Question

An infinitely long solid cylinder of radius $R$ has a uniform volume charge density $\rho$. It has a spherical cavity of radius $R/2$ with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point $P$, which is at a distance $2R$ from the axis of the cylinder, is given by the expression $\frac{23\rho R}{16n{\epsilon }_0}.$ The value of $n$ is

Answer 1

Answer: 6

Electric field due to cylinder $=\frac{2K\lambda }{2R}$
Electric field due to sphere $=\frac{KQ}{(2R{)}^2}$
From principle of superposition,
$E=\frac{2K(\lambda )}{2R}-\frac{K(Q)}{(2R{)}^2}=\frac{K\left(\rho \pi R^2\right)}{R}-\frac{K\left(\rho \times \frac{4\pi }{3}\left(\frac{R^3}{8}\right)\right)}{4R^2}$
$=K(\rho \pi R)-\frac{K(\rho )(\pi R)}{24}=\frac{23\rho \pi R}{4\pi {\epsilon }_0\times 24}=\frac{23\rho R}{16\times 6{\epsilon }_0}$
$\Rightarrow n=6$