Electric Field from Potential (1D)

Electric field component along any direction is the negative rate of change of potential in that direction. The negative sign: field points from high V to low V.

From the definition

The work done by the field moving charge $q_0$ by $d\vec{r}$:

$$dW = q_0\,\vec{E}\cdot d\vec{r} = -q_0\, dV$$

The last equality follows from $V = W/q_0$ and the sign convention: when the field does positive work, potential decreases.

For motion along a single direction $r$:

$$E_r\, dr = -dV \implies E_r = -\frac{dV}{dr}$$ $$\boxed{E = -\frac{dV}{dr}}$$

Interpretation

The electric field is the negative gradient of potential. Field lines point from high potential to low potential — downhill on the potential landscape.

Consequence for equipotentials

On an equipotential surface, $dV = 0$ along the surface. Therefore $E$ along the surface is zero — the field has no component tangent to the surface. Field lines are always perpendicular to equipotential surfaces.