All 6 Physics C: E&M units
AP Physics C: E & M · Unit 2 of 6

Electric Potential

15–21% of the exam4 lessons · 51 min19 terms

What this unit covers

The topics below follow the published Physics C: E&M course framework for Unit 2. This unit is worth 15–21% of the exam, so budget your time against that rather than against how long the unit takes to teach.

Potential energyPotential from fieldsEquipotentialsGradient

Lessons in this unit

Formulas in Unit 2

Potential energy from work
ΔU = U_b − U_a = −W_field = −∫ₐᵇ →F·d→l
The line integral is path-independent, so U depends only on the configuration, not on how the charges got there.
Potential energy of two point charges
U = k·q₁q₂ / r, k = 1/(4πε₀) ≈ 8.99 × 10⁹ N·m²/C²
Plug the charges in *with their signs*. Note the single power of r — energy goes as 1/r, force as 1/r².
Potential difference from the field
V_b − V_a = −∫ₐᵇ →E·d→l
Path-independent. Moving *with* the field, potential drops; moving against it, potential rises. For a uniform field along a straight displacement d: ΔV = −E·d.
Potential of point charges and distributions
V = k·q/r, V = Σ k·qᵢ/rᵢ, V = ∫ k·dq/r
All plain scalar sums — signed numbers, never components. This is why V is often far easier to compute than E.
Work and field from equipotentials
W_field = −qΔV = q(V_a − V_b), |E| ≈ |ΔV| / Δs
Δs is measured perpendicular to the surfaces. Along an equipotential, ΔV = 0 and the field does no work.
Field from potential (gradient)
Eₓ = −∂V/∂x, E_y = −∂V/∂y, E_z = −∂V/∂z; →E = −∇V
Units: 1 V/m = 1 N/C — the two field units are identical. For radial potentials, E_r = −dV/dr.

Every term in Unit 2

All 19 terms we publish for Electric Potential, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.

Electric potential of a point charge
V = kq/r, taking zero at infinity. A scalar, so potentials from several charges add algebraically.
Potential from a distribution
V = ∫k dq/r. Easier than the field integral because there are no vector components to resolve.
Relation between field and potential
V = −∫E·dl and E = −∇V, which in one dimension is E = −dV/dx.
Why field is the negative gradient
A positive charge accelerates toward lower potential, so the field points down the potential hill.
Equipotential surfaces
Always perpendicular to field lines. No work is required to move a charge along one, since ΔV is zero.
Potential energy of a charge pair
U = kq₁q₂/r, including sign. Opposite charges have negative U, which becomes more negative as they approach.
Potential energy of a charge configuration
Sum kq_iq_j/r_ij over all distinct pairs. Counting a pair twice is the standard error.
Potential of a charged conductor
Constant throughout the conductor and on its surface, since the internal field is zero. The surface is an equipotential.
Field and potential can be independently zero
At the midpoint between equal opposite charges, V = 0 but E ≠ 0; at the midpoint between equal like charges, E = 0 but V ≠ 0.
Work and potential difference
W = qΔV, and W = −ΔU. A positive charge released in a field moves toward lower potential.
Why potential is easier than field
V is a scalar, so contributions add without components. Compute V by integration, then get E by differentiating.
Sign of the potential integral
V_b − V_a = −∫E·dl from a to b. The minus sign is the difference between a right and a wrong answer.
Potential inside a conducting sphere
Constant and equal to the surface value kQ/R, since the interior field is zero and no work is done moving within it.
Potential inside a uniformly charged insulating sphere
Rises toward the center, reaching kQ(3R² − r²)/2R³ — unlike a conductor, where it is flat.
Potential from a graph of E vs x
The change in potential is the negative area under the field-position graph.
Field from a graph of V vs x
E is the negative slope. A flat V means zero field, and steeply falling V means a strong field in the positive direction.
Energy of assembling charges
Bring charges in one at a time from infinity, summing the work at each step. Each pair is counted once.
Accelerating a charge through a potential difference
qΔV = ½mv², which is how electron guns and accelerators set particle speeds.
Electron volt
The energy gained by one elementary charge through one volt, 1.6 × 10⁻¹⁹ J. A convenient unit at atomic scales.

What examiners penalize here

Practice Physics C: E&M

Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.

Questions about this unit

How much of the AP Physics C: E & M exam is Unit 2?

Unit 2, Electric Potential, is worth 15–21% of the Physics C: E&M multiple-choice section according to the published course framework. Across all 6 units that makes it one of the heaviest units on the exam, and worth front-loading.

What topics are covered in Physics C: E&M Unit 2?

Electric Potential covers Potential energy, Potential from fields, Equipotentials and Gradient. We publish 19 terms with definitions for this unit, all of them on this page.

How should I study Physics C: E&M Unit 2?

Read the 4 lessons below first — about 50 minutes — then drill the 19 terms in cram mode until you can produce each definition from memory rather than just recognize it. Recognition is what makes a unit feel finished when it is not. Finish with practice questions and read the explanation for every one you get right by elimination as well as the ones you miss.

All 6 units of AP Physics C: E & M

  1. Unit 1 · Electric Charges, Fields, and Gauss’s Law
  2. Unit 2 · Electric Potential
  3. Unit 3 · Conductors, Capacitors, and Dielectrics
  4. Unit 4 · Electric Circuits
  5. Unit 5 · Magnetic Fields
  6. Unit 6 · Electromagnetic Induction

Unit names, topics and exam weights follow the published College Board course framework for AP Physics C: E & M. AP® is a trademark registered by the College Board, which does not endorse this site.