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(D) The electric field inside a spherical cavity within a uniformly charged sphere can be calculated using the principle of superposition. We consider

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$\vec{E}$ is uniform and both its magnitude and direction depend on $\vec{a}$

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(D) The electric field inside a spherical cavity within a uniformly charged sphere can be calculated using the principle of superposition. We consider the sphere with the cavity as the sum of a solid sphere of charge density $\rho$ and a smaller sphere of charge density $-\rho$ filling the cavity.The electric field at any point inside the solid sphere of radius $R_1$ is $\vec{E}_1 = \frac{\rho \vec{r}_1}{3 \varepsilon_0}$,where $\vec{r}_1$ is the position vector from the center $O$.The electric field at any point inside the smaller sphere (the cavity) of radius $R_2$ is $\vec{E}_2 = \frac{-\rho \vec{r}_2}{3 \varepsilon_0}$,where $\vec{r}_2$ is the position vector from the center $P$.The net electric field inside the cavity is $\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\rho}{3 \varepsilon_0} (\vec{r}_1 - \vec{r}_2)$.Since $\vec{r}_1 - \vec{r}_2 = \vec{OP} = \vec{a}$,the electric field is $\vec{E} = \frac{\rho \vec{a}}{3 \varepsilon_0}$.This expression shows that the electric field inside the cavity is uniform (constant) and depends only on the vector $\vec{a}$ connecting the center of the large sphere to the center of the cavity. It is independent of the position $\vec{r}$ inside the cavity and independent of the radius $R_2$.

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