**6.2.12.** In a uniformly charged infinite plate, a spherical cavity was cut out as shown in the figure. The plate's thickness is $h$ and its bulk charge density is $\rho$.
What is the electric field strength at point $A$ (the center of the cavity)?
What is the electric field strength at point $B$ (on the surface of the plate directly above the center)?
Find the dependence of the electric field strength along the line $OA$ on the distance $r$ from the center $O$.
Solution
The key to solving this problem is the **principle of superposition**. We can treat the cavity as a combination of a solid, uniformly charged plate (density $+\rho$) and a sphere of the same size with an opposite charge density ($-\rho$).
Field at Point A
In a solid infinite plate of thickness $h$, the electric field at the exact center (point $A$) is zero due to symmetry. Therefore, the field at $A$ is generated solely by the "negative" sphere.
Using Gauss’s Law for the sphere at its surface ($r = h/2$): $$E_A \cdot 4\pi \left(\frac{h}{2}\right)^2 = \frac{Q_{enc}}{\varepsilon_0}$$
Given $V = \frac{4}{3}\pi\left(\frac{h}{2}\right)^3$ and $Q_{enc} = \rho \cdot V$: $$E_A \cdot 4\pi \left(\frac{h}{2}\right)^2 = \frac{\rho \cdot \frac{4}{3}\pi\left(\frac{h}{2}\right)^3}{\varepsilon_0}$$
**Field from the solid plate ($E_p$):** By Gauss’s Law, the total field emerging from both sides of a plate is $\frac{\rho h}{\varepsilon_0}$. For just one side (at point $B$): $$E_{plate} = \frac{\rho h}{2\varepsilon_0}$$
**Field from the "negative" sphere ($E_s$):** Point $B$ is on the surface of the sphere ($r = h/2$). As calculated in Part 1, the field from the sphere at its surface is: $$E_{sphere} = \frac{\rho h}{6\varepsilon_0}$$
The vectors point in opposite directions at $B$, so we subtract them: $$E_B = E_{plate} - E_{sphere} = \frac{\rho h}{2\varepsilon_0} - \frac{\rho h}{6\varepsilon_0}$$
Since line $OA$ is at the center of the infinite plate, the plate's own contribution to the field is zero at every point along this line. Thus, the field depends only on the charged sphere.
For any point at a distance $r$ inside the cavity ($0 \leq r \leq h/2$), we apply Gauss's Law to the sphere: $$E(r) \cdot 4\pi r^2 = \frac{\rho \cdot \frac{4}{3}\pi r^3}{\varepsilon_0}$$