Electric Field Inside a Spherical Cavity of Charged Sphere

To determine the electric field inside the spherical cavity, we can use the principle of superposition.

The given configuration of a uniformly charged solid sphere with a spherical cavity can be modeled as the superposition of:

1. A complete solid sphere with a uniform volume charge density $+\rho$ centered at point $O_1$.
2. A smaller sphere of the same size as the cavity with uniform volume charge density $-\rho$ centered at the center of the cavity, $O_2$.

By Gauss's Law, the electric field at an interior point of a uniformly charged solid sphere of charge density $\rho$ at a position vector $\vec{r}$ from its center is:
$\vec{E} = \frac{\rho}{3\varepsilon_0} \vec{r}$

Now, consider an arbitrary point $P$ inside the cavity:

• Let $\vec{r}_1$ be the position vector of point $P$ relative to the center of the large sphere $O_1$.
• Let $\vec{r}_2$ be the position vector of point $P$ relative to the center of the cavity $O_2$.
• Let $\vec{a}$ be the displacement vector from the center of the large sphere $O_1$ to the center of the cavity $O_2$, such that:
  $\vec{r}_1 = \vec{a} + \vec{r}_2 \implies \vec{r}_1 - \vec{r}_2 = \vec{a}$

Using the superposition principle, the net electric field $\vec{E}$ at point $P$ is the vector sum of the electric fields produced by the two spheres:
$\vec{E} = \vec{E}_1 + \vec{E}_2$
$\vec{E} = \frac{\rho}{3\varepsilon_0} \vec{r}_1 + \frac{-\rho}{3\varepsilon_0} \vec{r}_2 = \frac{\rho}{3\varepsilon_0} (\vec{r}_1 - \vec{r}_2)$

Substituting $\vec{r}_1 - \vec{r}_2 = \vec{a}$:
$\vec{E} = \frac{\rho}{3\varepsilon_0} \vec{a}$

Since $\vec{a}$ is a constant vector connecting the two centers and is non-zero (as the cavity is eccentric), the electric field inside the cavity is independent of the coordinates of point $P$.

Therefore, the electric field inside the emptied space is non-zero and uniform.

Hence, the correct option is (B).