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Reading Time: 6 min
Last Updated: March 13, 2026
Main Ideas: 4
Reading Time: 6 min
Last Updated: March 13, 2026
Main Ideas: 4

Topic 9.2 Notes – Electric Potential

Verified for 2027 AP® Physics C: Electricity and Magnetism Exam
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Electric potential describes how electric potential energy is distributed in space around charges. Instead of focusing on force like the electric field does, potential tells you the energy per unit charge at a point. In this topic, you connect potential to charge distributions, calculus, and the electric field itself.

1. What Electric Potential Is

Definition

Electric potential V V is electric potential energy per unit charge:

V=UEq V = \frac{U_E}{q}

  • Units: volts (V) where 1 V=1 J/C1\text{ V} = 1\text{ J/C}
  • It is a scalar, so no direction.

If a positive test charge has potential energy UEU_E at a point, dividing by its charge gives the potential of that location.

Reference Point

For isolated charge distributions, we define
V=0at infinity V = 0 \quad \text{at infinity}

Physically, VV tells you the work per unit charge required to bring a positive test charge from infinity to that point.

If it takes positive work to bring it in, the potential there is positive.

Potential Difference

Between two points:

ΔV=ΔUEqandΔUE=qΔV \Delta V = \frac{\Delta U_E}{q} \qquad\text{and}\qquad \Delta U_E = q \Delta V

Important sign ideas:

  • For positive charges: higher VV → higher UEU_E.
  • For negative charges: energy changes opposite the sign of ΔV \Delta V .

In circuits, batteries maintain a potential difference by using chemical reactions to separate charge. That separation creates electric potential energy.

2. Electric Potential from Charges

Point Charges

For a single point charge:

V=14πε0qr V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}

  • Depends only on distance rr.
  • The sign of VV matches the sign of qq.

If you double the charge, you double the potential. If you double the distance, potential is cut in half.

For multiple point charges:

Vtotal=∑i14πε0qiri V_{\text{total}} = \sum_i \frac{1}{4\pi\varepsilon_0}\frac{q_i}{r_i}

This is scalar superposition. You just add the values algebraically. No components. This is why potential problems are often easier than field problems.

Continuous Charge Distributions

When charge is spread out:

V=14πε0∫dqr V = \frac{1}{4\pi\varepsilon_0}\int \frac{dq}{r}

General setup:

  1. Choose coordinates.
  2. Replace dqdq with density:
    • dq=λ dldq = \lambda\,dl
    • dq=σ dAdq = \sigma\,dA
    • dq=ρ dVdq = \rho\,dV
  3. Express rr to the field point.
  4. Integrate over the distribution.

On the AP exam, calculus setups are limited to specific geometries:

  • Infinitely long wire or cylinder at distance rr
  • Thin ring along its axis
  • Semicircular arc at its center
  • Finite line charge
    • On its axis
    • On its perpendicular bisector

If you see something outside this list, it’s likely conceptual only.

3. Relationship Between Electric Potential and Electric Field

These two are tightly connected.

Field from Potential

E⃗=−∇V \vec{E} = -\nabla V

Component form:

Ex=−∂V∂x,Ey=−∂V∂y,Ez=−∂V∂z E_x = -\frac{\partial V}{\partial x}, \quad E_y = -\frac{\partial V}{\partial y}, \quad E_z = -\frac{\partial V}{\partial z}

The field points in the direction of steepest decrease in potential.

If V(x)=5x2V(x) = 5x^2, then
Ex=−10x E_x = -10x
The derivative tells you how fast potential changes in space.

On FRQs, they often give you V(x,y)V(x,y) and expect clean partial derivatives.

Potential from Field

ΔV=Vb−Va=−∫abE⃗⋅dr⃗ \Delta V = V_b - V_a = -\int_a^b \vec{E} \cdot d\vec{r}

Key pieces:

  • Dot product → only the component of E⃗ \vec{E} along the path matters.
  • Negative sign → field points toward decreasing potential.
  • Result is path independent because electrostatic fields are conservative.

If you move with the field, potential drops.
If you move against the field, potential increases.

4. Equipotential Lines and Field Maps

Here’s how potential looks in space. For a single positive point charge, the electric field lines radiate outward and the equipotential lines form concentric circles around the charge.

Study guide illustration

Field lines and equipotentials for a positive point charge

Equipotential Lines

Equipotential lines (or surfaces in 3D) connect points of equal VV.

Moving along one:

  • ΔV=0 \Delta V = 0
  • No work is done by the electric force.

Relationship to the Electric Field

  • Equipotentials are perpendicular to electric field vectors.
  • Field vectors point toward lower potential.
  • Closer spacing of equipotentials → stronger field.
  • There is no component of E⃗ \vec{E} along an equipotential.

In the diagram, notice how each field line crosses the circular equipotentials at right angles. If you’re given one map, you should be able to sketch the other. That shows up in both multiple choice and FRQs.

Key Takeaways

Electric potential is energy per charge V=UE/q V = U_E/q , and it is scalar.
Superposition for potential is algebraic, even when fields would require vectors.
For electrostatics, E⃗=−∇V \vec{E} = -\nabla V and ΔV=−∫E⃗⋅dr⃗ \Delta V = -\int \vec{E}\cdot d\vec{r} express the same physical relationship.
The electric field always points in the direction of decreasing potential.
Along an equipotential surface, ΔV=0 \Delta V = 0 and the electric field has no parallel component.

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Notes

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