Question

From a uniform solid sphere of charge $Q$ and radius $R$, a sphere of radius $\frac{R}{2}$ is cut out as shown. Find the electric field intensity at the common periphery of the cavity and solid sphere.

Answer 1

$\frac{Q}{8\pi\epsilon_0 R^2}$

Answer 2

The problem is solved using the superposition principle. The system is considered as a large uniformly charged sphere and a smaller sphere with negative charge density representing the cavity. The electric field at the common periphery point is the vector sum of fields from both. The formula for the electric field inside a uniformly charged sphere $\vec{E} = \frac{\rho \vec{r}}{3\epsilon_0}$ is applied. The point of interest is identified as $(R,0,0)$ and the center of the cavity at $(R/2,0,0)$. The individual fields are calculated and vectorially added to find the net field.