Last updated: 2026-06-15

Electric Potential of a Point Charge

The electric potential at a point in space tells you how much energy it takes to bring one unit of positive charge from infinity to that point. For a single point charge Q at distance r, the potential is:

$V = k \frac{Q}{r}$

where $k = 1/(4\pi\varepsilon_0) \approx 8.9876 \times 10^9\ \text{N·m}^2/\text{C}^2$ is Coulomb's constant. This calculator computes V given Q and r. Because V is a scalar (not a vector), the sign of V follows directly from the sign of Q.

Electric potential vs. electric field

The electric field $\mathbf{E}$ points from high potential to low potential and is related to V by $E = -dV/dr$ (in one dimension). The field is a vector; the potential is a scalar. For problems involving multiple charges, you can add potentials numerically:

$V_\text{total} = k \sum_i \frac{Q_i}{r_i}$

Adding electric fields from multiple charges requires vector addition, which is more involved. For this reason, potential is often the preferred route when computing forces or energies in multi-charge systems.

Formula

The reference convention is $V = 0$ at $r = \infty$. The potential falls off as $1/r$ (compared with the Coulomb force, which falls off as $1/r^2$).

Worked example — positive charge

A proton carries charge $Q = +1\ \mu\text{C} = 1 \times 10^{-6}\ \text{C}$. Find the electric potential at $r = 0.1\ \text{m}$:

A charge of just 1 µC — a quantity so small you cannot feel it on skin — creates nearly 90 kV at 10 cm range. This illustrates why electrostatic potentials can be enormous even for tiny charges.

Worked example — negative charge

A charge of $Q = -1\ \mu\text{C}$ at the same distance:

$V = 8.9876 \times 10^9 \times \frac{-1 \times 10^{-6}}{0.1} \approx -89\,876\ \text{V}$

The potential is negative, meaning you would gain energy (not expend it) by bringing a positive test charge from infinity to that point. Energy is released because the positive test charge is attracted to the negative source charge.

Superposition

For two or more point charges, the total potential is the algebraic sum of individual potentials. A dipole — charges $+Q$ and $-Q$ separated by distance $d$ — produces zero potential at any point equidistant from both, because the positive and negative contributions cancel. Along the axis of the dipole, the potentials add. This scalar superposition makes potential calculations far simpler than field calculations in many cases.

Connection to voltage in circuits

In circuit analysis, voltage is simply the electric potential difference between two points. The 1.5 V across a battery terminal means that the positive terminal is 1.5 V higher in potential than the negative terminal, so each coulomb of charge that passes through the battery gains 1.5 J of energy. The formula $V = kQ/r$ describes the potential from a static point charge in free space, which is the foundation of that same concept.