# Potential and Electric Potential Energy

> AP Physics 2 · Electric Force, Field, and Potential
> Source: https://www.owlsprep.com/study/ap-physics-2-u3-potential-and-electric-potential-energy/

This module covers key concepts for AP Physics 2: electric potential energy change, electric potential, potential difference, equipotential surfaces, potential of point charge systems, the field-potential relation, and energy conservation for moving charges.

**Prerequisites:** Coulomb's Law for point charges; Work and work-energy theorem from mechanics; Electric field definition for point charges

## Learning objectives

- Distinguish between electric potential energy and electric potential
- Calculate change in potential energy for moving charges
- Find total potential for systems of point charges
- Relate electric field to potential difference and identify equipotential properties
- Apply energy conservation to charged particle motion

## Core Definitions: Electric Potential Energy

**Electric Potential Energy** — Potential energy stored in a system of charges due to electrostatic interactions. It is a property of the entire system of charges, not just a single moving charge.

*Notation:* $U$

*Example:* Two opposite charges held a fixed distance apart have stored electric potential energy.

When a charge moves through an electric field, the change in electric potential energy equals the negative of the work done by the electric force on the charge: $
Delta U = -W_E$. If the charge moves at constant kinetic energy, this change equals the work done by an external force to move the charge: $
Delta U = W_{\text{ext}}$.

For any charge $q$ moving through a potential difference $
Delta V$, the change in potential energy simplifies to the core formula:

$$\Delta U = q \Delta V$$

For a system of two point charges $q_1$ and $q_2$ separated by distance $r$, with zero potential energy set at infinite separation, the absolute potential energy of the system is:

$$U = \frac{k q_1 q_2}{r} = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r}$$

If charges have the same sign, $U$ is positive: work must be done to bring repelling charges together from infinity. If charges have opposite signs, $U$ is negative: the system has less energy than when separated, so work must be done to pull them apart.

**Worked example:** An electron ($q = -1.6 \times 10^{-19} \text{C}$) moves from an initial position at $V_i = -100 \text{V}$ to a final position at $V_f = 300 \text{V}$. What is the change in electric potential energy of the electron-field system?

1. Calculate potential difference as final minus initial, by definition:
2. $$\Delta V = V_f - V_i = 300 \text{V} - (-100 \text{V}) = 400 \text{V}$$
3. Use the core relation $\Delta U = q \Delta V$, then substitute values:
4. $$\Delta U = (-1.6 \times 10^{-19} \text{C})(400 \text{V}) = -6.4 \times 10^{-17} \text{J}$$
5. Reasoning check: A negative electron accelerates toward higher potential, so it loses potential energy converted to kinetic energy, which matches the negative result.

> **tip**
>
> Always remember potential energy belongs to the entire system (charge + source charges), not just the moving charge. AP FRQ grading requires you to name the system correctly for full credit.

## Electric Potential and Potential of Point Charge Systems

**Electric Potential** — Electric potential energy per unit positive test charge at a point in an electric field. It is a scalar property of the field alone, independent of any test charge placed in it.

*Notation:* $V$

*Example:* A 1 nC positive point charge produces a positive potential at all points near it.

Potential difference (or voltage) $
Delta V = V_f - V_i$ is the difference in potential between two points, the measurable quantity used in circuits and experiments. For a single point charge $Q$, with $V=0$ at infinity, potential at distance $r$ is:

$$V = \frac{kQ}{r}$$

A key advantage of potential over electric field is that potential is a scalar quantity. For a system of multiple point charges, the total potential at any point is just the algebraic sum of potentials from each individual charge: $V_{\text{total}} = \sum \frac{k q_i}{r_i}$. No vector components are needed—just add signed values based on the sign of each charge.

**Worked example:** Two point charges are placed on the y-axis: $q_1 = +4 \text{nC}$ at $y=2 \text{m}$, and $q_2 = -2 \text{nC}$ at $y=-2 \text{m}$. What is the electric potential at the origin ($y=0$)?

1. Find the distance from each charge to the origin: $r_1 = 2 \text{m}$, $r_2 = 2 \text{m}$.
2. Use scalar addition for total potential:
3. $$V = \frac{k q_1}{r_1} + \frac{k q_2}{r_2}$$
4. Factor out common terms and substitute values:
5. $$V = \frac{k}{r}(q_1 + q_2) = \frac{9 \times 10^9 \text{Nm}^2/\text{C}^2}{2 \text{m}}(4 \times 10^{-9} \text{C} - 2 \times 10^{-9} \text{C})$$
6. Calculate and check the result:
7. $$V = 9 \text{V}$$
8. The positive charge contributes more positive potential than the negative charge contributes negative potential, so a net positive result makes sense.

> **tip**
>
> On multi-charge potential problems, always use the sign of the charge when adding potentials. Skipping the sign is the most common mistake on this type of question.

## Electric Field, Potential, and Equipotential Surfaces

Electric field and potential are closely related: the electric field points in the direction of maximum decreasing potential, and its magnitude equals the negative rate of change of potential with distance. For a uniform electric field aligned with the x-axis, this simplifies to:

$$E = - \frac{\Delta V}{\Delta x}$$

The magnitude of the field is $|E| = |\Delta V / d|$, where $d$ is the distance along the direction of the electric field between the two points.

**Equipotential Surface** — A surface where all points have the same electric potential, so $
Delta V = 0$ between any two points on the surface.

*Example:* Concentric spheres around a point charge, parallel planes between parallel charged plates

No work is done to move a charge along an equipotential, so electric field lines are always perpendicular to equipotential surfaces. For point charges, equipotentials are concentric spheres; for uniform fields between parallel plates, they are parallel planes perpendicular to field lines.

**Worked example:** A uniform electric field between two parallel conducting plates separated by 0.04 m has magnitude 120 N/C, pointing from the left plate to the right plate. If the left plate is set to $V = 0 \text{V}$, what is the potential of the right plate?

1. E points from left ($x=0$) to right ($x=0.04$ m), so $E = +120 \text{N/C}$, $
Delta x = 0.04 \text{m}$.
2. Rearrange the uniform field relation to solve for $
Delta V$:
3. $$\Delta V = V_{\text{right}} - V_{\text{left}} = - E \Delta x$$
4. Substitute values and conclude:
5. $$\Delta V = - (120 \text{N/C})(0.04 \text{m}) = -4.8 \text{V}$$
6. Reasoning check: Electric field points toward lower potential, so the right plate in the direction of the field must be at lower potential than the left plate, which matches the result.

> **tip**
>
> When asked to draw equipotential lines on a field diagram, confirm they cross field lines at 90 degrees—AP graders look for this to award credit.

## Applications: Energy Conservation for Charged Particles

Conservation of energy applies to conservative electric fields, allowing us to calculate the kinetic energy and speed of accelerated charged particles, a common AP Physics 2 problem type.

**Worked example:** A charge $q = -2 \mu\text{C}$ moves from point P at 50 V to point Q at 150 V. What is the change in electric potential energy of the system?

1. Calculate potential difference per definition:
2. $$\Delta V = V_Q - V_P = 150 \text{V} - 50 \text{V} = 100 \text{V}$$
3. Use the core relation $\Delta U = q \Delta V$ and substitute:
4. $$\Delta U = (-2 \times 10^{-6} \text{C})(100 \text{V}) = -2 \times 10^{-4} \text{J}$$
5. A negative charge loses potential energy when moving to higher potential, so the correct result is $-2 \times 10^{-4} \text{J}$.

**Worked example:** Two point charges, $q_1 = +2 \mu\text{C}$ and $q_2 = -2 \mu\text{C}$, are separated by 0.4 m on the x-axis. (a) Find total potential at the midpoint. (b) Find total potential energy of the system. (c) Explain why the energy is negative.

1. (a) Distance from each charge to midpoint is $r = 0.2 \text{m}$. Use scalar addition:
2. $$V = \frac{k q_1}{r} + \frac{k q_2}{r} = \frac{k}{r}(q_1 + q_2) = 0 \text{V}$$
3. (b) Use the two-charge potential energy formula:
4. $$U = \frac{k q_1 q_2}{r} = \frac{(9 \times 10^9)(2 \times 10^{-6})(-2 \times 10^{-6})}{0.4} = -0.09 \text{J}$$
5. (c) The total potential energy is negative because the charges have opposite signs, so the system has less energy than when infinitely separated. This means the force between the charges is attractive: energy must be added to pull them apart to infinity.

**Worked example:** A cathode ray tube accelerates electrons from rest across a 15,000 V potential difference. Find the electron's kinetic energy (in eV and J) and speed, given $m_e = 9.11 \times 10^{-31} \text{kg}$, $1 \text{eV} = 1.6 \times 10^{-19} \text{J}$.

1. Use conservation of energy: $\Delta K = - \Delta U = -q \Delta V$. $q = -e$, $
Delta V = 15,000 \text{V}$:
2. $$K_f = -(-1.6 \times 10^{-19} \text{C})(15,000 \text{V}) = 2.4 \times 10^{-15} \text{J} = 15,000 \text{eV} = 15 \text{keV}$$
3. Solve for speed from kinetic energy $K_f = \frac{1}{2} m_e v^2$:
4. $$v = \sqrt{\frac{2 K_f}{m_e}} = \sqrt{\frac{2(2.4 \times 10^{-15})}{9.11 \times 10^{-31}}} \approx 7.3 \times 10^7 \text{m/s}$$
5. This is ~24% the speed of light, matching typical operating speeds for cathode ray tubes.

## Common pitfalls

- **Wrong:** Confusing electric potential $V$ and electric potential energy $U$, solving for energy as $U = V$ instead of $U = qV$.
  - Why it fails: Similar names lead students to forget the definition of potential as "per unit charge".
  - Correct: Write down the definition $V = U/q$ at the start of every problem, then rearrange for your unknown before plugging in values.
- **Wrong:** Reversing the order of subtraction for potential difference, calculating $
Delta V = V_i - V_f$ instead of $V_f - V_i$.
  - Why it fails: Students know E points to lower potential and reverse subtraction to get a "reasonable" sign, leading to wrong energy changes.
  - Correct: Always follow $
Delta V = V_{\text{final}} - V_{\text{initial}}$ regardless of motion direction, and let the sign come out naturally.
- **Wrong:** Treating potential as a vector, adding magnitudes of potentials and adjusting for direction instead of adding signed scalar values.
  - Why it fails: Students just learned electric field is a vector, so they default to vector addition for potential.
  - Correct: Remind yourself before starting: "potential is scalar, add signed values" to avoid this mistake.
- **Wrong:** Using $V = kQ/r$ for problems where zero potential is set at a point other than infinity (e.g., parallel plates with $V=0$ at one plate).
  - Why it fails: The point charge formula is only defined with zero potential at infinity.
  - Correct: Only use $
Delta U = q \Delta V$ for problems with a given reference potential, and reserve $V = kQ/r$ for isolated point charge systems.
- **Wrong:** Drawing equipotential lines parallel to electric field lines.
  - Why it fails: Students rush and confuse the orientation of the two line types.
  - Correct: Always draw equipotentials crossing field lines at 90 degrees, and mark the right angle to confirm.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Change in Electric Potential Energy | $\Delta U = -W_E = q \Delta V$ | $\Delta V = V_{\text{final}} - V_{\text{initial}}$; $W_E$ = work done by electric field |
| Potential of a Point Charge | $V = \frac{kQ}{r}$ | Zero potential at $r \to \infty$; V is scalar, add signed values for multiple charges |
| Potential Energy of Two Point Charges | $U = \frac{k q_1 q_2}{r}$ | Zero potential energy at $r \to \infty$; positive for same-sign charges, negative for opposite-sign |
| Uniform E and Potential | $\|E\| = \left\| \frac{\Delta V}{d} \right\|$ | $d$ = distance along direction of E; E always points toward decreasing potential |
| Equipotential Surface Rule | $\Delta V = 0$ for all points | E is always perpendicular to equipotentials; no work done moving charge along an equipotential |
| Conservation of Energy for Charges | $K_i + U_i = K_f + U_f$ | Applies to conservative electric fields; used to find speed of accelerated charges |
| Potential of Multiple Charges | $V_{\text{total}} = \sum V_i$ | Scalar addition, no vector components required, unlike electric field |

## What's next

This topic is the foundational prerequisite for the next core topic in Unit 3: Gauss’s Law for electrostatics, where you will calculate potential for extended charged objects like conducting spheres. Mastery of the distinction between potential and potential energy is also required for all topics in Unit 4, which covers electric circuits, where potential difference (voltage) is the core quantity that drives current. This topic also connects directly to capacitor energy storage and to modern physics topics like the photoelectric effect. Without a solid grasp of sign conventions and scalar addition of potential, you will struggle to solve circuit problems and energy-based FRQs on the AP Physics 2 exam.

- [Capacitance for AP Physics 2](https://www.owlsprep.com/study/ap-physics-2-u3-capacitance/)
- [Electric Circuits Overview](https://www.owlsprep.com/study/ap-physics-2-u4-overview/)
- [Definition of a Circuit](https://www.owlsprep.com/study/ap-physics-2-u4-definition-of-a-circuit/)

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