JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Electrostatics

Using the principle of superposition, we can treat the cavity as:

• A uniformly charged solid sphere (positive charge)
• Plus a uniformly charged sphere of opposite charge density at the cavity location

The electric field due to a uniformly charged sphere at an internal point is $\vec{E}=\frac{\rho }{3{ϵ}_0}\vec{r}$ (proportional to position vector from center).

For the cavity:

• Field due to complete sphere at point P: $\overrightarrow{E_1}=\frac{\rho }{3{ϵ}_0}\overrightarrow{r_1}$ (from main sphere center)
• Field due to removed sphere: $\overrightarrow{E_2}=-\frac{\rho }{3{ϵ}_0}\overrightarrow{r_2}$ (from cavity center)

Net field: $\vec{E}=\overrightarrow{E_1}+\overrightarrow{E_2}=\frac{\rho }{3{ϵ}_0}(\overrightarrow{r_1}-\overrightarrow{r_2})=\frac{\rho }{3{ϵ}_0}\vec{d}$

where $\vec{d}$ is the constant vector from cavity center to main sphere center.

Therefore, the field is non-zero and uniform throughout the cavity.