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 $\text{P}$. [Take $K=\frac{1}{4\pi {ϵ}_0}$]

• A. $\frac{31}{27}K\rho \pi R$, towards right
• B. $\frac{35}{27}K\rho \pi R$, towards right
• C. $\frac{2}{7}K\rho \pi R$, towards left
• D. None of the above

Answer 1

The correct option is B $\frac{35}{27}K\rho \pi R$, towards right

Consider the cavity to be made up of material of charge density $-\rho {\text{C/m}}^3$.


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

Total charge on sphere of radius $4R$,
$Q_1=\frac{4}{3}\pi (4R{)}^3\rho$

So, electric field at point $\text{P}$ due to this sphere,

$|\overrightarrow{E_1}|=\frac{KQ}{(8R{)}^2}$

$\Rightarrow |{\overrightarrow{E}}_1|=\frac{K\times \rho (4\pi \times (4R{)}^3)}{3\times (8R{)}^2}=\frac{4}{3}K\rho \pi R$
[towards right due to positive charge]

Similarly, electric field at point $\text{P}$ due to material of spherical cavity,
$|{\overrightarrow{E}}_2|=\frac{K\times \rho (4\pi \times (R{)}^3)}{3\times (6R{)}^2}=\frac{1}{27}K\rho \pi R$
[towards left due to positive charge]

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

$\Rightarrow |{\overrightarrow{E}}_{net}|=\frac{4}{3}K\rho \pi R-\frac{1}{27}K\rho \pi R$

$\Rightarrow |{\overrightarrow{E}}_{net}|=\frac{35}{27}K\rho \pi R$, towards right.

Hence, option (c) is the correct answer.