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 inside the cavity of radius $a$ at a distance $r$ from the center of cavity.

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

Answer 1

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

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 $b$.

To find the electric field inside the cavity of radius $a$ at a distance $r$ from the center of cavity, let us draw a spherical gaussian surface of radius $r$ as shown below.


From the figure, it is clear that, charges $q_b$ and $q_a+q_b$ are outside the gaussian surface, So they cannot influence the electric field inside the spherical cavity of radius $a$.

Using Gauss's law,
$$\overrightarrow{E}\cdot \overrightarrow{ds}=\frac{q_{enc}}{{\epsilon }_0}$$

$\Rightarrow E\times 4\pi r^2=\frac{q_a}{{\epsilon }_0}$

$\Rightarrow E=\frac{q_a}{4\pi {\epsilon }_0{r}^2}$

Hence, option (a) is the correct answer.