Scalar Potential

\(\Phi\). \(\vec{E}=-\nabla\Phi\).

Recall, \(\nabla\times\nabla\psi=0\).

Relation to Energy

Lets say we move a charge q from a point A to a point B.

Remember, \(\vec{F}=q\vec{E}\) then \(W=-\int_A^B\vec{F}\cdot \vec{d\ell} = q\int_A^B\nabla \Phi\cdot \vec{d\ell} = q\int_{\Phi_A}^{\Phi_B}d\Phi=q(\Phi_B-\Phi_A)\).

Note: This is path independent (due to the curl of the electric field being zero).

Adding one path to the point and subtracting the path from the point to the original point gives the closed loop and verifies Stoke’s theorem which shows the curl is zero.

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:35