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 center 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:

• A. 6
• B. 7
• C. 8
• D. 9

Answer 1

Answer: (A)

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 $E_{\text{cyl}}=\frac{\lambda }{2\pi d{ϵ}_0}$

Here, $\lambda$ is charge per unit length $\lambda =\rho \times A=\pi R^2\rho$

and $d=2R$

Thus, $E_{\text{cyl}}=\frac{\pi R^2\rho }{2\pi \left(2R\right){ϵ}_0}=\frac{\rho R}{4{ϵ}_0}$

Field due to sphere is given by $E_{\text{sph}}=\frac{1}{4\pi {ϵ}_0}\frac{Q}{d^2}$

Here, $Q$ is the total charge in the sphere. $Q=-\rho \times V=-\frac{4}{3}\pi (\frac{R}{2}{)}^3\rho$

and $d=2R$

Thus, $E_{\text{sph}}=-\frac{4\pi (\frac{R}{2}{)}^3\rho }{3\times 4\pi {ϵ}_0(2R{)}^2}=-\frac{\rho R}{96{ϵ}_0}$

Thus, the net electric field is $E=\frac{\rho R}{{ϵ}_0}(\frac{1}{4}-\frac{1}{96})=\frac{23\rho R}{96{ϵ}_0}$

Thus $16k=96\Rightarrow k=6$