Method: Principle of Superposition

This problem can be solved by considering the sphere with a cavity as a superposition of two objects:

1. A large, solid sphere of radius R with uniform volume charge density $+\rho$.
2. A smaller, solid sphere (at the position of the cavity) with uniform volume charge density $-\rho$.

The vector sum of the electric fields produced by these two spheres at any point will give the electric field of the original object (the sphere with the cavity).

Step-by-Step Derivation

1. Define a coordinate system. Let the center of the large sphere be the origin $O$. Let the center of the spherical cavity be at a position vector $\vec{a}$ with respect to $O$. Let $P$ be an arbitrary point inside the cavity, with a position vector $\vec{r}$ with respect to $O$. The position vector of point $P$ with respect to the center of the cavity is then $\vec{r'} = \vec{r} - \vec{a}$.

2. Calculate the electric field due to the large sphere. The electric field inside a uniformly charged solid sphere at a point with position vector $\vec{r}$ from its center is given by the formula: $\vec{E} = \frac{\rho \vec{r}}{3\epsilon_0}$ So, the electric field at point $P$ due to the large sphere (with charge density $+\rho$) is: $\vec{E}_1 = \frac{\rho \vec{r}}{3\epsilon_0}$

3. Calculate the electric field due to the small sphere. We model the cavity as a smaller sphere with charge density $-\rho$. The electric field at point $P$ (which is inside this smaller sphere) due to this smaller sphere is calculated using the same formula, but with respect to its own center. The position vector of $P$ relative to the center of the cavity is $\vec{r'} = \vec{r} - \vec{a}$. The electric field at $P$ due to the smaller sphere is: $\vec{E}_2 = \frac{(-\rho) \vec{r'}}{3\epsilon_0} = -\frac{\rho (\vec{r} - \vec{a})}{3\epsilon_0}$

4. Apply the principle of superposition. The total electric field $\vec{E}_{total}$ at point $P$ inside the cavity is the vector sum of $\vec{E}_1$ and $\vec{E}_2$: $\vec{E}_{total} = \vec{E}_1 + \vec{E}_2$ $\vec{E}_{total} = \frac{\rho \vec{r}}{3\epsilon_0} + \left( -\frac{\rho (\vec{r} - \vec{a})}{3\epsilon_0} \right)$ $\vec{E}_{total} = \frac{\rho}{3\epsilon_0} [\vec{r} - (\vec{r} - \vec{a})]$ $\vec{E}_{total} = \frac{\rho}{3\epsilon_0} [\vec{r} - \vec{r} + \vec{a}]$ $\vec{E}_{total} = \frac{\rho \vec{a}}{3\epsilon_0}$

Conclusion

The resulting expression for the electric field, $\vec{E}_{total} = \frac{\rho \vec{a}}{3\epsilon_0}$, is a constant vector. It does not depend on the position vector $\vec{r}$ of the point $P$ within the cavity. This means the electric field has the same magnitude and direction at every point inside the emptied space.

• Non-zero: Since $\rho \neq 0$ and the cavity is displaced from the center ($\vec{a} \neq 0$), the field is non-zero.
• Uniform: Since the field is a constant vector, it is uniform.

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

Evaluation of Options

• A: zero everywhere - Incorrect. The field is $\frac{\rho \vec{a}}{3\epsilon_0}$, which is non-zero.
• B: non-zero and uniform - Correct. The derived field is a constant non-zero vector.
• C: non-uniform - Incorrect. The field is independent of the position within the cavity.
• D: zero only at its center - Incorrect. The field is uniform and non-zero everywhere inside the cavity.