Electric Potential of a Point Charge

Potential is the energy side of the same story the field tells with force. Carry a unit charge in from infinity against the field of a point charge $Q$, add up the work, and the total is $V = k_e Q/r$. One power of $r$ rather than two, and no direction — potential is a scalar. That single difference is why potential is usually the easier tool: fields from several charges have to be added as vectors, with components and angles, while potentials from several charges are simply added up like numbers, and the field can be recovered afterwards as the gradient.

A 1 nC charge holds the space 10 cm away at $V = (8.988 \times 10^9 \times 10^{-9})/0.1 = 89.9\ \text{V}$. Bring a second 1 nC charge from far away up to that point and it takes $qV = 8.99 \times 10^{-8}$ J of work — the potential energy of the pair. Note how much more gently the potential falls than the field: at double the distance the field is a quarter but the potential is only a half.

The zero matters, and here it is at infinity. That is the only reference for which $k_e Q/r$ is correct, and it is a mathematical convenience rather than a physical fact — potential differences are measurable, absolute potentials are not. In circuit work the zero is instead a chassis or an earth electrode, which is why a voltmeter has two leads and why "the voltage at this node" is always short for "the voltage between this node and the one we agreed to call zero".

The commonest confusions are all name collisions. Potential (volts, joules per coulomb) is not potential energy (joules); multiply by the charge to get from one to the other. Potential is also not the field: $V$ can be zero where $E$ is not — midway between equal and opposite charges, for instance — and $E$ can be zero where $V$ is not, as it is everywhere inside a hollow charged conductor. Beyond the naming, two practical points. Like the field, this form is exact outside a spherically symmetric charge and wrong inside it; inside a conductor the potential is constant, not falling as $1/r$. And superposition applies to $V$ as a signed sum, so a positive and a negative charge can cancel to zero potential at a point where the field is perfectly strong — which is the whole reason equipotential surfaces and field lines are drawn as two different pictures.