Question

A solid spherical region having a spherical cavity whose diameter $R$ is equal to the radius of the spherical region, has a total charge $Q$. The electric field at a point $P$ as shown is $\frac{Q}{28 \pi \varepsilon_{0}}\left[\frac{A}{x^{2}}-\frac{B}{\left(x-\frac{R}{2}\right)^{2}}\right]$. Find $(A-B)$

Answer 1

Volumetric charge density of the given structure,
$\rho=\frac{Q}{\frac{4}{3} \pi R^{3}-\frac{4}{3} \pi\left(\frac{R}{2}\right)^{3}} $
$\Rightarrow \rho=\frac{8}{7} \frac{Q}{\frac{4}{3} \pi R^{3}}$

$Q_{1}=\rho \frac{4}{3} \pi R^{3}=\frac{8}{7} Q ; Q_{2}=+\rho \frac{4}{3} \pi\left(\frac{R}{2}\right)^{3}=+\frac{1}{7} Q$
$E_{P}=E_{1}-E_{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q_{1}}{x^{2}}-\frac{1}{4 \pi \varepsilon_{0}} \frac{Q_{2}}{\left(x-\frac{R}{2}\right)^{2}}$
$E_{P}=\frac{Q}{4 \pi \varepsilon_{0}}\left[\frac{8}{7 x^{2}}-\frac{1}{7\left(x-\frac{R}{2}\right)^{2}}\right]$
$=\frac{Q}{28 \pi \varepsilon_{0}}\left[\frac{8}{x^{2}}-\frac{1}{\left(x-\frac{R}{2}\right)^{2}}\right]$