Physics - Electrostatics Question with Solution | TestHub

Let the center of the original sphere be the origin $O$. Let the center of the cavity be $O^{\prime }$. The electric field inside a uniformly charged sphere of radius $R_0$ and charge density $\rho$ at a position $\vec{r}$ from its center is $\vec{E}=\frac{\rho \vec{r}}{3{ϵ}_0}$.

By superposition, the electric field inside the cavity is the sum of the field due to the large sphere (radius $R$, charge density $\rho$) and the field due to a small sphere (radius $R/2$, charge density $-\rho$) placed at $O^{\prime }$. Let $\vec{r}$ be the position vector from $O$. The electric field inside the cavity is ${\vec{E}}_{cavity}(\vec{r})={\vec{E}}_{large}(\vec{r})+{\vec{E}}_{small}(\vec{r})=\frac{\rho \vec{r}}{3{ϵ}_0}+\frac{-\rho (\vec{r}-\overrightarrow{O^{\prime }})}{3{ϵ}_0}=\frac{\rho \overrightarrow{O^{\prime }}}{3{ϵ}_0}$.

Since $\overrightarrow{O^{\prime }}$ is a constant vector (from $O$ to $O^{\prime }$), the electric field inside the cavity is uniform. So, option A is correct. The magnitude of $\overrightarrow{O^{\prime }}$ is $R/2$. So, the magnitude of the electric field inside the cavity is $\frac{\rho (R/2)}{3{ϵ}_0}=\frac{\rho R}{6{ϵ}_0}$. So, option B is correct.

Let's choose $O$ as the origin $(0,0,0)$ and $O^{\prime }$ as $(R/2,0,0)$. The electric field inside the cavity is $\vec{E}=\frac{\rho R}{6{ϵ}_0}\hat{i}$. The potential difference $V_{O^{\prime }}-V_O=-{\int }_O^{O^{\prime }}\vec{E}\cdot d\vec{l}=-{\int }_0^{R/2}(\frac{\rho R}{6{ϵ}_0}\hat{i})\cdot (dx\hat{i})=-\frac{\rho R}{6{ϵ}_0}(R/2)=-\frac{\rho R^2}{12{ϵ}_0}$. Since $V_{O^{\prime }}-V_O<0$, $V_{O^{\prime }}$ is lower than $V_O$. So, option C is incorrect.

The point on the surface of the cavity farthest from the center of the original sphere is at $(R,0,0)$. At this point, the electric field is $\frac{\rho R}{6{ϵ}_0}\hat{i}$, which is not zero. So, option D is incorrect.