Charge and Electric Force — AP Physics C: E&M Study Guide

For: AP Physics C: E&M candidates sitting AP Physics C: E&M.

Covers: charge quantization, conservation of charge, Coulomb’s Law for point charges, the superposition principle for electric forces, vector addition of multiple forces, and equilibrium problems for systems of static point charges.

You should already know: Vector addition and component resolution, Newton's laws for static equilibrium, SI unit conventions for charge and distance.

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 C: E&M 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 Charge and Electric Force?

Charge is the fundamental property of matter that gives rise to electrostatic interactions, one of the four fundamental forces of nature. For AP Physics C: E&M, this topic is the foundation of all Unit 1: Electrostatics, and accounts for approximately 10-15% of the total exam score per the official College Board Course and Exam Description (CED). Concepts from this topic appear in both multiple choice (MCQ) and free response (FRQ) sections, most commonly as standalone MCQ or as the opening step of a longer FRQ on electric fields or Gauss's Law.

Standard notation uses $q$ or $Q$ for charge, with negative charge representing an excess of electrons and positive charge representing a deficit of electrons. The SI unit of charge is the coulomb (C). Electric force is the force one charged object exerts on another, acting at a distance without physical contact. Unlike gravity, electric force can be attractive (between opposite charges) or repulsive (between like charges), a key distinguishing feature. This topic focuses on stationary charge distributions, so magnetic effects are not considered here.

2. Fundamental Properties of Electric Charge

Two core properties of electric charge are universally tested on the AP exam: quantization and conservation. Quantization holds that all free net charge in nature exists as an integer multiple of the elementary charge $e$, where $e=1.602\times 10^{-19}\text{C}$. Electrons have charge $q_e=-e$, and protons have charge $q_p=+e$. For any charged object, net charge is given by: $$q=ne$$ where $n$ is a positive or negative integer equal to the net number of excess or missing electrons. While quantization is a fundamental rule, for macroscopic objects $n$ is so large that we often treat charge as a continuous quantity in calculations.

Conservation of charge states that the total net charge in an isolated system never changes. Charge can be transferred between objects in the system, but it cannot be created or destroyed. This property is most commonly tested in problems involving identical conducting spheres that are touched together to transfer charge: for identical conductors, total charge splits equally between the spheres after contact.

Worked Example

Three identical conducting spheres initially have charges $q_A=+6\mu \text{C}$, $q_B=-8\mu \text{C}$, and $q_C=0\mu \text{C}$. First, sphere A touches sphere B, then they are separated. Next, sphere A touches sphere C, then separated. What is the final charge on sphere C, and how many electrons are transferred to sphere C to reach this final state?

1. When identical conductors touch, total charge is conserved and splits equally between them. Total charge after A and B touch: $q_{A,B,\text{total}}=+6\mu \text{C}-8\mu \text{C}=-2\mu \text{C}$. After separation, $q_A=q_B=-1\mu \text{C}$.
2. Next, sphere A ($-1\mu \text{C}$) touches neutral sphere C. Total charge is $-1\mu \text{C}+0=-1\mu \text{C}$, which splits equally, giving $q_C=-0.5\mu \text{C}=-5\times 10^{-7}\text{C}$. This is the final charge on C.
3. The change in charge for C is $\Delta q_C=-5\times 10^{-7}\text{C}$. Each electron carries charge $-e$, so the number of electrons transferred is $n=\frac{|\Delta q_C|}{e}=\frac{5\times 10^{-7}\text{C}}{1.602\times 10^{-19}\text{C/electron}}\approx 3.1\times 10^{12}$ electrons.
4. Verify conservation: total final charge across all spheres is $-0.5\mu \text{C}-1\mu \text{C}-0.5\mu \text{C}=-2\mu \text{C}$, matching the initial total charge, so the calculation is consistent.

Exam tip: Only assume equal charge division after contact for identical conducting spheres. If spheres are different sizes or one is an insulator, the problem will explicitly state how charge divides, so never assume equal division without this information.

3. Coulomb's Law

Coulomb's Law describes the magnitude and direction of the electrostatic force between two stationary point charges. A point charge is any charged object whose size is much smaller than the distance between it and the other charge of interest, so it can be treated as a single point in space. The magnitude of the force between two point charges is: $$F=k\frac{|q_1{q}_2|}{r^2}$$ where $k=8.988\times 10^9{\text{Ncdotpm}}^2/{\text{C}}^2\approx 9\times 10^9{\text{Ncdotpm}}^2/{\text{C}}^2$ for most AP calculations, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them. The direction of the force is always along the line connecting the two charges: attractive for opposite charges, repulsive for like charges.

In vector form, for the force exerted by charge 1 on charge 2: $${\vec{F}}_{12}=k\frac{q_1{q}_2}{r_{12}^2}{\hat{r}}_{12}$$ where ${\hat{r}}_{12}$ is the unit vector pointing from charge 1 to charge 2. Coulomb's Law is an inverse-square law, meaning force is proportional to $1/r^2$, just like Newton's law of gravitation. The electric force is vastly stronger than gravity at atomic and macroscopic scales.

Worked Example

Two point charges are placed on the x-axis: $q_1=+2\text{nC}$ at $x=0$, and $q_2=-8\text{nC}$ at $x=3\text{m}$. What is the magnitude and direction of the force that $q_1$ exerts on $q_2$?

1. Convert all values to SI units: $q_1=2\times 10^{-9}\text{C}$, $q_2=-8\times 10^{-9}\text{C}$, distance $r=3\text{m}-0=3\text{m}$.
2. Calculate magnitude using Coulomb's Law: $F=k\frac{|q_1{q}_2|}{r^2}=\frac{(9\times 10^9)(2\times 10^{-9})(8\times 10^{-9})}{3^2}=1.6\times 10^{-8}\text{N}$.
3. Determine direction: $q_1$ and $q_2$ have opposite signs, so the force is attractive. $q_1$ pulls $q_2$ toward itself, which is the negative x-direction (from $x=3\text{m}$ toward $x=0$).
4. Verify with vector form: ${\hat{r}}_{12}$ points from $q_1$ to $q_2$, so it is in the +x direction. The product $q_1{q}_2$ is negative, so ${\vec{F}}_{12}$ points opposite to ${\hat{r}}_{12}$, confirming the negative x-direction.

Exam tip: Always use absolute values to calculate force magnitude, then assign direction separately based on whether charges are like or opposite. This avoids sign errors that come from misinterpreting the direction of the unit vector.

4. Superposition of Electric Forces

When more than two charges are present, the total force on any single charge is the vector sum of the forces exerted on it by each other individual charge. This is the superposition principle: pairwise interactions between charges are independent of the presence of other charges. Mathematically, for a charge $q_0$ acted on by $n$ other charges: $${\vec{F}}_{\text{total}}=\sum _{i=1}^n{\vec{F}}_{i0}$$ where ${\vec{F}}_{i0}$ is the force exerted by charge $i$ on $q_0$, calculated via Coulomb's Law for each pair.

The key mistake students make here is adding force magnitudes directly instead of adding vectors. For forces acting at different angles, you must break each force into x and y components, add the components separately, then calculate the magnitude and direction of the resultant total force. Common exam problems ask for the total force on a charge at the vertex of a triangle or square, or ask you to find the position where the total force on a third charge is zero (equilibrium).

Worked Example

Three charges are placed at the vertices of a right triangle: $q_1=+1\mu \text{C}$ at $(0,0)$, $q_2=+1\mu \text{C}$ at $(3\text{m},0)$, and $q_3=+2\mu \text{C}$ at $(0,4\text{m})$. What is the total force on $q_1$ from the other two charges?

1. Calculate ${\vec{F}}_{21}$ (force from $q_2$ on $q_1$): Distance $r_{21}=3\text{m}$, both charges are positive so force is repulsive, pushing $q_1$ in the negative x-direction. Magnitude: $F_{21}=\frac{k{q}_1{q}_2}{r_{21}^2}=\frac{(9\times 10^9)(1\times 10^{-6})(1\times 10^{-6})}{3^2}=1\times 10^{-3}\text{N}$. In components: ${\vec{F}}_{21}=(-1\times 10^{-3}\text{N})\hat{i}+0\hat{j}$.
2. Calculate ${\vec{F}}_{31}$ (force from $q_3$ on $q_1$): Distance $r_{31}=4\text{m}$, both charges positive so repulsive, pushing $q_1$ in the negative y-direction. Magnitude: $F_{31}=\frac{(9\times 10^9)(1\times 10^{-6})(2\times 10^{-6})}{4^2}=1.125\times 10^{-3}\text{N}$. In components: ${\vec{F}}_{31}=0\hat{i}+(-1.125\times 10^{-3}\text{N})\hat{j}$.
3. Add components: $F_x=-1\times 10^{-3}\text{N}$, $F_y=-1.125\times 10^{-3}\text{N}$.
4. Calculate total magnitude: $|{\vec{F}}_{\text{total}}|=\sqrt{F_x^2+F_y^2}\approx 1.5\times 10^{-3}\text{N}$, directed 48 degrees below the negative x-axis.

Exam tip: Always draw a free-body diagram for the charge of interest before adding forces. This makes it easy to get the direction of each force right before you calculate components.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Assuming charge divides equally between any two conducting spheres that touch. Why: Students memorize equal division for identical spheres and incorrectly apply it to non-identical spheres. Correct move: Only assume equal division if the problem explicitly states the spheres are identical. For non-identical spheres, use the ratio given in the problem.
• Wrong move: Forgetting that $r$ is squared in Coulomb's Law, calculating $F\propto 1/r$ instead of $1/r^2$. Why: Students confuse inverse-square behavior with inverse proportionality from other topics, or miss the exponent in rushed calculations. Correct move: Circle the exponent 2 in Coulomb's Law as you write it down for every problem, to remind yourself it is squared.
• Wrong move: Adding force magnitudes directly instead of using vector components when forces act at different angles. Why: Superposition is described as "adding forces", so students add magnitudes without accounting for direction. Correct move: Always break each force into x and y components before adding, regardless of how small the angle is.
• Wrong move: Reversing attraction and repulsion when assigning force direction. Why: Students rely on sign conventions instead of checking the physical interaction. Correct move: After calculating magnitude, always explicitly ask "are these charges like (repel) or opposite (attract)?" to confirm direction.
• Wrong move: Forgetting to convert micro/nanocoulombs to coulombs or centimeters to meters before plugging into Coulomb's Law. Why: Unit prefixes are easy to miss in rushed work, and $k$ uses SI units for charge and distance. Correct move: Write all values converted to SI units before substituting into the formula, and circle unit conversions for a final check.

6. Practice Questions (AP Physics C: E&M Style)

Question 1 (Multiple Choice)

Two identical positive point charges $+Q$ are fixed at positions $x=-a$ and $x=+a$ on the x-axis. A third point charge $+q$ is placed at the origin, free to move along the x-axis. What is the magnitude of the net force on $+q$, and what is the stability of the equilibrium? A) Net force zero, stable equilibrium B) Net force zero, unstable equilibrium C) Net force $2kQq/a^2$, stable equilibrium D) Net force $2kQq/a^2$, unstable equilibrium

Worked Solution: By the superposition principle, the left $+Q$ exerts a repulsive force of magnitude $kQq/a^2$ pushing $+q$ to the right, and the right $+Q$ exerts an equal magnitude repulsive force pushing $+q$ to the left. The forces cancel, so net force is zero, eliminating options C and D. To check stability: if $+q$ is displaced a small distance to the right, the force from the right $+Q$ becomes larger than the force from the left $+Q$, pushing $+q$ further right. Any displacement away from the origin results in a net force pushing further away, so equilibrium is unstable. The correct answer is B.

Question 2 (Free Response)

Three point charges are arranged along the x-axis: $q_1=-4\mu \text{C}$ at $x=0$, $q_2=+9\mu \text{C}$ at $x=4\text{m}$. (a) Find all positions along the x-axis where the net force on a third point charge $+q$ would be zero. (b) Is the equilibrium you found stable or unstable for motion along the x-axis? Justify your answer. (c) If the third charge has $q=+1\text{nC}$ and mass $m=1\times 10^{-6}\text{kg}$, what is the initial acceleration of the third charge if it is placed at $x=2\text{m}$?

Worked Solution: (a) For net force to be zero, forces must have opposite directions and equal magnitudes. Opposite directions only occur outside the two charges (between 0 and 4 m, both forces point left, so they cannot cancel). Let $x$ be the equilibrium position, set force magnitudes equal: $$\frac{k|q_{1}q|}{x^2}=\frac{k|q_{2}q|}{(x-4{)}^2}⟹\frac{4}{x^2}=\frac{9}{(x-4{)}^2}⟹\frac{2}{x}=\frac{3}{|x-4|}$$ Solving gives the only valid solution $x=-8\text{m}$ (the other solution is in the region between charges, where directions do not oppose, so it is discarded). The only equilibrium position is $\boxed{x=-8\text{m}}$.

(b) If $+q$ is displaced a small distance right toward $x=0$, the attractive force from $q_1$ increases, pulling $+q$ further right. If displaced left away from $x=0$, the force from $q_2$ dominates, pulling $+q$ further left. Any displacement leads to net force away from equilibrium, so equilibrium is unstable.

(c) At $x=2\text{m}$, $r_1=r_2=2\text{m}$. Both forces point left (negative x): $$F_{\text{total}}=-\frac{k|q_{1}q|}{r_1^2}-\frac{k{q}_{2}q}{r_2^2}=-\frac{(9e9)(4e-6)(1e-9)+(9e9)(9e-6)(1e-9)}{4}=-2.925\times 10^{-2}\text{N}$$ Acceleration $a=F/m=-2.925\times 10^{-2}/1\times 10^{-6}=\boxed{-2.9\times 10^4{\text{m/s}}^2}$ (negative sign indicates direction toward negative x).

Question 3 (Application / Real-World Style)

In a hydrogen atom, the electron (charge $-e$, mass $m_e=9.11\times 10^{-31}\text{kg}$) orbits the proton (charge $+e$, mass $m_p=1.67\times 10^{-27}\text{kg}$) at an average orbital radius of $5.29\times 10^{-11}\text{m}$. Compare the magnitude of the electric force between the electron and proton to the magnitude of the gravitational force between them, and comment on what this means for atomic structure. Use $e=1.602\times 10^{-19}\text{C}$, $G=6.67\times 10^{-11}{\text{Ncdotpm}}^2/{\text{kg}}^2$.

Worked Solution: First calculate electric force magnitude: $$F_e=k\frac{e^2}{r^2}=\frac{(9\times 10^9)(1.602\times 10^{-19}{)}^2}{(5.29\times 10^{-11}{)}^2}\approx 8.2\times 10^{-8}\text{N}$$ Next calculate gravitational force magnitude: $$F_g=G\frac{m_e{m}_p}{r^2}=\frac{(6.67\times 10^{-11})(9.11\times 10^{-31})(1.67\times 10^{-27})}{(5.29\times 10^{-11}{)}^2}\approx 3.6\times 10^{-47}\text{N}$$ The ratio of forces is $F_e/F_g\approx 2.3\times 10^{39}$. This means the electric force is more than $10^{39}$ times stronger than gravity in a hydrogen atom, so gravitational forces are completely negligible for atomic structure, and only electric forces hold atoms together.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational building block for all of electrostatics, and all of E&M more broadly. Immediately after this, you will extend the concept of electric force to electric fields, where we describe the force per unit charge produced by any charge distribution, rather than just the force between two discrete charges. Without mastering Coulomb's Law and superposition for forces, you will not be able to correctly calculate electric fields for discrete or continuous charge distributions, which is a core skill for AP C: E&M FRQs. This topic also underpins all later concepts including Gauss's Law, electric potential, capacitance, and even electromagnetic induction, since all electric interactions originate from the electric force between charges.

Electric Fields, Continuous Charge Distributions, Gauss's Law