Field from Potential: E = −∇V

Getting the field from the potential simply means this: if you know the voltage at every point, the electric field is the downhill slope of that voltage landscape. The upside-down triangle $\nabla$ is read "del", and $\nabla V$ is called the gradient of $V$ because it measures how steeply $V$ climbs, the way a road's gradient measures how steep it is. It's useful because voltages are plain numbers and easy to add up, while fields are arrows. The equation $\mathbf{E} = -\nabla V$ is just two pieces: $\nabla V$ (the uphill direction, with length equal to the steepness) and the minus sign (flip it, so the field points downhill). It looks fancy, but it's only "the field points downhill", so bear with the symbols.

Reading a map. You already know this picture from hiking. On a topographic map, contour lines join points at equal height. Water runs downhill straight across the contours, fastest where they crowd together. Equipotentials are the contour lines of $V$, and the electric field is the downhill direction (for a positive charge), strongest where the equipotentials bunch up.

Why $\mathbf{E} = -\nabla V$. For a tiny step $d\mathbf{l}$, the definition of potential says $dV = -\mathbf{E}\cdot d\mathbf{l}$. The chain rule says $dV = \nabla V\cdot d\mathbf{l}$, the directional derivative times the step's length. These agree for every step, so $\mathbf{E} = -\nabla V$. In one dimension, $E_x = -\frac{dV}{dx}$; between parallel plates, where $V$ drops steadily by $\Delta V$ across a gap $d$, that's $E = \Delta V/d$.

Equipotentials are perpendicular to field lines. Think about walking along a contour line on a hillside: you stay at the same height, so the downhill direction must be straight off to the side. Same here. Step along an equipotential and $dV = 0$, so $\mathbf{E}\cdot d\mathbf{l} = 0$: the field is perpendicular to every direction within the equipotential surface. It's the fact from The Gradient, Tangent Planes and Normal Lines that gradients are normal to level sets. A conductor's surface is an equipotential (with no field inside, moving charge around it takes no work), so field lines always meet conductors at right angles.

A built-in test. Since $\mathbf{E}$ is a gradient, $\nabla\times\mathbf{E} = \nabla\times(-\nabla V) = \mathbf{0}$ automatically (Curl and Divergence). A proposed static field with nonzero curl can't be electrostatic.

The dipole, the easy way. Remember the dipole from The Electric Field? Getting its field meant adding two vectors and approximating. Watch how much easier it is with potential. Charges $+q$ and $-q$ sit a distance $d$ apart, and $\theta$ is measured from the axis, which points from $-q$ to $+q$. At a distance $r \gg d$ the charges are $r_\pm \approx r \mp \frac d2\cos\theta$ away, so

Now take the gradient in polar coordinates, $\nabla V = \frac{\partial V}{\partial r}\hat{\mathbf{r}} + \frac1r\frac{\partial V}{\partial\theta}\hat{\boldsymbol{\theta}}$:

On the axis that's $\frac{2kp}{r^3}$, the dipole field from The Electric Field, with no vectors added at all. The $1/r^3$ falloff is faster than a single charge's because the two fields nearly cancel far away.

A common mistake. It's tempting to think that where $V = 0$ the field must be zero too, since zero potential sounds like "nothing there". It isn't. The field depends on how $V$ changes, not on its value, just as a hillside at sea level can still be steep. On the plane midway between a dipole's charges $V = 0$, yet the field is strong. It works the other way too: $\mathbf{E} = \mathbf{0}$ doesn't mean $V = 0$, and inside a charged metal sphere $V$ is constant but not zero. Two smaller slips: don't drop the minus sign (the field points toward lower potential), and don't forget the $\frac1r$ in the $\theta$ part of the gradient.

Field solvers for circuit boards, high-voltage bushings and electron optics compute the scalar potential on a grid and then take $-\nabla V$, because one scalar is cheaper to find than three vector components. Medicine reads potentials directly: an ECG or EEG measures potential differences between electrodes on the skin, and from outside, the heart's electrical activity looks much like a rotating dipole. Power engineers use $E = -dV/dr$ to set safe distances around grounding rods and fallen lines.