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(A) The electric field inside a uniformly charged solid sphere at a point $P$ is given by $\vec{E} = \frac{\rho \vec{r}}{3 \epsilon_0}$,where $\rho$ i

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in the direction parallel to vector $\vec{l}$.

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(A) The electric field inside a uniformly charged solid sphere at a point $P$ is given by $\vec{E} = \frac{\rho \vec{r}}{3 \epsilon_0}$,where $\rho$ is the volume charge density and $\vec{r}$ is the position vector of point $P$ relative to the center of the sphere.Using the principle of superposition,we can treat the cavity as a sphere with charge density $-\rho$ superimposed on the original sphere with charge density $+\rho$.The net electric field $\vec{E}$ at any point inside the cavity is the sum of the field due to the solid sphere and the field due to the negative charge density of the cavity:$\vec{E} = \vec{E}_{sphere} + \vec{E}_{cavity} = \frac{\rho \vec{r}_1}{3 \epsilon_0} + \frac{-\rho \vec{r}_2}{3 \epsilon_0}$Here,$\vec{r}_1$ is the position vector from the center of the sphere and $\vec{r}_2$ is the position vector from the center of the cavity. The vector difference $\vec{r}_1 - \vec{r}_2$ is equal to the displacement vector $\vec{l}$ between the centers.Therefore,$\vec{E} = \frac{\rho}{3 \epsilon_0} (\vec{r}_1 - \vec{r}_2) = \frac{\rho \vec{l}}{3 \epsilon_0}$.Since the force on charge $q$ is $\vec{F} = q\vec{E}$,the force is $\vec{F} = q \frac{\rho \vec{l}}{3 \epsilon_0}$.Thus,the force is in the direction parallel to the displacement vector $\vec{l}$.

$A$ positive charge $q$ is placed in a spherical cavity made in a positively charged sphere. The centres of the sphere and the cavity are displaced by a small distance $\vec{l}$. The force on charge $q$ is:

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