Module 2 : Electrostatics
Lecture 7 : Electric Flux
 

The direction of the area vector $\vec {dS}$, is also radial at each point of the surface $ \vec{dS} = dS\hat r$. The flux

\begin{eqnarray*} \phi &=& \int \vec E\cdot\vec{dS}\\ &=& \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\int dS \end{eqnarray*}


The integral over $dS$is equal to the surface area of the sphere, which is, $4\pi R^2$. Thus the flux out of the surface of the sphere is

\begin{displaymath}\phi = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\cdot 4\pi R^2 = \frac{Q}{\epsilon_0}\end{displaymath}

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