# Charge and Electric Force

> AP Physics 2 · Unit 3: Electric Force, Field, and Potential
> Source: https://www.owlsprep.com/study/ap-physics-2-u3-charge-and-electric-force/

This sub-topic covers core properties of electric charge, common charging processes, Coulomb's law of electrostatic force, and vector superposition of multiple forces. It is the foundation for all electrostatics topics on the AP Physics 2 exam.

**Prerequisites:** Basic 2D vector addition and decomposition; Newton's laws of motion; Properties of conductors and insulators

## Learning objectives

- Explain the fundamental properties of charge (conservation, quantization)
- Distinguish between charging by conduction and induction
- Apply Coulomb's law to calculate electrostatic force between point charges
- Use the principle of superposition to find net force on a charge from multiple charges

## Fundamental Properties of Charge

Charge is a fundamental intrinsic property of matter that causes it to experience force in an electromagnetic field. There are two types: positive and negative. Like charges repel, opposite charges attract. The SI unit of charge is the coulomb (C), and the elementary charge $e = 1.6 \times 10^{-19}\ \text{C}$ (the magnitude of charge on one proton or electron).

**Conservation and Quantization of Charge** — Conservation of charge states that total net charge in an isolated system is constant; charge is only transferred, not created or destroyed. Quantization of charge states that all free charge is an integer multiple of the elementary charge $e$, so $q = ne$ where $n$ is any non-zero positive or negative integer.

*Notation:* $q = ne$

*Example:* An electron has $n=-1$, a proton has $n=+1$.

**Worked example:** 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.

   $$Q_{\text{total}} = 0 + 9.6 \times 10^{-18}\ \text{C} = 9.6 \times 10^{-18}\ \text{C}$$
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:

   $$\boxed{-30}$$

> **Exam tip:** When asked for the number of excess electrons, remember positive net charge means a deficit, which corresponds to a negative number of excess electrons. Always check question wording.

## Charging by Conduction and Induction

AP Physics 2 regularly tests conceptual understanding of the two most common charging processes. Conductors allow free electrons to move through the material, while insulators bind electrons to their atoms, a key distinction for understanding charging.

- **Conduction (contact):** Requires physical contact between a charged and neutral object. Both objects end up with the same sign of net charge after charge transfer.
- **Induction:** Charging without physical contact, relying on polarization and a ground connection to remove excess charge of one sign. The charged object ends up with the opposite sign of net charge to the original charged object.

**Worked example:** 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 copper are repelled, leaving the side near the rod positive and the far side negative.
2. When the far side is grounded, excess negative charge flows out of the sphere into the ground, leaving only trapped positive charge attracted to the rod.
3. When the ground connection is broken, negative charge cannot return. Removing the rod leaves the sphere with net positive charge, opposite the sign of the original rod.

> **Exam tip:** AP multiple-choice almost always tests the sign difference between conduction and induction. The shortcut: contact = same sign, no-contact induction = opposite sign.

## Coulomb's Law and Superposition of Electric Force

Coulomb's law describes the magnitude of the electrostatic force between two stationary point charges. When multiple charges act on a single charge, the principle of superposition applies: forces add as vectors.

**Coulomb's Law** — The magnitude of the electrostatic force between two point charges $q_1$ and $q_2$ separated by distance $r$, where $k = 8.99 \times 10^9\ \text{N m}^2/\text{C}^2 \approx 9 \times 10^9$ (provided on the AP formula sheet). Force direction follows the rule: like charges repel, opposite charges attract.

*Notation:* F = k\frac{|q_1 q_2|}{r^2}

*Example:* Electric force is an inverse-square law, similar to gravitational force, but can be attractive or repulsive unlike gravity.

For systems of more than two charges, the net force on any charge is the vector sum of the individual forces exerted by each other charge. This requires decomposing all forces into $x$ and $y$ components, adding components, then finding the magnitude and direction of the net force.

**Worked example:** 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 $q_1$: $r = 2\ \text{m}$, like charges repel so force points along negative y-axis:

   $$F_{21} = k \frac{|q_1 q_2|}{r^2} = (9 \times 10^9) \frac{(1 \times 10^{-6})(1 \times 10^{-6})}{2^2} = 2.25 \times 10^{-3}\ \text{N}, \quad \vec{F}_{21} = (0, -2.25 \times 10^{-3}\ \text{N})$$
2. Calculate the force from $q_3$ on $q_1$: $r = 2\ \text{m}$, opposite charges attract so force points along positive x-axis:

   $$F_{31} = k \frac{|q_1 q_3|}{r^2} = (9 \times 10^9) \frac{(1 \times 10^{-6})(2 \times 10^{-6})}{2^2} = 4.5 \times 10^{-3}\ \text{N}, \quad \vec{F}_{31} = (4.5 \times 10^{-3}, 0\ \text{N})$$
3. Add the x and y components of force:

   $$F_{\text{net},x} = 4.5 \times 10^{-3}\ \text{N}, \quad F_{\text{net},y} = -2.25 \times 10^{-3}\ \text{N}$$
4. Calculate the magnitude of the net force:

   $$|F_{\text{net}}| = \sqrt{F_x^2 + F_y^2} = \sqrt{(4.5 \times 10^{-3})^2 + (2.25 \times 10^{-3})^2} \approx 5.0 \times 10^{-3}\ \text{N}$$

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

## AP-Style Worked Practice Problems

**Worked example:** 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?<br>Options: A) $\frac{25F}{32}$, B) $\frac{9F}{8}$, C) $\frac{25F}{16}$, D) $\frac{9F}{4}$

1. Write the original force from Coulomb's law:

   $$F = k \frac{(2Q)(4Q)}{d^2} = \frac{8kQ^2}{d^2}$$
2. When identical spheres touch, total charge is conserved and splits equally. Total charge: $2Q + 4Q = 6Q$, so each sphere has $3Q$ after contact.
3. Calculate the new force and substitute the original force relation:

   $$F_{\text{new}} = k \frac{(3Q)(3Q)}{d^2} = \frac{9kQ^2}{d^2} = 9 \left(\frac{F}{8}\right) = \frac{9F}{8}$$
4. The correct answer is option B.

**Worked example:** 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 $x = 3.0\ \text{m}$.<br>(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.<br>(b) Explain why your answer cannot be located between the two charges.<br>(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).

1. (a) A zero net force point must lie outside the two charges, on the side of the smaller magnitude charge. Let $d$ be the position of the point, so $d > 3.0\ \text{m}$. Set force magnitudes equal:

   $$k \frac{|q_A q_C|}{d^2} = k \frac{|q_B q_C|}{(d-3)^2}$$
2. Cancel common terms, substitute charge magnitudes and solve:

   $$\frac{4}{d^2} = \frac{1}{(d-3)^2} \implies \frac{2}{d} = \frac{1}{d-3} \implies 2(d-3) = d \implies d = 6.0\ \text{m}$$
3. (b) Between the two charges ($0 < x < 3\ \text{m}$), any charge $q_C$ will experience forces in the same direction from both charges, so they cannot cancel to give zero net force. For example: positive $q_C$ is repelled right by $q_A$ and attracted right by $q_B$, so both forces add in the same direction.
4. (c) Calculate individual forces at $x=6.0\ \text{m}$:

   $$F_A = k \frac{q_A q_C}{(6.0)^2} = 2.0 \times 10^{-3}\ \text{N (right)}, \quad F_B = k \frac{|q_B q_C|}{(3.0)^2} = 2.0 \times 10^{-3}\ \text{N (left)}$$
5. Net force is the sum of the two opposing forces:

   $$F_{\text{net}} = 2.0 \times 10^{-3} - 2.0 \times 10^{-3} = \boxed{0\ \text{N}}$$

**Worked example:** 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.

1. Convert all units to SI:

   $$r = 1.0\ \text{cm} = 0.01\ \text{m}, \quad |q_1| = |q_2| = 1.0 \times 10^{-6}\ \text{C}$$
2. Apply Coulomb's law to find the force magnitude:

   $$F = k \frac{|q_1 q_2|}{r^2} = (9 \times 10^9) \frac{(1.0 \times 10^{-6})^2}{(0.01)^2} = 90\ \text{N}$$
3. Calculate the weight of the 10 g paperclip:

   $$W = mg = (0.010\ \text{kg})(9.8\ \text{m/s}^2) \approx 0.1\ \text{N}$$
4. The electric force is ~900 times larger than the weight of the paperclip. This 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.

## Common pitfalls

- **Wrong:** For an induction problem, you conclude the sphere has the same sign charge as the original rod.
  - Why it fails: Students confuse induction with conduction, since both start with a charged object brought near a neutral object.
  - Correct: 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:** 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 it fails: For uniform spherical charge distributions, we treat them as point charges at the center, and students confuse this with problems involving sphere radii.
  - Correct: 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:** When finding the number of electrons for a net charge $q$, you calculate $n = q e$ instead of $n = q/e$.
  - Why it fails: Students mix up the rearrangement of $q = ne$, especially when working with small scientific notation exponents.
  - Correct: Always write the original formula $q = ne$ first, then rearrange step by step instead of solving for $n$ in your head.
- **Wrong:** When adding multiple electric forces, you add the magnitudes directly instead of adding as vectors.
  - Why it fails: Superposition of force is often misinterpreted as 'add the numbers', so students forget force is a vector quantity.
  - Correct: 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:** 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 it fails: Students get so focused on different charge sizes that they forget Coulomb's law is symmetric and follows Newton's third law.
  - Correct: Always confirm that the force pair between two charges has equal magnitude and opposite direction, regardless of the charge magnitudes.

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Elementary charge | $e = 1.60 \times 10^{-19}\ \text{C}$ | Electrons: $-e$, Protons: $+e$ |
| Quantization of charge | $q = ne$ | $n$ is non-zero integer; applies to all free charge |
| Conservation of charge | $Q_{\text{total, initial}} = Q_{\text{total, final}}$ | Applies to all isolated systems; charge is transferred, not created |
| Coulomb's Law (magnitude) | $F = k\frac{\|q_1 q_2\|}{r^2}$ | $k = 9.0 \times 10^9\ \text{Nm}^2/\text{C}^2$; $r$ = distance between charge centers |
| Superposition of electric force | $\vec{F}_{\text{net}} = \sum \vec{F}_i$ | Add as vectors, decompose into components before summing |
| Charging by conduction | Same sign charge as original object | Requires physical contact between objects |
| Charging by induction | Opposite sign charge as original object | No contact required; uses grounding to remove excess charge |
| Force direction | Like charges repel, opposite charges attract | Use this to assign direction after calculating magnitude |

## What's next

Charge and electric force is the foundational concept for all electrostatics in AP Physics 2. The next step is to build on this to understand electric fields, which describe the effect of a charge on the space around it, a core topic for Unit 3 that appears frequently in both multiple-choice and free-response sections. From electric fields, you will move on to electric potential, Gauss's law for electrostatics, and eventually capacitance and simple DC circuits, all of which rely on a solid understanding of charge interactions and Coulomb's law. Mastering the concepts and problem-solving skills in this sub-topic will make all subsequent electrostatics topics much easier to grasp.

- [Unit 3 Overview](https://www.owlsprep.com/study/ap-physics-2-u3-overview/)
- [Electric Field](https://www.owlsprep.com/study/ap-physics-2-u3-electric-field/)
- [Potential and Electric Potential Energy](https://www.owlsprep.com/study/ap-physics-2-u3-potential-and-electric-potential-energy/)

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