Question

Consider a uniform spherical charge distribution of radius $R_1$ centred at the origin O. In this distribution, a spherical cavity of radius $R_2$, centre at P with distance $OP=\alpha =R_1-R_2$ is made. If the electric field inside the cavity at position $\overrightarrow{r}$ is $\overrightarrow{E}\left(\overrightarrow{r}\right)$, then the correct statement(s) is (are)

• A. $\overrightarrow{E}$ is uniform, its magnitude is independent of $R_2$ but its direction depends on $\overrightarrow{r}$
• B. $\overrightarrow{E}$ is uniform, its magnitude is depends on $R_2$ and its direction depends on $\overrightarrow{r}$
• C. $\overrightarrow{E}$ is uniform, its magnitude is independent of $\alpha$ but its direction depends on $\overrightarrow{\alpha }$
• D. $\overrightarrow{E}$ is uniform and both its magnitude and direction depend on $\overrightarrow{\alpha }$

Answer 1

The correct option is D $\overrightarrow{E}$ is uniform and both its magnitude and direction depend on $\overrightarrow{\alpha }$

The problem can be considered as a superposition of electric field E due to the larger sphere without the cavity and a smaller sphere with opposite charge density.

Inside a spherical charge distribution for an internal point

$E\left(r\right)=\frac{\rho r}{3{ϵ}_r}$ which is a result of the application of Gauss's law.

Let the vector to the point P inside the cavity be $\overrightarrow{r_1}$ taking origin at the center of the large sphere and $\overrightarrow{r_2}$ be the vector from center O' of the smaller sphere.

$\therefore \overrightarrow{E}=\frac{\rho \overrightarrow{r_1}}{3{ϵ}_0}+\left(\frac{-\rho \overrightarrow{r_2}}{3{ϵ}_0}\right)=\frac{\rho }{3{ϵ}_0}(\overrightarrow{r_1}-\overrightarrow{r_2})=\frac{\rho }{3{ϵ}_0}\overrightarrow{a}$ where $\overrightarrow{a}$ is the vector joining the centres O and O' of the large and small spheres which is a constant vector.

$\therefore$ electric field is uniform both in magnitude and direction.