Potential is a scalar, and the field is its slope

Define the potential at a point as the potential energy a charge would have there, divided by that charge:

$$V = \frac{U}{q_0}, \qquad U = qV, \qquad 1\ \text{V} = 1\ \text{J}/\text{C}.$$

For a single point charge, with $V \to 0$ infinitely far away,

$$V = \frac{kq}{r}.$$

One power of $r$, and $q$ enters with its sign: a negative charge surrounds itself with negative potential. For several charges, superposition is arithmetic:

$$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2} + \cdots$$

That is the entire procedure. No components, no angles between contributions, no partial cancellation from geometry. Students arriving from Unit 8, where every field problem was a component sum, often keep the vector machinery running and report two contributions "at $30$ degrees and $150$ degrees" as partially cancelling. There is nothing to resolve: two charges each contributing $300$ V give $600$ V, full stop. Angles enter a potential problem only by setting the distances $r$.

The one thing that survives from the vector work is the sign. Cancellation in potential happens when a positive and a negative contribution are equal in size, and only then. Directions are not involved.

Going the other way, the field is recovered by differentiating:

$$E_x = -\frac{dV}{dx}, \qquad \vec{E} = -\nabla V.$$

Everything hangs on that being a derivative. $\vec{E}$ reports how fast $V$ is changing and in which direction it falls fastest; it has no access to what $V$ actually is. So $V = 0$ and $E = 0$ are independent claims, and here are the two standard demonstrations.

1. $V = 0$ with $E$ large. Put $+q$ at $x = 0$ and $-q$ at $x = d$. At the midpoint the two contributions are $+2kq/d$ and $-2kq/d$, so $V = 0$. But both field contributions point the same way, from the positive charge toward the negative one, giving $E = 8kq/d^2$. The potential is plunging steeply through zero, and that steepness IS the field.
2. $E = 0$ with $V$ large. Put $+q$ at both ends instead. At the midpoint the fields cancel exactly, $E = 0$, while $V = 4kq/d$, nowhere near zero. The midpoint is a stationary point of $V$: a minimum along the line joining the charges, a maximum across it, so the slope vanishes in every direction while the value does not.

The habit worth building is to stop reading the number and start reading the graph. A zero of $V$ is a crossing; a zero of $E$ is a flat spot. They have no reason to occur together.

Between two parallel plates a distance $d$ apart at a potential difference $V$, the field is uniform and the derivative collapses to a ratio:

$$E = \frac{V}{d} \qquad \text{(uniform field only)}.$$

That is a genuine result and it is heavily used. It is also the most over-exported formula in the unit. Away from a uniform field, dividing some available voltage by some available length is not an approximation to $-dV/dx$; it is an unrelated calculation.

The point charge is the trap that hides itself, because $V = kq/r$ and $E = kq/r^2$ happen to satisfy $E = V/r$ exactly. The number comes out right and the reasoning is still wrong, which means it fails silently the first time the potential is not a pure $1/r$. Take $V(x) = 6x^2$ volts. At $x = 2.0$ m the potential is $24$ V, so $V/x$ gives $12$ V/m, while the field is

$$E_x = -\frac{dV}{dx} = -12x = -24 \ \text{V/m},$$

twice as large and pointing in $-x$. Differentiate FIRST and substitute afterward. Substituting first turns $V(x)$ into a constant, and the derivative of a constant is zero.

Going the other way, from field to potential, the general statement is an integral:

$$\Delta V = V_B - V_A = -\int_A^B \vec{E}\cdot d\vec{l}.$$

An equipotential is a surface on which $V$ is constant. Two consequences follow directly from $W = q\Delta V$ and $\vec{E} = -\nabla V$.

1. Moving a charge along an equipotential costs zero work. $\Delta V = 0$, so $W = q\Delta V = 0$ regardless of how far the charge travelled. A half-circuit of a $50$ V contour is free whether the contour is a metre around or a kilometre.
2. Field lines cross equipotentials at right angles. If $\vec{E}$ had any component along the surface, moving that way would change $V$, and the surface would not be an equipotential. If a sketch shows a field line meeting a contour at $60$ degrees, one of the two is drawn wrong.

Behind both is the fact that the electrostatic field is conservative: $\oint \vec{E}\cdot d\vec{l} = 0$ around any closed loop, so $\Delta V$ between two points is fixed by the endpoints alone. The long way round does not accumulate extra voltage, and passing through a region of stronger field on the way does not either. That path independence is what allows $V$ to be a function of position at all.

It is also a licence to be lazy in the best way. To evaluate $-\int \vec{E}\cdot d\vec{l}$, choose whatever route makes the integral trivial, usually straight along a field line or straight along an equipotential, and the answer you get is the answer for every other route, including the one the charge actually took.