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(B) Consider a solid sphere of charge density $\rho$ with a spherical cavity. The electric field at any point $P$ inside the cavity can be found using

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non-zero and uniform

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(B) Consider a solid sphere of charge density $\rho$ with a spherical cavity. The electric field at any point $P$ inside the cavity can be found using the principle of superposition. We can treat the system as a large solid sphere of charge density $\rho$ and a smaller sphere of charge density $-\rho$ that fills the cavity.The electric field at a point $P$ inside a uniformly charged sphere at a distance $\vec{r}$ from its center is given by $\vec{E} = \frac{\rho \vec{r}}{3 \epsilon_0}$.Let $\vec{b}$ be the position vector of point $P$ from the center of the large sphere $(O)$ and $\vec{a}$ be the position vector of point $P$ from the center of the cavity $(Q)$. The net electric field at $P$ is the sum of the fields due to the large sphere and the negative charge of the cavity:$\vec{E}_{net} = \vec{E}_{large} + \vec{E}_{cavity} = \frac{\rho \vec{b}}{3 \epsilon_0} + \frac{-\rho \vec{a}}{3 \epsilon_0} = \frac{\rho}{3 \epsilon_0} (\vec{b} - \vec{a})$.From the geometry,$\vec{b} - \vec{a} = \vec{r}$,where $\vec{r}$ is the constant vector from the center of the large sphere to the center of the cavity. Thus,$\vec{E}_{net} = \frac{\rho \vec{r}}{3 \epsilon_0}$.Since $\rho$,$\vec{r}$,and $\epsilon_0$ are constants,the electric field inside the cavity is non-zero and uniform.

$A$ spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

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