Question

An infinitely long cylinder of radius $R$ has an infinitely long cylindrical cavity of radius $\frac{R}{2}$ as shown in figure. The renaming portion has uniform volume charge density $\rho$. The magnitude of electrical field at the centre of the cavity is-

• A. $\frac{\rho R}{6{\epsilon }_0}$
• B. $\frac{\rho R}{2{\epsilon }_0}$
• C. $\frac{\rho R}{3{\epsilon }_0}$
• D. $\frac{\rho R}{4{\epsilon }_0}$

Answer 1

The correct option is A $\frac{\rho R}{6{\epsilon }_0}$

The given system of cylinder with cavity can be expressed as superposition of infinite cylinder with charge density $+\rho$ and a sphere with charge field due to infinite cylinder is given by $E_{cyl}=\frac{\lambda }{2\pi d{ϵ}_0}$

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

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

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

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

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, $16K=96$

or, $K=6$

hence, magnitude of electrical field at the centre of the cavity is $=\frac{\rho R}{6{ϵ}_0}$