Question

An infinitely long solid cylinder of radius $R$ has a uniform volume charge density $\rho$ . It has a spherical cavity of radius $\frac{R}{2}$ with it's 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 $\frac{23\rho R}{16k{ϵ}_0}$. The value of $k$ is

Answer 1

The given system of cylinder with cavity can be expressed as superposition of Infinite cylinder with charge density $+\rho$ and a sphere with charge density $-\rho$.


Field due to infinite cylinder is given by $E_{cyl}=\frac{\lambda }{2\pi {ϵ}_{0}d}$​
Here, λ is charge per unit length $\lambda =\rho \times A=\pi R^2\rho$
and $d=2R$
Thus, $E_{cyl}=\frac{\pi R^2\rho }{2\pi \left(2R\right){ϵ}_o}=\frac{\rho R}{4{ϵ}_o}$

Field due to sphere is given by $E_{sph}=\frac{1}{4\pi {ϵ}_o}\frac{Q}{d^2}$
Here, $Q$ is the total charge in the sphere. $Q=-\rho \times V=-\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3\rho$ ​
and $d=2R$

Thus, $E_{sph}=-\frac{4\pi {\left(\frac{R}{2}\right)}^3\rho }{3\times 4\pi {ϵ}_o{\left(2R\right)}^2}=-\frac{\rho R}{96{ϵ}_o}$​​
Thus, the net electric field is $E=\frac{\rho R}{{ϵ}_o}(\frac{1}{4}-\frac{1}{96})=\frac{23\rho R}{96{ϵ}_o}$
On comparing with the relation given, we get
$16k=96\Rightarrow k=6$