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 $\frac{Q}{28\pi {\epsilon }_0}\left[\frac{A}{x^2}-\frac{B}{{\left(x-\frac{R}{2}\right)}^2}\right]$. Find $(A-B)$

Answer 1

Answer: 7

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

$Q_1=\rho \frac{4}{3}\pi R^3=\frac{8}{7}Q;Q_2=+\rho \frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3=+\frac{1}{7}Q$
$E_P=E_1-E_2=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1}{x^2}-\frac{1}{4\pi {\epsilon }_0}\frac{Q_2}{{\left(x-\frac{R}{2}\right)}^2}$
$E_P=\frac{Q}{4\pi {\epsilon }_0}\left[\frac{8}{7x^2}-\frac{1}{7{\left(x-\frac{R}{2}\right)}^2}\right]$
$=\frac{Q}{28\pi {\epsilon }_0}\left[\frac{8}{x^2}-\frac{1}{{\left(x-\frac{R}{2}\right)}^2}\right]$