Question

A spherical cavity of radius $R$ is cut from a non-conducting sphere of volulme charge density $\rho {\text{C/m}}^3$ as shown in figure. Find the magnitude of the net electric field at point $\text{P}$.

• A. $\frac{\rho R}{2{\epsilon }_o}$
• B. $\frac{3\rho l}{{\epsilon }_o}$
• C. $\frac{\rho (r_1+r_2)}{3{\epsilon }_o}$
• D. $\frac{\rho l}{3{\epsilon }_o}$

Answer 1

The correct option is D $\frac{\rho l}{3{\epsilon }_o}$

The above problem can be solved by the superposition of electric fields due to the complete sphere and the cavity (considered to be of charge density $-\rho {\text{C/m}}^3$).

Let the electric field at point $\text{P}$ due to the whole sphere of radius $4R$ is ${\overrightarrow{E}}_1$ as shown in figure.



From figure,
${\overrightarrow{E}}_1=\frac{\rho {\overrightarrow{r}}_1}{3{\epsilon }_o}$

Similarly electric field at point $\text{P}$ due to cavity of unform charge density $-\rho {\text{C/m}}^3$ is,
${\overrightarrow{E}}_2=-\frac{\rho {\overrightarrow{r}}_2}{3{\epsilon }_o}$

Now, from superposition principle net electric field at point $\text{P}$ will be,
${\overrightarrow{E}}_{net}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2$

$\Rightarrow {\overrightarrow{E}}_{net}=\frac{\rho ({\overrightarrow{r}}_1-{\overrightarrow{r}}_2)}{3{\epsilon }_o}$

From triangle law,
$\overrightarrow{l}+{\overrightarrow{r}}_2={\overrightarrow{r}}_1$

Substituting it in the above equation, we get
${\overrightarrow{E}}_{net}=\frac{\rho \left(\overrightarrow{l}\right)}{3{\epsilon }_o}$

Thus, magnitude of the net electric field at point $\text{P}$, will be
$\Rightarrow |{\overrightarrow{E}}_{net}|=\frac{\rho l}{3{\epsilon }_o}$

Hence option (a) is the correct answer.

Why this question? Note: Inside a non-conducting cavity electric field is always zero.