[NOTES/EM-02015] Proof of curl free nature of \(\vec E\)

To prove \(\nabla \times \vec E=0\), it is sufficient to prove it for a point charge and use superposition principle. We use the notation \(\vec r=(x,y,z)\) and \(\vec R=\vec r-\vec r_0=(X,Y,Z)\). Then the electric field due to a point charge located at \(\vec r_0\) is \begin{equation} \vec E = \frac{q}{4\pi\epsilon_0} \frac{\vec r-\vec r_0}{|\vec r-\vec r_0|^3}. \end{equation} The components of \(\vec R\) are donated by \(X,Y,Z\), therefore \begin{eqnarray} (\nabla \times \vec E)_x &=& \pp[E_y]{z} -\pp[E_z]{y} =\pp[E_y]{Z} -\pp[E_z]{Y}\\ &=&\frac{q}{4\pi\epsilon_0}\left[\pp{Z}\Big(\frac{Y}{R^2}\Big) - \pp{Y}\Big(\frac{Z}{R^2}\Big)\right] \end{eqnarray} Note that \(R=\sqrt{X^2+Y^2+Z^2}\) and therefore \begin{equation} \pp{Z}\Big(\frac{Y}{R^2}\Big) - \pp{Y}\Big(\frac{Z}{R^2}\Big) = Y\frac{-2Z}{R^3}-Z\frac{-2Y}{R^3}=0. \end{equation} Therefore \(\nabla \times \vec E_x=0\). Similarly other components of \(\nabla \times \vec E\) vanish.