Charge and Electric Force — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Conservation of charge, quantization of charge, charging by conduction and induction, Coulomb’s law of electrostatic force, vector superposition of multiple electric forces, and problem-solving for electrostatic point charge systems on the AP Physics 2 exam.

You should already know: Basic vector addition and decomposition for 2D systems, Newton’s laws of motion for force systems, and basic properties of conductors and insulators.

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 Charge and Electric Force?

Charge is a fundamental intrinsic property of matter that causes it to experience a force when placed in an electromagnetic field, one of the four fundamental forces of nature. There are exactly two types of charge, labeled positive and negative per convention, with like charges repelling and opposite charges attracting. The SI unit of charge is the coulomb (C), with the elementary charge (magnitude of charge on a single proton or electron) equal to $e=1.6\times 10^{-19}\text{C}$.

Per the AP Physics 2 Course and Exam Description (CED), this topic is part of Unit 3, which makes up 11-16% of the total exam score. Charge and electric force concepts appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often paired with electric field or potential questions. This topic forms the foundational framework for all electrostatics. Neutral objects have equal numbers of positive protons and negative electrons, so their net charge is zero. Unlike mass, charge can be positive or negative, allowing net charge to cancel out when positive and negative charges are combined.

2. Fundamental Properties of Charge

The two most fundamental rules governing all charge interactions tested on AP Physics 2 are conservation of charge and quantization of charge. Conservation of charge states that for any isolated system, the total net charge (sum of all positive charge minus the sum of all negative charge) is constant. Charge cannot be created or destroyed, only transferred between objects or rearranged within a system. This rule applies to all charging processes: when you rub a balloon on your hair, charge is transferred from one material to the other, not created from nothing.

Quantization of charge states that all free charge found in nature exists as integer multiples of the elementary charge $e$, meaning any net charge can be written as: $$q=ne$$ where $n$ is a non-zero positive or negative integer, and $e=1.6\times 10^{-19}\text{C}$ (this value is provided on the AP Physics 2 formula sheet). Electrons carry a charge of $-e$, and protons carry a charge of $+e$. For macroscopic objects, $n$ is so large that we often treat charge as a continuous quantity, but AP problems regularly test understanding of quantization by asking for the number of excess or deficit electrons in a charged object.

Worked Example

Problem: A neutral copper sphere is touched to a second identical aluminum sphere that has an initial net charge of $+9.6\times 10^{-18}\text{C}$. Assuming charge spreads equally over both identical spheres after contact, how many excess electrons does the copper sphere have?

1. Apply conservation of charge: the total net charge of the isolated two-sphere system is constant, equal to the initial total charge: $Q_{\text{total}}=0+9.6\times 10^{-18}\text{C}=9.6\times 10^{-18}\text{C}$.
2. Charge spreads equally over identical conductors in contact, so each sphere gets half the total charge: $q_{\text{copper}}=Q_{\text{total}}/2=4.8\times 10^{-18}\text{C}$.
3. Use the quantization rule $q=ne$ to solve for $n$, the number of excess electrons: $n=q/e=(4.8\times 10^{-18}\text{C})/(1.6\times 10^{-19}\text{C per electron})=30$.
4. Positive net charge means the copper sphere has a deficit of 30 electrons, so the number of excess electrons is $\boxed{-30}$ (or 0 excess electrons, 30 deficit, depending on question wording).

Exam tip: When a problem asks for the number of excess electrons, remember that positive net charge means a negative number of excess electrons (i.e., a deficit); double-check the question wording to avoid mixing up excess vs. deficit.

3. Charging by Conduction and Induction

AP Physics 2 regularly tests conceptual understanding of the two common charging processes, and the resulting sign of net charge on the charged object. To understand these processes, recall that conductors allow free electrons to move through the material, while insulators bind electrons to their atoms.

Charging by conduction (or charging by contact) requires physical contact between a charged object and a neutral object. When contact is made, charge transfers between the two objects, and both end up with the same sign of net charge. For example, if you touch a negatively charged rubber rod to a neutral metal sphere, excess electrons flow from the rod to the sphere, leaving the sphere negatively charged (same sign as the rod).

Charging by induction is charging without any physical contact between the original charged object and the neutral object being charged. Induction relies on polarizing the conductor and using a ground connection (a large reservoir of charge) to remove charge of one sign. The resulting net charge on the neutral object is always opposite in sign to the original charged object.

Worked Example

Problem: A negatively charged rubber rod is brought near (but does not touch) a neutral copper sphere mounted on an insulating stand. The sphere is briefly connected to ground on the side opposite the rod, the ground connection is broken, then the rod is removed. What is the sign of the net charge on the sphere after all steps?

1. When the negative rod is brought near the neutral sphere, free electrons in the copper are repelled by the rod’s negative charge. This leaves the side of the sphere near the rod with net positive charge, and the far side with net negative charge.
2. When the far side is grounded, the excess negative charge on the far side can flow out of the sphere into the ground. Positive charge near the rod is trapped by electrostatic attraction to the rod and cannot leave.
3. When the ground connection is broken, the negative charge that left cannot return. Removing the rod leaves the sphere with a net positive charge, opposite the sign of the original rod.

Answer: The sphere has a net positive charge.

Exam tip: AP MCQ almost always tests the sign difference between conduction and induction. Memorize the shortcut: contact = same sign, no contact induction = opposite sign.

4. Coulomb's Law and Superposition of Electric Force

Coulomb’s law describes the magnitude of the electrostatic force between two stationary point charges. For two point charges $q_1$ and $q_2$ separated by a distance $r$, the magnitude of the force that each charge exerts on the other is: $$F=k\frac{|q_1{q}_2|}{r^2}$$ where $k=8.99\times 10^9{\text{N m}}^2/{\text{C}}^2\approx 9\times 10^9$ (the approximate value is acceptable for all AP calculations, and $k$ is provided on the formula sheet). Electric force is an inverse-square law, meaning force decreases with the square of the distance between charges, just like gravitational force. Unlike gravity, electric force can be attractive (opposite charges) or repulsive (like charges).

When more than two charges are present, the total net force on any charge is the vector sum of the individual forces exerted by each other charge. This is the principle of superposition: forces add as vectors, so you must decompose each force into x and y components, add the components, then calculate the magnitude and direction of the net force.

Worked Example

Problem: Three point charges are placed on an xy-plane: $q_1=+1\mu \text{C}$ at $(0,0)$, $q_2=+1\mu \text{C}$ at $(0,2\text{m})$, and $q_3=-2\mu \text{C}$ at $(2\text{m},0)$. Find the magnitude of the net force on $q_1$.

1. Calculate the force from $q_2$ on $q_1$: $r=2\text{m}$, so $F_{21}=k\frac{|q_1{q}_2|}{r^2}=(9\times 10^9)\frac{(1\times 10^{-6})(1\times 10^{-6})}{2^2}=2.25\times 10^{-3}\text{N}$. Like charges repel, so $F_{21}$ points along the negative y-axis: ${\vec{F}}_{21}=(0,-2.25\times 10^{-3}\text{N})$.
2. Calculate the force from $q_3$ on $q_1$: $r=2\text{m}$, so $F_{31}=k\frac{|q_1{q}_3|}{r^2}=(9\times 10^9)\frac{(1\times 10^{-6})(2\times 10^{-6})}{2^2}=4.5\times 10^{-3}\text{N}$. Opposite charges attract, so $F_{31}$ points along the positive x-axis: ${\vec{F}}_{31}=(4.5\times 10^{-3},0\text{N})$.
3. Add the components: $F_{\text{net},x}=4.5\times 10^{-3}\text{N}$, $F_{\text{net},y}=-2.25\times 10^{-3}\text{N}$.
4. Calculate the magnitude: $|F_{\text{net}}|=\sqrt{F_x^2+F_y^2}=\sqrt{(4.5\times 10^{-3}{)}^2+(2.25\times 10^{-3}{)}^2}\approx 5.0\times 10^{-3}\text{N}$.

Exam tip: Always calculate magnitudes of individual forces with Coulomb's law first, then assign direction based on charge signs, instead of plugging negative signs into the magnitude formula. This avoids common sign errors in vector addition.

5. Common Pitfalls (and how to avoid them)

• Wrong move: For an induction problem, you conclude the sphere has the same sign charge as the original rod. Why: Students confuse induction with conduction, since both start with a charged object brought near a neutral object. Correct move: Memorize the rule: conduction = contact = same sign, induction = no contact (uses grounding) = opposite sign; write the rule down at the start of any charging problem if you are unsure.
• Wrong move: When calculating Coulomb's law force between two charged spheres, you use the distance between the spheres' surfaces instead of the distance between their centers. Why: For uniform spherical charge distributions, we treat them as point charges at the center, and students confuse this with problems involving sphere radii. Correct move: For any uniform spherical charge, $r$ in Coulomb's law is always the distance between the centers of the spheres, regardless of sphere size.
• Wrong move: When finding the number of electrons for a net charge $q$, you calculate $n=qe$ instead of $n=q/e$. Why: Students mix up the rearrangement of $q=ne$, especially when working with small scientific notation exponents. Correct move: Always write the original formula $q=ne$ first, then rearrange step by step instead of solving for $n$ in your head.
• Wrong move: When adding multiple electric forces, you add the magnitudes directly instead of adding as vectors. Why: Superposition of force is often misinterpreted as "add the numbers", so students forget force is a vector quantity. Correct move: Every time you have more than one force on a charge, immediately draw a coordinate system and decompose all forces into components before adding.
• Wrong move: You claim the force on $q_1$ from $q_2$ has a different magnitude than the force on $q_2$ from $q_1$, when the charges have different magnitudes. Why: Students get so focused on different charge sizes that they forget Coulomb's law is symmetric and follows Newton's third law. Correct move: Always confirm that the force pair between two charges has equal magnitude and opposite direction, regardless of the charge magnitudes.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

Two identical conducting spheres A and B carry net charge $+2Q$ and $+4Q$ respectively, separated by a distance $d$ much larger than the sphere radius, so they can be treated as point charges. The magnitude of the electrostatic force between them is $F$. The spheres are brought into contact, then returned to their original separation $d$. What is the new magnitude of the force between them? A) $\frac{25F}{32}$ B) $\frac{9F}{8}$ C) $\frac{25F}{16}$ D) $\frac{9F}{4}$

Worked Solution: First, write the original force from Coulomb's law: $F=k\frac{(2Q)(4Q)}{d^2}=\frac{8k{Q}^2}{d^2}$. When identical spheres are brought into contact, charge is conserved and splits equally between the two spheres. Total charge is $2Q+4Q=6Q$, so each sphere has $3Q$ after contact. The new force is $F_{\text{new}}=k\frac{(3Q)(3Q)}{d^2}=\frac{9k{Q}^2}{d^2}$. Substitute $\frac{k{Q}^2}{d^2}=\frac{F}{8}$ from the original force equation to get $F_{\text{new}}=\frac{9F}{8}$. The correct answer is B.

Question 2 (Free Response)

Two point charges are fixed on the x-axis: $q_A=+4.0\times 10^{-6}\text{C}$ at $x=0$, and $q_B=-1.0\times 10^{-6}\text{C}$ at $x=3.0\text{m}$. (a) Find the location on the x-axis where the net electric force on a third point charge $q_C$ of any sign is zero. (b) Explain why your answer cannot be located between the two charges. (c) If $q_C=+2.0\times 10^{-6}\text{C}$, calculate the net force on $q_C$ when placed at the location you found in part (a).

Worked Solution: (a) A zero net force point must lie outside the two charges, on the side of the smaller magnitude charge, so $d>3.0\text{m}$ where $d$ is the position of the point. Set magnitudes of force from each charge equal: $$k\frac{|q_A{q}_C|}{d^2}=k\frac{|q_B{q}_C|}{(d-3{)}^2}$$ Cancel common terms, substitute values: $\frac{4}{d^2}=\frac{1}{(d-3{)}^2}$. Take square roots (distances are positive): $\frac{2}{d}=\frac{1}{d-3}$. Solve for $d$: $2(d-3)=d⟹d=6.0\text{m}$ on the positive x-axis.

(b) Between the two charges ($0<x<3\text{m}$), any charge $q_C$ will experience forces in the same direction from both charges. For positive $q_C$, $q_A$ repels right and $q_B$ attracts right; for negative $q_C$, $q_A$ attracts left and $q_B$ repels left. Since both forces point the same direction, they cannot cancel to give zero net force.

(c) Substitute $q_C$ at $x=6.0\text{m}$: $F_A=k\frac{q_A{q}_C}{(6.0{)}^2}=2.0\times 10^{-3}\text{N}$ right (repulsion), $F_B=k\frac{|q_B{q}_C|}{(3.0{)}^2}=2.0\times 10^{-3}\text{N}$ left (attraction). Net force $F_{\text{net}}=2.0\times 10^{-3}-2.0\times 10^{-3}=\boxed{0\text{N}}$, matching the result from part (a).

Question 3 (Application / Real-World Style)

A typical static shock when touching a metal doorknob occurs when your body carries a net charge of approximately $-1.0\mu \text{C}$, and the doorknob acts like a point charge of $+1.0\mu \text{C}$ at a distance of 1.0 cm from your finger just before contact. What is the magnitude of the attractive electric force between your finger and the doorknob at this distance? Compare this force to the weight of a 10 gram paperclip ($g=9.8{\text{m/s}}^2$), and comment on the strength of electrostatic force.

Worked Solution: Convert units: $r=1.0\text{cm}=0.01\text{m}$, $|q_1|=|q_2|=1.0\times 10^{-6}\text{C}$. Apply Coulomb's law: $$F=k\frac{|q_1{q}_2|}{r^2}=(9\times 10^9)\frac{(1.0\times 10^{-6}{)}^2}{(0.01{)}^2}=90\text{N}$$ Weight of the 10 g paperclip is $mg=(0.010\text{kg})(9.8{\text{m/s}}^2)\approx 0.1\text{N}$. The electric force is ~900 times larger than the weight of the paperclip. This result shows that even small net static charges produce very large forces at close distances, which is why electrostatic effects like static shocks are easily detectable by humans.

7. Quick Reference Cheatsheet

8. What's Next

Charge and electric force is the absolute foundation for all of Unit 3: Electric Force, Field, and Potential. Next, you will extend the idea of force between discrete charges to the concept of the electric field, which describes the force per unit charge that a test charge experiences at any point in space. Without mastering Coulomb's law, superposition of force, and the fundamental properties of charge, you will not be able to correctly calculate electric fields from point charges or continuous charge distributions, which is a core skill for both MCQ and FRQ questions on the AP exam. This topic also feeds into later units, including electric circuits and electromagnetic induction, where the motion of charge drives all circuit behavior.

• Electric Field
• Electric Potential and Potential Difference
• Gauss's Law for Electrostatics