Question

A neutral conducting sphere of radius $R$ contains two spherical cavities of radii $a$ and $b$ that contains point charges $q_a$ and $q_b$ placed at the center of each cavity as shown in the figure below . Find the electric field at a distance $r$ outside the conductor.

• A. $\frac{q_b}{4\pi {\epsilon }_0{r}^2}$
• B. $\frac{q_a+q_b}{4\pi {\epsilon }_0{r}^2}$
• C. Zero
• D. $\frac{q_a}{4\pi {\epsilon }_0{r}^2}$

Answer 1

The correct option is B $\frac{q_a+q_b}{4\pi {\epsilon }_0{r}^2}$

Given,

Radius of the conducting sphere be $R$

Radii of two spherical cavities be $a$ and $b$.

A point charge $q_a$ is placed in a spherical cavity of radius $a$ and point charge $q_b$ is placed in the spherical cavity of radius $b$.

Since, the electric field intensity is zero inside the conducting region we can say that,



Induced charge on the surface of the spherical cavity of radius $a$ be$"-q_a"$



Similarly, Induced charge on the surface of the spherical cavity of radius $b$ be $"-q_b"$

We know that, Surface charge density $\sigma =\frac{\text{Total charge}}{\text{Area}}$

$\therefore$ Surface charge density on spherical cavity $a$ be $${\sigma }_a=\frac{-q_a}{4\pi a^2}$$
Surface charge density on spherical cavity $b$ be $${\sigma }_b=\frac{-q_b}{4\pi b^2}$$
Total charge inside the conducting sphere be $q=q_a+q_b$ . These charges behave as though they are concentrated at the center.



Now, using Gauss Law we have
$E\times 4\pi r^2=\frac{Q_{\text{Inclosed}}}{{ϵ}_0}$
$\Rightarrow E=\frac{q_a+q_b}{4\pi {ϵ}_0{r}^2}$