Module 2 : Electrostatics
Lecture 7 : Electric Flux
 

Consider a circular strip of radius $r$at a depth $h$from the apex of the cone. The angle between the electric field through the strip and the vector $\vec {dS}$ is $\pi-\theta$, where $\theta$ is the semi-angle of the cone. If $dl$ is the length element along the slope, the area of the strip is $2\pi r dl$. Thus,

\begin{displaymath}\vec E\cdot \vec{dS} = 2\pi r dl \mid E\mid \sin\theta\end{displaymath}

We have, $l = h/\cos\theta$, so that $dl = dh/\cos\theta$. Further, r = h tan $\theta$ Substituting, we get

\begin{displaymath}\vec E\cdot \vec{dS} = 2\pi h\tan^2\theta \mid E\mid dh\end{displaymath}

Integrating from $h=0$to $h=H$, the height of the cone, the outward flux is .

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