Module 2 : Electrostatics
Lecture 7 : Electric Flux
  We define the flux of the electric field through an area $\vec {dS}$to be given by the scalar product
 
\begin{displaymath}d\phi = \vec E\cdot\vec{dS}\end{displaymath}   If $\theta$is the angle between the electric field and the area vector   \begin{displaymath}d\phi = \mid E\mid \mid dS\mid \cos\theta\end{displaymath} 
  For an arbitrary surface S, the flux is obtainted by integrating over all the surface elements
  \begin{displaymath}\phi = \int d\phi = \int_S \vec E\cdot\vec {dS}\end{displaymath}
  If the electric field is uniform, the angle $\theta$is constant and we have
  \begin{displaymath}\phi = ES\cos\theta= E (S\cos\theta)\end{displaymath}
  Thus the flux is equal to the product of magnitude of the electric field and the projection of area perpendicular to the field.
  \includegraphics{fig9.eps}
 
  Unit of flux is N-m $^2$/C. Flux is positive if the field lines come out of the surface and is negative if they go into it.
   
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