A number attached to a place

Keep the two quantities and their units apart:

$$V = \frac{kq_{\text{source}}}{r} \ \ [\text{volts}], \qquad U = qV \ \ [\text{joules}], \qquad 1\ \text{V} = 1\ \text{J/C}.$$

So "the potential at P is $40$ V" does not mean there are $40$ joules at P. It means every coulomb placed at P carries $40$ J. A proton at P therefore carries

$$U = (1.6\times10^{-19}\ \text{C})(40\ \text{V}) = 6.4\times10^{-18}\ \text{J}.$$

Ask which one the question wants: the property of the place, or the energy of the charge you put there. The units settle every case.

Potential is a scalar, so contributions from several charges add term by term:

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

The signs of the charges do everything that direction does elsewhere. Geometry enters a potential problem only through the distances $r$, and nothing is ever resolved into components.

Two positive charges each contributing $250$ V put the point at $500$ V, even though they sit on opposite sides of it. A negative charge subtracts, and subtraction is the only kind of cancellation potential has. Coming straight from field superposition, the temptation is to leave the vector machinery running: take components of each $kq/r$, or refuse to add contributions arriving from opposite directions. Switch it off.

Dividing a potential difference by a distance gives the average field over that interval, along the direction you measured:

$$E_{\text{avg}} = \frac{|\Delta V|}{d}, \qquad \text{measured along the field, not across it}.$$

Where the field is uniform, the average is the value everywhere, so between closely spaced parallel plates $E = V/d$ is exact and holds at every interior point. That is the case worth memorizing.

Near a charged sphere it is not. If the potential drops $60$ V over the first $0.10$ m outward from the surface, then $600$ N/C is the average over that interval and the field right at the surface is considerably stronger, because the field falls off steeply. There, $E = kq/r^2$ is what gives the value at a point.

The direction detail matters too: $d$ has to be measured along the field. Move across the field and $\Delta V$ is zero however far you go, which is the next section.

An equipotential is a surface on which $V$ is constant. Two consequences:

1. No work to move along one. $W = q\,\Delta V$ and $\Delta V = 0$ between any two points on it, however far apart they are and however long the path.
2. Field lines cross equipotentials at right angles, pointing from higher potential toward lower. A sketch showing lines slicing a contour at a slant is wrong: any along-the-contour component of $\vec{E}$ would make $V$ change along it.

Now the pair of zeros. $V$ is a number at a point; $\vec{E}$ depends on how that number changes from point to point. So they are independent, and the two standard cases prove it:

1. Halfway between $+q$ and $-q$: the potentials cancel, $V = 0$, while the field is at its strongest there. $V$ is dropping steeply through zero, and that steep change is the field.
2. Halfway between two $+q$ charges: the field contributions cancel, $\vec{E} = 0$, while $V$ is large and positive.

A charge released where $V = 0$ is generally not in equilibrium. Check whether $V$ changes as you step across the point; the reading at the point settles nothing.