Physics - Electrostatics Question with Solution | TestHub

The problem can be solved using the superposition principle. A sphere with a cavity is equivalent to a complete sphere of radius $R$ and charge density $\rho$ superimposed with a sphere of radius $R/2$ and charge density $-\rho$, whose center is at the cavity's center. Let the center of the main sphere be $O$ and the center of the cavity be $C$. The electric field at the center of the cavity ($C$) is the vector sum of the field due to the complete sphere at $C$ ($E_1$) and the field due to the negative sphere at $C$ ($E_2$). The electric field due to a uniformly charged non-conducting sphere at an internal point at distance $r^{\prime }$ from its center is $E=\frac{{\rho }^{\prime }{r}^{\prime }}{3{ϵ}_0}$. For the complete sphere (radius $R$, charge density $\rho$), the field at $C$ (distance $R/2$ from $O$) is $E_1=\frac{\rho (R/2)}{3{ϵ}_0}=\frac{\rho R}{6{ϵ}_0}$. This field points from $O$ towards $C$. For the negative sphere (radius $R/2$, charge density $-\rho$), the field at its own center $C$ is $E_2=0$. Therefore, the net electric field at the center of the cavity is $E_{net}=E_1+E_2=\frac{\rho R}{6{ϵ}_0}$.