Electrostatic electric fields are conservative, so the electric potential can be connected directly to the spatial structure of the electric field.

For electric potential $V$ and electric field $\mathbf E$,

$$\boxed{\mathbf E=-\nabla V}.$$

The minus sign is physical: the electric field points in the direction in which electric potential decreases most rapidly.

From work to potential difference

A charge $q$ moving from $A$ to $B$ experiences electric work

$$W=q\int_A^B\mathbf E\cdot d\mathbf r.$$

Electric potential energy changes by the negative of the work done by the electrostatic force,

$$\Delta U=-W.$$

Since $V=U/q$,

$$\boxed{ V(B)-V(A) =-\int_A^B\mathbf E\cdot d\mathbf r}. $$

Because the electrostatic field is conservative, this result depends only on the endpoints. A different path between the same points gives the same potential difference.

For an infinitesimal displacement $d\mathbf r$,

$$dV=-\mathbf E\cdot d\mathbf r,$$

which is the local form of the same relationship and leads to $\mathbf E=-\nabla V$.

Equipotential surfaces

An equipotential surface is a surface on which $V$ is constant. A displacement tangent to such a surface has

$$dV=0,$$

so

$$\mathbf E\cdot d\mathbf r=0.$$

Therefore the electric field is perpendicular to a regular equipotential surface. Moving a charge along an equipotential requires no work from the electrostatic field.

The spacing of equipotential surfaces also carries information: where a given change in $V$ occurs over a shorter distance, $\lVert\nabla V\rVert$ is larger and the electric field is stronger.

Example: recovering the field of a point charge

With zero potential chosen at infinity, a point charge $Q$ has

$$V(x,y,z)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}, \qquad r=\sqrt{x^2+y^2+z^2}.$$

Writing $k=1/(4\pi\varepsilon_0)$, differentiation gives

$$\frac{\partial V}{\partial x}=-kQ\frac{x}{r^3},\qquad \frac{\partial V}{\partial y}=-kQ\frac{y}{r^3},\qquad \frac{\partial V}{\partial z}=-kQ\frac{z}{r^3}.$$

Therefore

$$\nabla V =-kQ\frac{(x,y,z)}{r^3} =-kQ\frac{\hat{\mathbf r}}{r^2},$$

and hence

$$-\nabla V =kQ\frac{\hat{\mathbf r}}{r^2} =\mathbf E.$$

The familiar point-charge electric field is recovered from one scalar function.

The potential description replaces a vector field with one scalar function while retaining the complete electrostatic field through its gradient. This often makes electrostatic calculations and geometric reasoning substantially simpler.