This AP Physics C lesson on electric potential notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on electric potential and check yourself with the quiz.
Quick answer: Electric potential is potential energy per unit charge, V = U/q, measured in volts. The potential difference between two points is the negative line integral of the field, ΔV = −∫E·dl, and the field is the negative gradient of potential, E = −dV/dx. Also, conductors in equilibrium are equipotentials with zero internal field.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:E&M Unit 9
Electric potential notes: the key ideas
What is electric potential and how is it linked to the field?
A charge placed in an electric field has potential energy that depends on where it sits. Dividing that energy by the charge gives a property of the point itself, which is the electric potential. Because potential is a scalar, adding the contributions of many charges needs no vectors, and the field can always be recovered from it afterwards.
Electric potential at a point is the work per unit positive charge done by an external agent in bringing a small test charge from infinity to that point at constant speed: V = U/q. its unit is the volt, 1 V = 1 J/C.
V = (1/4πε₀) q/r = kq/r ΔV = Vb − Va = −∫ E · dl Ex = −dV/dx
V is potential (V), q is source charge (C), r is distance (m), E is field (V/m = N/C), k = 8.99 × 10⁹ N·m²/C². A charge q moved through ΔV changes its energy by ΔU = qΔV.
How do you find potential by integrating the field?
Choose a path from the reference point to the point of interest and integrate E · dl along it. Since the electrostatic force is conservative, every path gives the same answer, so you may pick the easiest one. If you move against the field the potential rises, which is why V is highest next to a positive source.
Data: along the x-axis the field is E = (600 V/m³)x² in the +x direction. Find V(0.20 m) − V(0).
Method: ΔV = −∫ from 0 to 0.20 of 600x² dx = −200x³ evaluated at 0.20 m = −200 × 0.0080.
Answer: ΔV = −1.6 V. Moreover, the potential falls in the direction of the field.
Also, for continuous charge, add the potentials of small pieces: V = k∫dq/s, where s is the distance from the piece to the point. This is easier than adding vector fields because every piece contributes with the same sign.
Data: a ring of radius R = 0.30 m carries Q = 2.0 nC. Find V and Ex on the axis at x = 0.40 m.
Method: s = √(0.30² + 0.40²) = 0.50 m, so V = kQ/s = (8.99 × 10⁹)(2.0 × 10⁻⁹)/0.50. Then Ex = −dV/dx = kQx/(x² + R²)^(3/2).
Answer: V = 36 V and Ex = 57.5 V/m, using Ex = (17.98 × 0.40)/0.125.
What are equipotentials and why are conductors special?
In addition, an equipotential is a surface on which V is constant. No work is done moving a charge along it, so the field must be perpendicular to it everywhere. Closely spaced equipotentials mean a strong field, because |E| = |dV/dn| is the steepness of the potential.
| Source | Potential V (V = 0 at infinity) |
|---|---|
| Point charge q | kq/r |
| Ring, on axis at x | kQ/√(x² + R²) |
| Uniform field E, distance d along it | V₀ − Ed |
| Conducting sphere (R, Q), r ≥ R | kQ/r |
| Conducting sphere, r ≤ R | kQ/R (constant) |
| Infinite line, charge λ per metre | −2kλ ln r + constant |
In electrostatic equilibrium a conductor has E = 0 inside, so the whole conductor, surface included, is at one potential. Charge sits on the surface and gathers where the surface is most sharply curved, which is why points discharge first. Moreover, two conductors joined by a wire reach the same potential, not the same charge.
Data: a conducting sphere of radius 0.15 m is at 3000 V. Find its charge and the potential 0.45 m from its centre.
Method: Q = VR/k = (3000)(0.15)/(8.99 × 10⁹) and V(r) = V·R/r.
Answer: Q = 50 nC and V = 1000 V at 0.45 m.
Also, zero potential does not mean zero field. In addition, at the midpoint of two equal and opposite charges V = 0 yet E is large. Moreover, the field depends on the slope of V, not its value.
To get E from a V–x graph, read the slope, then flip its sign. A graph falling to the right means a field pointing right.
- V = U/q (1 V = 1 J/C); point charge V = kq/r, with V = 0 at infinity.
- ΔV = −∫E·dl is path-independent; E = −dV/dx gives the field from a slope.
- Potentials add as scalars: V = k∫dq/s for continuous charge.
- Also, equipotentials are perpendicular to field lines; E points toward falling V.
- A conductor in equilibrium has E = 0 inside and is one equipotential.
Potential from fields and equipotentials: frequently asked questions
Is potential the same as potential energy?
No. Potential energy U belongs to a charge in a field, while potential V = U/q belongs to a point in space. A 2 μC charge at a 100 V point has U = 2 × 10⁻⁴ J.
Why does the field point toward lower potential?
Since ΔV = −∫E·dl, moving along the field gives a negative ΔV. A positive charge released from rest therefore accelerates toward lower potential, gaining kinetic energy as it loses potential energy.
Can V be zero where E is not zero?
Yes. V = 0 only tells you the value, while E depends on how fast V changes. At the midpoint of equal and opposite charges V = 0 but E is non-zero.
Why is the surface of a conductor an equipotential?
If two surface points had different V, a field would exist along the surface and push charges until it vanished. In equilibrium the tangential E is zero, so V is constant.
When do I use calculus for potential?
Use integration when the field varies with position or the charge is continuous. Use ΔV = −∫E·dl for a known field and V = k∫dq/s for a charge distribution, then differentiate to get E.
