Question

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

• A. In the direction parallel to vector $\overrightarrow{l}$
• B. In radial direction
• C. In a direction which depends on the magnitude of charge density in sphere
• D. Direction cannot be determined

Answer 1

Answer: (A)

In the direction parallel to vector $\overrightarrow{l}$

Electric field inside solid sphere is given as: $E=\frac{\rho \overrightarrow{r}}{3{ϵ}_{\circ }}$

where $\rho$ is charge density and $\overrightarrow{r}$ is position vector of point P inside sphere w.r.t. it's center.

Using superposition principle for resulting electric field:

$E=E_{sphere}-E_{cavity}$

$=\frac{\rho \overrightarrow{r_{p/sphere}}}{3{ϵ}_{\circ }}-\frac{\rho \overrightarrow{r_{p/cavity}}}{3{ϵ}_{\circ }}$

$=\frac{\rho (\overrightarrow{r_{p/sphere}}-\overrightarrow{r_{p/cavity}})}{3{ϵ}_{\circ }}$

$=\frac{\rho \overrightarrow{l}}{3{ϵ}_{\circ }}$

Thus the force will be parallel to the direction of displacement $\overrightarrow{l}$.