Question

An insulating sphere of radius $R$ having a cavity of radius $\frac{R}{4}$ at the periphery is charged uniformly with charge $Q$. If the magnitude of electric field at point 'P' due to the above charge distribution is $\frac{8\alpha KQ}{63R^2}$, then the value of $\alpha$ is (Answer upto two digits after the decimal point)

Answer 1

Let $\rho$ be the surface charge density.
$\rho =\frac{Q}{\frac{4}{3}\pi \left(R^3-\frac{R^3}{64}\right)}$
Let $E_1$ is the electric field at point $P$ due to $Q_1$ and $E_2$ is the electric field at point $P$ due to $Q_2$
$$\begin{array}{cc}E& =E_1-E_2\\ & =\frac{k{Q}_1}{R^2}-\frac{k{Q}_2}{(R/4{)}^2}\\ & =\frac{k}{R^2}\frac{4}{3}\pi R^3\rho -\frac{16k}{R^2}\frac{4}{3}\pi {\left(\frac{R}{4}\right)}^3\rho \end{array}$$$$\begin{array}{c}=kR\pi \rho [\frac{4}{3}-\frac{1}{3}]=k\pi R\rho \\ =kR\pi \frac{Q}{\frac{4}{3}\pi (R^3-\frac{R^3}{64})}=\frac{kQR}{R^3}\frac{3}{4}\frac{64}{63}=\frac{16}{21}\frac{kQ}{R^2}\end{array}$$