Module 2 : Electrostatics
Lecture 8 : Electrostatic Potential
  Consider the line integral $\int_a^b\vec E\cdot \vec{dl}$. Since the integral is independent of the path of integration, we have
 
\begin{displaymath}\int_a^b\vec E\cdot \vec{dl}\;\; {\rm path\;\; 1} = \int_a^b\vec E\cdot \vec{dl}\;\; {\rm path\;\; 2} \end{displaymath}
 
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  Since $\int_a^b \vec E\cdot\vec{dl} = - \int_b^a\vec E\cdot\vec{dl}$along any particular path,
 
\begin{displaymath}\int_a^b\vec E\cdot \vec{dl}\;\; {\rm along path\;\; I} + \int_a^b\vec E\cdot \vec{dl}\;\; {\rm along path\;\; II} = 0 \end{displaymath}
  L.H.S. is the line integral of the electric field along the closed loop,
 
\begin{displaymath}\oint \vec E\cdot\vec{dl} = 0\end{displaymath}

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