Statement
\\$6.2.12.\\$ In a uniformly charged infinite plate, a spherical cavity was cut out as shown in the figure. Plate's thickness is \\$h\\$ and its bulk charge density is \\$\\\\rho\\$. What is the electric field strength at point \\$A\\$? At point \\$B\\$? Find the dependence of the electric field strength along the line \\$OA\\$ on the distance to the point \\$O\\$.
Solution
The idea for solving this problem is to consider superposition principle: as cavity is uncharged, we can suppose that it exists a superposition of a plate (with charge density \\$\\\\rho\\$) and a sphere with charge density \\$-\\\\rho\\$. So, if we analyze the field in \\$A\\$ without cavity, we have two plates of thickness \\$\\\\frac{h}{2}\\$, one over \\$A\\$ and the other one under A, both with charge density \\$\\\\rho\\$, then field in \\$A\\$ is null. While for \\$A\\$ in the sphere's surface, applying Gauss's Law,
\\$\\$
-E_A\\\\cdot 4\\\\pi\\\\left(\\\\frac{h}{2}\\\\right)^2 = -\\\\frac{q_{enc}}{\\\\varepsilon_0}
\\$\\$
as \\$V = \\\\frac{4}{3}\\\\pi\\\\left(\\\\frac{h}{2}\\\\right)^3\\$, so \\$6\\\\frac{V}{h} = 4\\\\pi\\\\left(\\\\frac{h}{2}\\\\right)^2\\$, and taking in account that \\$\\\\rho = \\\\frac{q_{enc}}{V}\\$
Answer 1
\\$\\$
E_A = \\\\frac{\\\\rho h}{6\\\\varepsilon_0}
\\$\\$
For point \\$B\\$, the way is similar: in this case the plate without cavity generates certain field in \\$B\\$, located on its surface. From Gauss's Law
\\$\\$
E_p\\\\cdot S = \\\\frac{q_t}{\\\\varepsilon_0}
\\$\\$
as \\$S = \\\\frac{V}{h}\\$, and \\$\\\\rho = \\\\frac{q_t}{V}\\$
\\$\\$
E_p = \\\\frac{\\\\rho h}{\\\\varepsilon_0}
\\$\\$
but, for one of the side of plate, field is \\$E_B' = \\\\frac{E_p}{2} = \\\\frac{\\\\rho h}{2\\\\varepsilon_0}\\$ Now, overlapping with field generated by sphere,
\\$\\$
E_B = E_B' - E_c
\\$\\$
where \\$E_c = E_A\\$,
Answer 2
\\$\\$
E_B = \\\\frac{\\\\rho h}{3\\\\varepsilon_0}
\\$\\$
As \\$A\\$ is on line \\$OA\\$, all points over that line "feel" only the electric field generated by sphere. From Gauss's Law
\\$\\$
-E(r) \\\\cdot 4\\\\pi r^2 = -\\\\frac{q_{enc}}{\\\\varepsilon_0}
\\$\\$
as \\$V = \\\\frac{4}{3}\\\\pi r^3\\$, so \\$3\\\\frac{V}{r} = 4\\\\pi r^2\\$, and \\$\\\\rho = \\\\frac{q_{enc}}{V}\\$
Answer 3
\\$\\$
E(r) = \\\\frac{\\\\rho r}{3\\\\varepsilon_0}
\\$\\$
for \\$0\\\\leq r \\\\leq \\\\frac{h}{2}\\$.