Electric potential

Two charges can pull or push on each other with a force that depends on direction — but describing every point around a charge with an arrow (magnitude and direction) is hard work. Electric potential trades that vector picture for a single number at each point: the work done per unit positive charge in bringing it from far away, against the field, without speeding it up. Because it is just one number, potentials from several charges add by ordinary arithmetic, not vector algebra — no angles, no components. This article builds the definition of potential, the formula for a point charge and a system of charges, the special case of a dipole, the idea of an equipotential surface, and the compact link between potential and field, $E = -dV/dr$ , that ties this topic to everything that follows in electrostatics.

The core insight

Electric potential is the scalar work-per-unit-charge needed to bring a test charge from infinity to a point, $V = W/q_0$ , so potentials from multiple charges add algebraically and the field itself is just $E = -dV/dr$ , the negative gradient of this potential.

The core idea

Definition. Electric potential $V$ at a point is the work done per unit positive test charge in bringing it from infinity (the zero-potential reference) to that point, quasi-statically (no change in kinetic energy):

V = \frac{W}{q_0}

where: $W$ is the work done by an external agent against the electric force, and $q_0$ is the (small, positive) test charge. $V$ is a scalar with SI unit volt, $1\ \text{V} = 1\ \text{J/C}$ . Potential difference between two points A and B is $V_A - V_B = W_{B\to A}/q_0$ , the work per unit charge to move it from B to A.

Point charge. For a source charge $q$ , the potential at distance $r$ (with $V = 0$ at $r = \infty$ ) is

V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}

where: $\varepsilon_0$ is the permittivity of free space, so $1/4\pi\varepsilon_0 = k = 9\times10^9\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$ . Unlike the field, which is a vector falling as $1/r^2$ , potential is a scalar falling as $1/r$ , and it can be positive or negative depending on the sign of $q$ .

Superposition. Because potential is scalar, the net potential due to several point charges is the plain algebraic sum of the individual potentials — no components:

V_{\text{net}} = \frac{1}{4\pi\varepsilon_0}\sum_i \frac{q_i}{r_i}

This is far simpler than superposing fields, where directions matter.

Dipole potential. At a point P at distance $r$ from the centre of a dipole of moment $p$ , making angle $\theta$ with the dipole axis ( $r \gg$ dipole length):

V = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}

where: $p = q \times 2a$ is the dipole moment. On the axial line ( $\theta = 0°$ ), $V = kp/r^2$ (maximum); on the equatorial line ( $\theta = 90°$ ), $V = 0$ — every point there is equidistant from $+q$ and $-q$ , so the two contributions $+kq/r$ and $-kq/r$ cancel exactly, even though the field there is nonzero.

Equipotential surfaces join points of equal $V$ . Four properties matter for exams: no work is done moving a charge along one, since $V_A = V_B$ forces $W = q_0(V_A - V_B) = 0$ ; the field is always perpendicular to the surface, because any component along it would do work and break the constant- $V$ condition; surfaces crowd together where the field is strong and spread out where it is weak; and for a point charge they are concentric spheres, while for a uniform field they are parallel planes.

Figure

Equipotential spheres around a point charge with the electric field drawn perpendicular to each surface and spacing that widens with distance

Field from potential. The electric field equals the negative rate of change of potential with distance, taken along the direction of steepest decrease of $V$ :

E = -\frac{dV}{dr}

The minus sign shows $\vec{E}$ points toward decreasing potential, so a free positive charge accelerates from high to low potential. This also gives volt per metre as a second unit for field, identical to N/C.

Derivation ADeriving the potential of a point charge from the work integral

The goal is to derive $V = kq/r$ directly from the definition $V = W/q_0$ by computing, with a single integral, the work done bringing a small positive test charge $q_0$ from infinity up to a distance $r$ from a source charge $q$ against the repulsive or attractive Coulomb force, then dividing by $q_0$ to strip away the arbitrary size of the test charge and leave a property of the source alone.

Step 1. Set up the force. At a distance $x$ from $q$ , the Coulomb force on the test charge is $F(x) = kqq_0/x^2$ , directed radially. Moving the test charge inward by a tiny step $dx$ (note $dx$ is negative as $x$ decreases toward $r$ ), the external agent must supply a force equal and opposite to the electric force at every instant to keep the motion quasi-static, so the work done by the agent over that step is $dW = -F(x)\,dx$ .

Step 2. Integrate from infinity to $r$ . Total work is $W = -\int_{\infty}^{r} \dfrac{kqq_0}{x^2}\,dx$ . Using $\int x^{-2}\,dx = -x^{-1}$ , this evaluates to $W = -kqq_0\left[-\dfrac{1}{x}\right]_{\infty}^{r} = kqq_0\left(\dfrac{1}{r} - \dfrac{1}{\infty}\right) = \dfrac{kqq_0}{r}$ .

Step 3. Divide by $q_0$ to get the potential. By definition $V = W/q_0 = \dfrac{kqq_0}{r\,q_0} = \dfrac{kq}{r} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ . The test charge $q_0$ has cancelled completely, confirming that $V$ is a property of the source charge and the field point alone, not of whatever test charge is used to probe it.

Step 4. Check the sign and the reference. If $q$ is positive, $W > 0$ for the agent bringing in a positive $q_0$ (repulsion must be overcome), so $V > 0$ — consistent with the point being a source of higher potential. If $q$ is negative, $V < 0$ , matching the direct-substitution result used in the worked example below. Setting $r \to \infty$ correctly recovers $V \to 0$ , confirming the choice of reference.

Worked example

A point charge of $-5\ \mu\text{C}$ sits alone in space. Find the potential at a distance of $25\ \text{cm}$ from it, and the work done by an external agent bringing a $+2\ \mu\text{C}$ charge from infinity to that point without gaining kinetic energy.

Step 1. Apply $V = kq/r$ with $k = 9\times10^9\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$ , $q = -5\times10^{-6}\ \text{C}$ , $r = 0.25\ \text{m}$ :

V = (9\times10^9) \times \frac{-5\times10^{-6}}{0.25} = (9\times10^9)\times(-2\times10^{-5}) = -1.8\times10^{5}\ \text{V}

Step 2. Use $W = q_0 V$ with the test charge $q_0 = +2\times10^{-6}\ \text{C}$ :

W = (2\times10^{-6})\times(-1.8\times10^{5}) = -0.36\ \text{J}

Step 3. Interpret the sign. The external agent's work is negative because the positive test charge is attracted toward the negative source — the electric force itself does positive work of $0.36\ \text{J}$ pulling the charge in, so the agent must actually hold it back (do negative work) to prevent it from gaining kinetic energy on the way in. The potential is $-1.8\times10^{5}\ \text{V}$ , and the external agent does $-0.36\ \text{J}$ of work.

Competitive stretch

Two charges $q_1 = +4\ \mu\text{C}$ and $q_2 = +6\ \mu\text{C}$ sit $60\ \text{cm}$ apart. At the midpoint (each charge $0.30\ \text{m}$ away), superposition gives $V_{\text{mid}} = k(q_1 + q_2)/0.30 = 1.2\times10^{5} + 1.8\times10^{5} = 3.0\times10^{5}\ \text{V}$ — added algebraically, with no angles needed since potential is scalar even though the two source charges are not both aligned along a single line through the observation point in general. At a point $20\ \text{cm}$ from $q_1$ on the far side (so $0.80\ \text{m}$ from $q_2$ ): $V = k(4\times10^{-6}/0.20) + k(6\times10^{-6}/0.80) = 1.8\times10^{5} + 0.675\times10^{5} = 2.475\times10^{5}\ \text{V}$ .

Common traps

Adding potentials from several charges as vectors, with angles and components — potential is scalar; superposition is plain algebraic addition, even off-axis.
Confusing $E = -dV/dr$ with $E = V/r$ — the relation needs the rate of change (gradient) of $V$ , not $V$ divided by distance, except for a uniform field measured along a straight line.
Forgetting potential can be negative near negative charges, unlike some students' instinct that "potential" should always feel positive like energy magnitudes.
Thinking equipotential surfaces let charge "flow freely" in a dynamic sense — the correct statement is only that no work is needed to move a charge along the surface in a static field.

Ask the tutor: want to see why the equatorial line of a dipole stays at zero potential while its field does not, or get more practice combining potentials from three or more point charges?

Why is potential a scalar when the electric field it comes from is a vector?

Potential is defined as work done per unit charge, and work itself is a scalar (force dotted with displacement, or here obtained by integrating force over a straight radial path). The field's directional information gets absorbed into that integration, leaving a single number, $V$ , at each point. This is exactly why superposing potentials from several charges is simple algebraic addition, while superposing fields needs vector components and angles — the price of the vector field's extra directional detail is paid at the superposition stage, not at the potential stage.

Why does the negative sign appear in E = -dV/dr, and what does it mean physically?

The sign encodes the fact that the field points from high potential toward low potential, i.e. in the direction potential is decreasing. If $V$ increases as $r$ increases, $dV/dr$ is positive, and the minus sign flips it so $E$ points inward, toward smaller $r$ — consistent with a positive source charge where potential is highest at the source and falls with distance, yet the field itself points outward, away from the source, which is the direction of decreasing potential when you stand on the far side. Physically, a free positive charge always accelerates toward lower potential, the same way a ball rolls downhill toward lower height.

Can potential be zero at a point where the electric field is not zero?

Yes, and the dipole's equatorial line is the standard example: the field there is nonzero (it takes a finite value pointing antiparallel to the dipole axis), yet the potential is exactly zero because the equal-and-opposite contributions from $+q$ and $-q$ cancel algebraically at every equatorial point. This shows potential and field are genuinely different quantities — one being zero says nothing about the other, since $V$ is an integrated, scalar quantity while $E$ is the local rate of change of $V$ in a particular direction.

Electric potential

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads