac{\left(-p\right)\left(\overrightarrow{r}-\overrightarrow{a}\right)}{3{\epsilon }_0} \\ =\rho \alpha ⃗/3{\epsilon }_0 \end{gathered}$$

Hence, the value of the electric field is $\rho \alpha ⃗/3{\epsilon }_0$.