Module 2 : Electrostatics
Lecture 7 : GAUSS'S LAW
i.e. a total outward flux of $2\pi r^2 \mid E\mid$. Hence
 
\begin{displaymath}2\pi r^2\mid E\mid = \frac{Q}{\epsilon_0} =\frac{ \pi r^2\sigma}{\epsilon_0}\end{displaymath}
  so that
 
\begin{displaymath}\vec E = \frac{\sigma}{2\epsilon_0}\hat n\end{displaymath}
  where $\hat n$is a unit vector perpendicular to the sheet, directed upward for points above and downwards for points below (opposite, if the charge density is negative).
  Field due to a uniformly charged sphere of radius $R$with a charge $Q$
  By symmetry, the field is radial. Gaussian surface is a concentric sphere of radius $r$. The outward normals to the Gaussian surface is parallel to the field $\vec E$at every point. Hence $\int \vec E\cdot \vec{dS} = 4\pi r^2\mid E\mid$
   
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