Gujarati
Hindi
1. Electric Charges and Fields
medium

A cylinder of radius $R$ and length $L$ is placed in a uniform electric field $E$ parallel to  the cylinder axis. The total flux for the surface of the cylinder is given by-

A

$2 \pi R^2E$

B

$\pi R^2/E$

C

$(\pi R^2/ \pi R)/E$

D

zero

Solution

Flux through surface $A$

$\phi_{A}=E \times \pi R^{2}$

$\phi_{B}=-E \times \pi R^{2}$

Flux through curved surface, $C=\int E . d s$

$=\int E d s \cos 90^{\circ}=0$

$\therefore$ Total flux through cylinder $=\phi_{A}+\phi_{B}+\phi_{C}=0$

Standard 12
Physics

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