Potential and Electric Potential Energy — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: electric potential energy change, electric potential and potential difference, equipotential surfaces, potential of point charge systems, the relation between electric field and potential, and energy conservation for moving charged particles.

You should already know: Coulomb’s law for the force between two point charges; The definition of work and the work-energy theorem from mechanics; The relationship between force and electric field for point charges.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Potential and Electric Potential Energy?

This topic accounts for ~20% of Unit 3 (Electric Force, Field, and Potential), which contributes 12-16% of the total AP Physics 2 exam weight, per the official Course and Exam Description (CED). It appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with electric field analysis, circuit problems, or energy conservation questions.

Electric potential energy (denoted $U$) is the potential energy stored in a system of charges due to their electrostatic interaction, analogous to gravitational potential energy in mass systems. Electric potential (denoted $V$, often shortened to "potential") is a separate, field-specific quantity: it is defined as electric potential energy per unit positive test charge, $V=U/q$, so it is a property of the electric field itself, independent of any test charge placed in it.

This distinction between $U$ (which depends on both the field and the test charge) and $V$ (which depends only on the source charges creating the field) is the core of the topic. AP exam questions regularly test this distinction directly and rely on it implicitly for all energy-based electrostatics problems.

2. Electric Potential Energy

Electric potential energy is a property of the entire system of charges, not just a single moving charge. 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 $\Delta V$, this 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 $r$, we set the zero of potential energy at infinite separation ($r\to \infty$), so the absolute potential energy of the system is: $$U=\frac{k{q}_1{q}_2}{r}=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r}$$ Intuition: If charges have the same sign, $U$ is positive—work must be done on the system to bring the repelling charges together from infinity, and the system will release potential energy when the charges separate. If charges have opposite signs, $U$ is negative—the system has less potential energy than when separated, so work must be done to pull them apart. Only changes in potential energy matter for motion, but absolute values are useful for comparing total system energy of charge configurations.

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

Exam 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.

3. Electric Potential and Potential of Charge Systems

Electric potential $V$ at a point is defined as potential energy per unit positive test charge: $V=U/q$, with units of volts ($1\text{V}=1\text{J/C}$). Potential difference (or voltage) $\Delta V=V_f-V_i$ is the difference in potential between two points, which is the measurable quantity used in circuits and experiments.

For a single point charge $Q$, with $V=0$ set at infinity, potential at distance $r$ is: $$V=\frac{kQ}{r}$$ A key advantage of working with potential instead of 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 the potentials from each individual charge: $V_{\text{total}}=\sum \frac{k{q}_i}{r_i}$, where $r_i$ is the distance from charge $q_i$ to the point of interest. No vector components are needed—you just add the signed values based on the sign of each charge.

Intuition: Potential is positive and decreases with distance around a positive point charge, and negative and becomes more negative closer to a negative point 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: $V=\frac{k{q}_1}{r_1}+\frac{k{q}_2}{r_2}$.
3. Factor out common terms: $V=\frac{k}{r}(q_1+q_2)=\frac{9\times 10^9{\text{Nm}}^2/{\text{C}}^2}{2\text{m}}(4\times 10^{-9}\text{C}-2\times 10^{-9}\text{C})$.
4. Calculate: $V=(4.5\times 10^9)(2\times 10^{-9})=9\text{V}$.
5. Check: The positive charge contributes more positive potential than the negative charge contributes negative potential, so a net positive result makes sense.

Exam 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.

4. 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 the common case of 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.

An equipotential surface is a surface where all points have the same potential, so $\Delta V=0$ between any two points on the surface. This means no work is done to move a charge along an equipotential, so the electric force (and thus the electric field) has no component parallel to the equipotential. By definition, 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 the 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. Use the uniform field relation $E=-\Delta V/\Delta x$. E points from left (x=0) to right (x=0.04 m), so $E=+120\text{N/C}$, $\Delta x=0.04\text{m}-0=0.04\text{m}$.
2. Rearrange to solve for $\Delta V=V_{\text{right}}-V_{\text{left}}=-E\Delta x$.
3. Substitute values: $\Delta V=-(120\text{N/C})(0.04\text{m})=-4.8\text{V}$. Since $V_{\text{left}}=0$, $V_{\text{right}}=-4.8\text{V}$.
4. 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.

Exam tip: When asked to draw equipotential lines on a field diagram, mark a small right angle at every intersection with a field line to confirm your orientation—this is what AP graders look for to award credit.

Common Pitfalls (and how to avoid them)

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

Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

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? A) $-2\times 10^{-4}\text{J}$ B) $-4\times 10^{-4}\text{J}$ C) $+2\times 10^{-4}\text{J}$ D) $+4\times 10^{-4}\text{J}$

Worked Solution: First, calculate potential difference per definition: $\Delta V=V_Q-V_P=150\text{V}-50\text{V}=100\text{V}$. Use the relation $\Delta U=q\Delta V$, substitute $q=-2\times 10^{-6}\text{C}$: $\Delta U=(-2\times 10^{-6}\text{C})(100\text{V})=-2\times 10^{-4}\text{J}$. A negative charge loses potential energy when moving to higher potential, which matches the negative sign. The correct answer is A.

Question 2 (Free Response)

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) Calculate the total electric potential at the midpoint between the two charges. (b) Calculate the total electric potential energy of the two-charge system. (c) Explain why the system has negative total potential energy, and what this means for the force between the charges.

Worked Solution: (a) The distance from each charge to the midpoint is $r=0.2\text{m}$. Total potential is scalar sum: $V=\frac{k{q}_1}{r}+\frac{k{q}_2}{r}=\frac{k}{r}(q_1+q_2)=\frac{9\times 10^9}{0.2}(2\times 10^{-6}-2\times 10^{-6})=0\text{V}$. (b) Total potential energy for two charges: $U=\frac{k{q}_1{q}_2}{r}=\frac{(9\times 10^9)(2\times 10^{-6})(-2\times 10^{-6})}{0.4}=-0.09\text{J}$. (c) The total potential energy is negative because the charges have opposite signs, so the system has less energy than when the charges are infinitely separated. A negative potential energy means the force between the charges is attractive: energy must be added to the system to pull the charges apart to infinite separation.

Question 3 (Application / Real-World Style)

A television cathode ray tube accelerates electrons from rest across a potential difference of 15,000 V between the cathode and the screen. What is the kinetic energy of an electron when it reaches the screen, in electron-volts (eV) and in joules? What is the speed of the electron (mass of electron $m_e=9.11\times 10^{-31}\text{kg}$, $1\text{eV}=1.6\times 10^{-19}\text{J}$)?

Worked Solution: Use conservation of energy: initial kinetic energy $K_i=0$, so $\Delta K=-\Delta U=-q\Delta V$. The potential difference between cathode (initial) and screen (final) is $\Delta V=15,000\text{V}$, and $q=-e=-1.6\times 10^{-19}\text{C}$. Substitute: $K_f=-(-1.6\times 10^{-19}\text{C})(15,000\text{V})=2.4\times 10^{-15}\text{J}=15,000\text{eV}=15\text{keV}$. Solve for speed: $K_f=\frac{1}{2}{m}_e{v}^2⟹v=\sqrt{\frac{2K_f}{m_e}}=\sqrt{\frac{2(2.4\times 10^{-15})}{9.11\times 10^{-31}}}\approx 7.3\times 10^7\text{m/s}$, about 24% of the speed of light. This matches the typical operating speed of electrons in cathode ray tubes.

Quick Reference Cheatsheet

What's Next

This chapter is the foundational prerequisite for the next core topic in Unit 3: continuous charge distributions and Gauss’s Law, 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 the sign conventions and scalar addition of potential, you will struggle to solve circuit problems and energy-based FRQs on the exam.

• Gauss's Law for Electrostatics
• Capacitance and Dielectrics
• DC Circuits and Ohm's Law
• Photoelectric Effect and Photon Energy