Electric field and potential are closely related. (As seen in the equation below) If we know $\vec{E}$ at various points, we can calculate potential differences.
$$V_{a} – V_{b} = \int\limits_{a}^{b} \vec{E}.d\vec{l}$$
If we know the potential V at various points, we can use it to determine $\vec{E}$:
$$\begin{aligned} E_{x} \, &= \, – \frac{\partial V}{\partial x} \\ E_{y} \, &= \, – \frac{\partial V}{\partial y} \\ E_{z} \, &= \, – \frac{\partial V}{\partial z} \end{aligned}$$
Short Proof:
$$\begin{aligned} V_{a} – V_{b} &= \int\limits_{a}^{b} \vec{E}.d\vec{l} \\ \int\limits_{b}^{a} dV &= \int\limits_{a}^{b} \vec{E}.d\vec{l} \\ – \int\limits_{a}^{b} dV &= \int\limits_{a}^{b} \vec{E}.d\vec{l} \\ -dV &= \vec{E}.d\vec{l} \end{aligned}$$
Note that $d\vec{l} = dx \, \hat{i} + dy \, \hat{j} + dz \, \hat{k}$. Hence,
$$ -dV = E_{x} \, dx + E_{y} \, dy + E_{z}$$
Since $-dV = – \left( \frac{\partial V}{\partial x} \, dx + \frac{\partial V}{\partial y} \, dy + \frac{\partial V}{\partial z} \, dz \right)$ by definition, you can just equate the components of $\vec{E}$ with the derivatives counterparts.
In general: $\vec{E} = – \left( \hat{i} \frac{\partial V}{\partial x} + \hat{j} \frac{\partial V}{\partial y} + \hat{k} \frac{\partial V}{\partial z} \right) = – \vec{\nabla} V$.
Note: If you are not familiar with $\vec{\nabla}$, it would not matter at this level of electromagnetism. You can ignore it for now. It is a gradient operator. (Just like $\frac{d}{dx}$ is an operator except that $\vec{\nabla}$ gives the gradient of a function when applied to the function.
If $\vec{E}$ is radial with respect to a point or an axis and r is the distance from the point or the axis, then
$$E_{r} = \, – \frac{\partial V}{\partial r}$$
Some notes on electric field:
Direction of the electric field is in the direction of
- decreasing V
- increasing kinetic energy
