Question

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

• A. $\frac{27}{48}\frac{\rho R}{{\epsilon }_o}$, towards left
• B. $48\frac{\rho R}{{\epsilon }_o}$, towards left
• C. $\frac{31}{48}\frac{\rho R}{{\epsilon }_o}$, towards left
• D. $\frac{12}{29}\frac{\rho R}{{\epsilon }_o}$, towards right

Answer 1

The correct option is C $\frac{31}{48}\frac{\rho R}{{\epsilon }_o}$, towards left

Consider the cavity to be made up material of charge density $-\rho {\text{C/m}}^3$. The directions of the electric field are as shown in the figure.


Net electric field due to the above combination can be calculated by superposition.
${\overrightarrow{E}}_{net}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2$

Electric field at point $\text{P}$ due to complete sphere,

$|{\overrightarrow{E}}_1|=\frac{\rho \left(2R\right)}{3{\epsilon }_o}$
(towards left)

Similarly, electric field at point $\text{P}$ due to spherical cavity,

$|{\overrightarrow{E}}_2|=\frac{1}{4\pi {\epsilon }_o}\frac{q}{(4R{)}^2}...\left(2\right)$

where, $q$ is the total charge in the cavity which will be,
$q=\frac{4}{3}\pi (R{)}^3\rho$ (negative)

From equation $\left(2\right)$,
$|{\overrightarrow{E}}_2|=\frac{1}{4\pi {\epsilon }_o}\frac{\rho \times 4\pi \times (R{)}^3}{3\times (4R{)}^2}=\frac{\rho R}{48{\epsilon }_o}$
(towards right)

Now, the net electric field,
$|{\overrightarrow{E}}_{net}|=|{\overrightarrow{E}}_1|-|{\overrightarrow{E}}_2|$

$\Rightarrow |{\overrightarrow{E}}_{net}|=\frac{\rho \left(2R\right)}{3{\epsilon }_o}-\frac{\rho R}{48{\epsilon }_o}$

$\Rightarrow |{\overrightarrow{E}}_{net}|=\frac{31\rho R}{48{\epsilon }_o}$, towards left

Hence, option (c) is the correct answer.