Question

A long cylindrical volume contains a uniformly distributed charge of density rho.Find flux due to electric field through curved surface of the small cylinder whose axis is op,and whose radius is a.O lies on axis ab of main cylinder and its axis op is perpendicular to ab

Answer 1

volume charge density of cylinder is $\rho$

if a is radius of base of cylinder and l is the length of cylinder then,

charge on cylinder , q = $\rho\pi a^2l$

from Gaussian theorem,

electric flux through cylinder is the ratio of charge enclosed inside the gaussian surface to permittivity of medium.

e.g., $\phi=\frac{\rho\pi a^2l}{\epsilon_0}$

we also know, $\phi=E(2\pi al)$

[ here you should noticed that electric flux will be cancelled in case of lateral surface , only flux through curve surface will appear ]

now, E(2πal) = $\frac{\rho\pi a^2l}{\epsilon_0}$

or, E = $\frac{\rho a}{2\epsilon_0}$