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(B) Consider a solid sphere of radius $R$ with uniform charge density $\rho$. The electric field at a point $\vec{r}$ inside the sphere is given by $\

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nonzero and uniform

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(B) Consider a solid sphere of radius $R$ with uniform charge density $\rho$. The electric field at a point $\vec{r}$ inside the sphere is given by $\vec{E} = \frac{\rho \vec{r}}{3\epsilon_0}$.When a smaller spherical cavity is created,the electric field in the cavity is the superposition of the field due to the original large sphere and the field due to a smaller sphere of opposite charge density $-\rho$ that fills the cavity.Let $\vec{r}_1$ be the position vector from the center of the large sphere and $\vec{r}_2$ be the position vector from the center of the cavity.The electric field in the cavity is $\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\rho \vec{r}_1}{3\epsilon_0} + \frac{-\rho \vec{r}_2}{3\epsilon_0} = \frac{\rho}{3\epsilon_0} (\vec{r}_1 - \vec{r}_2)$.Since $(\vec{r}_1 - \vec{r}_2) = \vec{d}$,where $\vec{d}$ is the constant vector connecting the center of the large sphere to the center of the cavity,the electric field $\vec{E} = \frac{\rho \vec{d}}{3\epsilon_0}$ is constant (uniform) and non-zero throughout the cavity.

$A$ 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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