Module 2 : Electrostatics
Lecture 8 : Electrostatic Potential
 

Comparing the above with eqn. (A) above,

\begin{displaymath}\vec E = -\nabla\phi\end{displaymath}

In cartesian coordinates,

\begin{displaymath}\vec E = -\hat i\frac{\partial\phi}{\partial x} -\hat j \frac{\partial\phi}{\partial y} -\hat k\frac{\partial\phi}{\partial z}\end{displaymath}
  Example 14
  Exercise 2
  Exercise 3
   
Electric Field is Irrotational, i.e. Curl$\vec E = 0$
  This follows from Stoke's theorem.
 
\begin{displaymath}\oint \vec E\cdot\vec{dl} = \int_S (\vec\nabla\times\vec E)\cdot \vec{dS}\end{displaymath}
  where the surface integral is over any surface bounded by the closed curve. As the surface $S$ is arbitrary (as long as it is bounded by the same curve) , the integrand must vanish. Hence,
 
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