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(B) Consider a solid sphere of radius $R$ with uniform charge density $\rho$. Let the center of the large sphere be at the origin $O_1$ and the center

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non-zero and uniform

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(B) Consider a solid sphere of radius $R$ with uniform charge density $\rho$. Let the center of the large sphere be at the origin $O_1$ and the center of the removed spherical cavity be at $O_2$. Let $\vec{a}$ be the position vector of $O_2$ relative to $O_1$.The electric field at any point $P$ inside the cavity can be calculated using the principle of superposition. We can treat the system as a large positively charged sphere and a smaller negatively charged sphere (the cavity) superimposed on it.The electric field due to the large sphere at point $P$ (position vector $\vec{r}_1$ from $O_1$) is $\vec{E}_1 = \frac{\rho \vec{r}_1}{3 \epsilon_0}$.The electric field due to the smaller sphere (cavity) at point $P$ (position vector $\vec{r}_2$ from $O_2$) is $\vec{E}_2 = -\frac{\rho \vec{r}_2}{3 \epsilon_0}$.The net electric field $\vec{E}$ at point $P$ is $\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\rho}{3 \epsilon_0} (\vec{r}_1 - \vec{r}_2)$.Since $\vec{r}_1 = \vec{a} + \vec{r}_2$,we have $\vec{r}_1 - \vec{r}_2 = \vec{a}$.Thus,$\vec{E} = \frac{\rho \vec{a}}{3 \epsilon_0}$.Since $\rho$,$\vec{a}$,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. The electric field inside the emptied space is

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