Integrating the field gives the potential. Running that backward, the field is the slope of the potential. In three dimensions "slope" means the gradient, built from partial derivatives. This is often the fastest way to a field: find the scalar $V$ (no components), then differentiate.
🔑 Key idea
$$E_x = -\frac{\partial V}{\partial x}, \quad E_y = -\frac{\partial V}{\partial y}, \quad E_z = -\frac{\partial V}{\partial z}, \qquad \text{that is}\quad \vec{E} = -\nabla V.$$
For a potential that depends only on $r$: $E_r = -dV/dr$. The field points downhill in $V$, at right angles to the equipotentials, and is strongest where they crowd together.
💡 Why it works
Take a tiny step $d\ell$ in some direction. From $dV = -\vec{E}\cdot d\vec{\ell}$, the component of $\vec{E}$ along the step is $-dV/d\ell$. Step along $x$ and you get $E_x = -\partial V/\partial x$; the same for $y$ and $z$. Stepping along an equipotential changes nothing, so the field has no component along it: field lines cross equipotentials at right angles.
A dipole with its equipotentials (dashed). Field lines cross them at right angles and are densest where the equipotentials crowd. The black dashed line, $V = 0$, runs between the charges, where the field is strong.
Think of a hiking map. The equipotentials are the contour lines, $V$ is the height, and $\vec{E}$ points straight downhill, steepest where the contours bunch up. A conductor in equilibrium has $E = 0$ inside, so the whole conductor sits at one potential and its surface is an equipotential. That is why field lines always meet a metal surface at right angles.
Worked examples
✏️ Worked example: One dimension
Along the $x$-axis, $V(x) = 5x^2 - 3x + 2$ (volts, $x$ in meters). Find $E_x$ at $x = 1.0\ \text{m}$.
$E_x = -\dfrac{dV}{dx} = -(10x - 3)$.
At $x = 1.0$: $E_x = -(10 - 3) = -7\ \text{V/m}$. The field points along $-x$, toward lower $V$.
On a ring's axis $V = kQ/\sqrt{z^2 + R^2}$. Find $E_z$.
$E_z = -\dfrac{dV}{dz} = -kQ\cdot\left(-\tfrac{1}{2}\right)(z^2 + R^2)^{-3/2}\cdot 2z$ by the chain rule.
$E_z = \dfrac{kQz}{(z^2 + R^2)^{3/2}}$: the same result as adding field vectors, with no components to track.
Answer: $E_z = \dfrac{kQz}{(z^2 + R^2)^{3/2}}$
⚠️ Common mistake
"Where $V = 0$, the field must be zero too." The field depends on how fast $V$ changes, not on its value. Midway between $+q$ and $-q$, $V = 0$ but the potential is dropping steeply from the positive charge toward the negative one, so $E$ is large there. The reverse happens inside a charged metal sphere: $V$ is not zero but it is flat, so $E = 0$. Check: look at the slope, never the height.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.