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Physics C: E&M · Electrostatics · 14 min read · Updated 2026-05-11

Electric Potential — AP Physics C: Electricity and Magnetism

AP Physics C: Electricity and Magnetism · Electrostatics · 14 min read

1. What is Electric Potential? ★★☆☆☆ ⏱ 3 min

Electric potential is a scalar quantity that describes the electric potential energy per unit test charge at a point in space, created by a source charge distribution. Denoted $V$, it has units of volts ($1\ \text{V} = 1\ \text{J/C} = 1\ \text{N·m/C}$), and is often called "voltage" when referring to potential difference between two points.

Unlike electric field, which is a vector, electric potential is a signed scalar. This simplifies calculations for charge distributions because you only add algebraic values instead of resolving vector components. Electric potential is defined relative to a reference point; for finite charge distributions, we almost always take $V = 0$ at infinity, which aligns with the convention for electric potential energy. Because the electric force is conservative, potential difference between two points is path-independent, meaning it only depends on the endpoints of any path between them.

2. Potential Difference and Potential from Point Charges ★★☆☆☆ ⏱ 4 min

Potential difference $\Delta V$ between two points $a$ and $b$ is defined as the change in electric potential energy per unit test charge moving between the points:

ΔV = V_b - V_a = ΔU/q_0 = -W_{a→b}/q_0

where $W_{a\to b}$ is the work done by the electric field on the test charge $q_0$, and $\Delta U$ is the change in potential energy. If we take the reference potential $V_\infty = 0$, we can derive the absolute potential at a distance $r$ from a point charge $Q$:

V(r) = \frac{kQ}{r} = \frac{1}{4πε_0} \frac{Q}{r}

Potential is positive for positive source charges and negative for negative source charges. For a system of point charges, potential follows scalar superposition: the total potential is just the algebraic sum of potentials from each individual charge, with no vector components required: $V_{total} = \sum_i \frac{kQ_i}{r_i}$.

3. Potential from Continuous Charge Distributions ★★★☆☆ ⏱ 5 min

For a continuous charge distribution, we split the distribution into infinitesimal point charges $dq$, use the point charge potential for each $dq$, then integrate to find the total potential:

V = \int \frac{1}{4πε_0} \frac{dq}{r}

where $r$ is the distance from the infinitesimal charge $dq$ to the point where we calculate potential. Because this is a scalar integral, it is almost always simpler than integrating to find electric field, which requires resolving vector components first.

4. Relation Between Electric Potential and Electric Field ★★★☆☆ ⏱ 5 min

Potential difference is the negative integral of electric field over a path, so we can invert this relationship to find electric field from the derivative of potential. For the one-dimensional case where $V$ only depends on $x$, this gives:

E_x = -\frac{dV}{dx}

In three dimensions, this generalizes to $\vec{E} = -\nabla V$, meaning electric field is the negative gradient of potential. This tells us that electric field always points in the direction of decreasing potential, and its magnitude equals the rate of change of potential with distance. Equipotential surfaces (surfaces of constant potential) are always perpendicular to electric field lines, since no work is done moving a charge along an equipotential.

5. Additional Exam-Style Worked Examples ★★★★☆ ⏱ 6 min

Common Pitfalls

Why: Students confuse scalar superposition for potential with vector addition for electric field, where they add magnitudes of components.

Why: Students assume that if the sum of potential is zero, the derivative (which gives E) must also be zero, which is not true.

Why: Students get confused between potential difference (which can use any reference) and absolute potential for finite charge distributions.

Why: Students mix up $E = 0$ inside a conductor with the derivative relation $E = -dV/dx = 0$, which implies $V$ is constant, not zero.

Why: Students are used to calculating E from continuous distributions which requires components, so they carry that habit over to potential.

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