An equipotential surface is a three-dimensional surface in which the electric potential V is the same at every point.
Using the example of a single positive charge q, the expression for V is:
$$V = \frac{1}{4 \pi \epsilon_{0}} \frac{q}{r}$$
If a test charge $q_{0}$ is moved from point to point on an equipotential surface, the electric potential energy $q_{0}V$ will remain constant. In equation form, this means that the work done is 0:
$$\begin{aligned} W &= \, – \Delta U \\ &= \, – q_{0} \, \Delta V \\ &= 0 \end{aligned}$$
It follows that $\vec{E}$ must be perpendicular to the equipotential surface at every point. How do you reach this conclusion? Recall that:
$$\begin{aligned} dV &= \frac{\partial V}{\partial x} \, dx + \frac{\partial V}{\partial y} \, dy + \frac{\partial V}{\partial z} \, dz \\ &= \vec{\nabla} V . d\vec{l} \\ &= \, – \vec{E}.d\vec{l} \end{aligned}$$
Since dV = 0, $\vec{E}.d\vec{l}$ is 0 $\rightarrow$ perpendicular.
At each point, the direction of $\vec{E}$ is the direction in which V decreases most rapidly.
$$\begin{aligned} dV &= \vec{\nabla}V . d\vec{l} \\ &= \, – \vec{E} . d\vec{l} \\ &= \, – E \, dl \, \text{cos} \, \phi \end{aligned}$$
where $\phi$ is the angle between electric field and displacement vector (direction a point charge move).
In a region where an electric field is present, we can construct an equipotential surface through any point.
Note: Equipontial surfaces for different potentials can never touch or interact.
When all the charges are at rest (equilibrium), the electric field just outside a conductor must be perpendicular to the surface at every point. If the electric field contains a non-zero parallel component, there will be a force on the charges at the surface which will cause the charges to move and distribute themselves.
When all charges are at rest, the surface of a conductor is always an equipotential surface.
Next: Gauss’s Law (Simple Version)
