Q. A cylinder of radius $r$ and length $l$ is placed in an uniform electric field E parallel to the axis of the cylinder. The total flux for the surface of the cylinder is given by
A) Zero
B) $2\pi r^{2}E$
C) $\pi r^{2}E$
D) $\left( \pi r^{2} + \pi l^{2} \right)E$
Explanation
(a) Let us take a small area $\mathrm{\Delta}S$ on the cylindrical surface.
[image:1_image369.emf]
The normal to this area will be perpendicular to the axis of the cylinder and electric field is parallel to axis of cylinder.
So, flux $\mathrm{\Delta}\phi = \mathbf{E.}\mathrm{\Delta}\mathbf{S}$
$\mathbf{=}E\mathrm{\Delta}S\cos{90{^\circ} = 0}$
Summing over the entire surface, the total flux from the surface of cylinder is zero.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.