🎯⭐ INTERACTIVE LESSON

Electric Charge and Coulomb's Law

Learn step-by-step with interactive practice!

← Back to Standard Lesson

Electric Charge and Coulomb's Law - Complete Interactive Lesson

Part 1: Introduction & Charge

⚡ Electric Charge & Coulomb's Law

Part 1 of 7

Everything in electrostatics starts with one idea: matter carries electric charge, and charges exert forces on each other.

In this part you'll learn:

What electric charge actually is
The two types of charge and how they interact
Why charge is always conserved
The fundamental unit of charge

By the end, you'll have the vocabulary and intuition for the rest of electrostatics.

What Is Electric Charge?

Electric charge is a fundamental property of matter — just like mass, but for electromagnetic interactions.

Two Types of Charge

Positive (+): carried by protons
Negative (−): carried by electrons

Key Facts

Property	Value
Electron charge	$e = 1.6 \times 10^{-19}$ C
Proton charge	$+e = 1.6 \times 10^{-19}$ C

One coulomb is an enormous amount of charge. A typical static shock involves only about $10^{-6}$ C (a microcoulomb).

The Interaction Rules

Like charges repel. Opposite charges attract.

This is the single most important rule in electrostatics. Every force, every field, every circuit behavior traces back to this.

Conservation & Quantization of Charge

Conservation of Charge

Charge is never created or destroyed — it can only be transferred from one object to another.

When you rub a balloon on your hair:

Electrons transfer from hair → balloon
Balloon becomes negative, hair becomes positive
Total charge of the system is unchanged

$q_{\text{total, before}} = q_{\text{total, after}}$

Why Does This Matter?

Electrostatics isn't just theory — it's everywhere:

🖨️ Laser printers — use static charge to attract toner to paper in precise patterns

⚡ Lightning — massive charge separation in clouds, then sudden discharge

🚗 Spray painting — charged paint droplets are attracted to grounded car bodies for even coating

🧪 Particle accelerators — use electric forces to accelerate charged particles to near light speed

📱 Touchscreens — detect changes in electric field when your charged finger approaches

Every one of these applications relies on the same simple rule: charges exert forces on each other.

Check Your Understanding — Charge basics.

Before You Move On — Two common misconceptions about charge.

Part 2: Coulomb's Law

📐 Coulomb's Law — The Equation

Part 2 of 7

Now that you know what charge is, let's quantify the force between charges. Coulomb's Law is the foundation equation of electrostatics — every other concept builds on it.

Coulomb's Law

$F_E = k\frac{|q_1 q_2|}{r^2}$

Part 3: Problem Solving

🧮 Coulomb's Law — Problem Solving

Part 3 of 7

Time to use Coulomb's Law with real numbers. You'll practice the 5-step problem-solving workflow and build confidence computing electrostatic forces.

By the end of this part, you'll solve Coulomb's Law calculations in a few clean lines.

The 5-Step Workflow

Every Coulomb's Law problem follows the same pattern:

Step 1 — Given: List charges, distances, and what you need to find

Step 2 — Formula: Write Coulomb's Law

$F_E = k\frac{|q_1 q_2|}{r^2}$

Part 4: Conductors & Charging

🔌 Conductors, Insulators & Charging Methods

Part 4 of 7

Not all materials respond to charge the same way. Understanding the difference between conductors and insulators — and the three methods of charging — is essential for AP Physics 2.

Conductors vs. Insulators

Conductors

Materials where electrons move freely throughout the material.

Metals (copper, aluminum, gold)
Saltwater, plasma
Charge distributes itself on the outer surface
Electric field inside a conductor is zero (in electrostatic equilibrium)

Insulators

Materials where electrons are locked in place — charge stays where you put it.

Rubber, glass, plastic, wood
Charge can be deposited locally and stays put
Electric field can exist inside an insulator

Semiconductors

In between — conductivity can be controlled. Think silicon chips. (Not heavily tested on AP Physics 2, but good to know.)

The Key Insight

In a conductor, excess charge always migrates to the outer surface and distributes itself to minimize repulsion. Inside, $E = 0$ .

Part 5: Conservation & Quantization

⚖️ Charge Conservation & Quantization

Part 5 of 7

Conservation of charge isn't just a rule — it's a law of nature as fundamental as conservation of energy. In this part you'll practice applying it to real situations and build the quantitative reasoning the AP exam demands.

Conservation of Charge — Deep Dive

The Law

The total electric charge in an isolated system is always conserved.

$\sum q_{\text{before}} = \sum q_{\text{after}}$

Part 6: Superposition

🎯 Multi-Charge Systems & Superposition of Forces

Part 6 of 7

Real problems rarely involve just two charges. When three or more charges interact, you need superposition — the principle that forces add as vectors.

This is where Coulomb's Law gets powerful — and where most students start making mistakes.

The Principle of Superposition

The net force on any charge is the vector sum of the individual Coulomb forces from every other charge.

$\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \cdots$

Part 7: Synthesis & AP Strategies

🏆 Synthesis & AP Exam Strategies

Part 7 of 7 — The Grand Finale

You've built every tool you need. Now let's put it all together with multi-step problems and the strategies that top-scoring AP students use.

AP Exam Strategy Guide

What the Exam Tests

The AP Physics 2 exam tests Coulomb's Law in three main ways:

1. Proportional Reasoning (most common) "What happens to the force when ___?" → Use scaling, not calculation.

2. Vector Superposition Three or more charges → find net force. Usually 1D or with symmetric 2D geometry.

3. Conceptual Understanding Conductors vs insulators, charging methods, conservation of charge.

Time-Saving Tips

🚀 Don't plug in numbers for scaling questions — use ratios
🎯 Symmetry first — if the geometry is symmetric, exploit it to eliminate components
📐 Draw before you solve — a quick diagram prevents sign errors
⚡ Watch units — μC → C is the #1 calculation error
🧪 Sanity check — lab-scale forces should be in the range of 0.001–10 N

Synthesis Problem 1 — Combined Concepts

Problem: Two identical metal spheres, A and B, are on insulating stands.

A has charge $+12\ \mu\text{C}$

Quantization of Charge

Charge comes in discrete packets. You can't have half an electron!

where $n$ is an integer and $e = 1.6 \times 10^{-19}$ C.

This means any charged object has a charge that's a whole-number multiple of $e$ .

$F_E$ = electrostatic force (N)
$k = 8.99 \times 10^9$ N·m²/C² (Coulomb's constant)
$q_1, q_2$ = the two charges (C)
$r$ = distance between charge centers (m)

What This Equation Tells You

Force is proportional to each charge — double one charge, double the force
Force follows an inverse-square law — double the distance, force drops to 1/4
The equation gives magnitude only — direction comes from the charge signs

$q_1$	$q_2$	Force
+	+	Repulsive (push apart)
−	−	Repulsive (push apart)
+	−	Attractive (pull together)
−	+	Attractive (pull together)

Comparison with Gravity

$F_g = G\frac{m_1 m_2}{r^2}$

Same mathematical form! But key differences:

Gravity is always attractive; electric force can be attractive or repulsive
Electric force is vastly stronger ( $\sim 10^{36}$ times for an electron-proton pair)

Proportional Reasoning — The AP Shortcut

Most AP Physics 2 Coulomb's Law questions don't ask you to plug in numbers. They ask: "What happens to the force if..."

The Key Relationships

$F \propto q_1 \qquad F \propto q_2 \qquad F \propto \frac{1}{r^2}$

Quick Scaling Examples

Change	Effect on Force
Double $q_1$	$F \times 2$
Triple $r$

The trick: multiply all the individual factors together.

Proportional Reasoning Check 🎯

Scaling Drill ⚡

Enter the multiplier for force in each case:

Distance is doubled (charges unchanged)
Distance is halved (charges unchanged)
One charge is doubled AND distance is doubled

Use fractions like $21/4` or whole numbers like$ 24`.

Coulomb's Law Concept Check

Before You Move On — Common Coulomb's Law traps.

Step 3 — Convert: Make sure everything is in SI units (C, m, N)

Prefix	Conversion
$\mu$ C (micro)	$\times 10^{-6}$ C
nC (nano)	$\times 10^{-9}$ C
cm	$\times 10^{-2}$ m
mm	$\times 10^{-3}$ m

Step 4 — Substitute & Solve: Plug in and compute

Step 5 — Interpret: State magnitude AND direction

Worked Example 1

Problem: Two charges, $q_1 = +3.0\ \mu\text{C}$ and $q_2 = -5.0\ \mu\text{C}$ , are separated by 0.20 m. Find the force.

Step 1 — Given:

$q_1 = +3.0 \times 10^{-6}$ C
$q_2 = -5.0 \times 10^{-6}$ C

Step 2 — Formula:

$F_E = k\frac{|q_1 q_2|}{r^2}$

Step 3 — Substitute:

$F_E = (8.99 \times 10^9)\frac{|(3.0 \times 10^{-6})(-5.0 \times 10^{-6})|}{(0.20)^2}$

Step 4 — Solve:

$F_E = (8.99 \times 10^9)\frac{15.0 \times 10^{-12}}{0.04}$

$F_E = (8.99 \times 10^9)(3.75 \times 10^{-10})$

$F_E = 3.37 \text{ N}$

Step 5 — Interpret:

Magnitude: 3.37 N
Direction: Attractive (opposite charges)
The charges pull toward each other with 3.37 N of force

Worked Example 2

Problem: Two identical charges of $+4.0\ \mu\text{C}$ are 0.30 m apart. Find the force on each.

Given:

$q_1 = q_2 = +4.0 \times 10^{-6}$ C
$r = 0.30$ m

Solution:

$F_E = (8.99 \times 10^9)\frac{(4.0 \times 10^{-6})^2}{(0.30)^2}$

$F_E = (8.99 \times 10^9)\frac{16.0 \times 10^{-12}}{0.09}$

$F_E = (8.99 \times 10^9)(1.78 \times 10^{-10})$

$F_E = 1.60 \text{ N}$

Direction: Repulsive — each charge pushes the other away with 1.60 N.

Key point: By Newton's Third Law, both charges experience the same magnitude of force, even if they had different charges!

Your Turn!

$q_1 = +2.0\ \mu\text{C}$ , $q_2 = +8.0\ \mu\text{C}$ , separated by $r = 0.40$ m.

Use $k = 9.0 \times 10^9$ N·m²/C² (rounded for cleaner math).

Enter in order:

$|q_1 q_2|$ in C² (use scientific notation like $216e-12`)
$r^2$ in m²
Force magnitude in N (to 3 significant figures)

Unit Conversion Check — These trip up students constantly on the AP exam.

Problem-Solving Check

Before You Move On — The biggest calculation traps.

This is why a car protects you during a lightning strike — it's a conducting shell (Faraday cage).

Three Methods of Charging

1. Charging by Friction (Triboelectric)

Rub two insulators together
Electrons transfer from one to the other
Both objects end up with equal and opposite charges
Example: balloon on hair, glass rod on silk

2. Charging by Conduction (Contact)

Touch a charged object to a neutral conductor
Charge flows until both reach the same potential
Both objects end up with the same sign of charge
Example: touching a charged rod to a metal sphere

3. Charging by Induction

Bring a charged object near (but not touching) a conductor
Charge in the conductor redistributes (polarizes)
Ground the conductor → one sign of charge drains away
Remove ground, then remove charged object
Result: conductor has charge opposite to the inducing object
No contact needed!

Summary Table

Method	Contact?	Resulting Sign	Works On
Friction	Yes	Opposite on each	Insulators
Conduction	Yes	Same as source	Conductors
Induction	No	Opposite to source	Conductors

Polarization — Why Neutral Objects Are Attracted

Even a neutral object can be attracted to a charged object! Here's why:

When a charged rod approaches a neutral conductor:

Opposite charges in the conductor are pulled closer to the rod
Like charges are pushed farther away
The nearby opposite charges feel a stronger force (closer = stronger, by Coulomb's Law)
Net result: attraction

This also works with insulators, but by a different mechanism — the electron clouds of individual atoms shift slightly, creating tiny dipoles. This is called polarization.

A charged object attracts all neutral objects — conductors and insulators alike — through polarization.

This is why a charged balloon sticks to a neutral wall!

Conductor & Insulator Check 🎯

Charging Methods Quiz

Before You Move On — Conductor misconceptions.

This works at every scale:

Rubbing a balloon (electrons transfer, total = 0)
Nuclear reactions ( $\beta$ decay: neutron → proton + electron + antineutrino — charge conserved)
Pair production ( $\gamma \to e^+ + e^-$ — charge conserved: 0 → +1 + (−1))

Sharing Charge Between Conductors

When two identical conducting spheres touch, they share charge equally:

$q_{\text{each}} = \frac{q_1 + q_2}{2}$

Example: Sphere A has +6 μC, Sphere B has −2 μC. After touching: $q_{\text{each}} = \frac{+6 + (-2)}{2} = \frac{+4}{2} = +2\ \mu\text{C each}$

For non-identical spheres, charge distributes based on capacitance (surface area), but the AP exam usually gives identical spheres.

Quantization Problems

Since $q = ne$ , where $e = 1.6 \times 10^{-19}$ C:

Finding Number of Electrons

Problem: An object has charge $-4.8 \times 10^{-19}$ C. How many excess electrons?

$n = \frac{|q|}{e} = \frac{4.8 \times 10^{-19}}{1.6 \times 10^{-19}} = 3 \text{ electrons}$

Finding Charge from Electron Count

Problem: An object gains $10^{12}$ electrons. What is its charge?

$q = -ne = -(10^{12})(1.6 \times 10^{-19}) = -1.6 \times 10^{-7} \text{ C} = -0.16\ \mu\text{C}$

Checking if a Charge is Possible

Any measured charge must satisfy $q/e =$ integer. If not, the measurement has an error.

Charge Sharing Drill

Two identical metal spheres: A has $+8\ \mu\text{C}$ and B has $-4\ \mu\text{C}$ .

They touch and then separate. Enter:

Total charge of the system (in μC)
Charge on each sphere after separation (in μC)
If sphere A then touches a third neutral identical sphere C, what is A's final charge? (in μC)

Quantization Drill

Use $e = 1.6 \times 10^{-19}$ C.

An object has charge $-8.0 \times 10^{-19}$ C. How many excess electrons? (integer)
An object gains $5 \times 10^{6}$ electrons. What is its charge in C? (use scientific notation like -8e-13)
Is a charge of $2.0 \times 10^{-19}$ C possible? (yes or no)

Conservation & Quantization Quiz

Real-World Check — Applying conservation to a circuit scenario.

What This Means in Practice

Calculate the force from each other charge separately using Coulomb's Law
Determine the direction of each force (attract/repel)
Add as vectors — break into components if needed

Each pairwise force is independent — charge A's force on charge C is not affected by the presence of charge B.

1D Superposition — Three Charges in a Line

Problem: Three charges on the x-axis:

$q_A = +2\ \mu\text{C}$ at $x = 0$
$q_B = -3\ \mu\text{C}$ at $x = 0.3$ m
$q_C = +1\ \mu\text{C}$ at $x = 0.5$ m

Find the net force on $q_B$ .

Solution

Force from A on B: $F_{AB} = k\frac{|q_A q_B|}{r_{AB}^2} = (9 \times 10^9)\frac{(2 \times 10^{-6})(3 \times 10^{-6})}{(0.3)^2} = 0.60 \text{ N}$ Direction: A is positive, B is negative →

Force from C on B: $F_{CB} = k\frac{|q_C q_B|}{r_{CB}^2} = (9 \times 10^9)\frac{(1 \times 10^{-6})(3 \times 10^{-6})}{(0.2)^2} = 0.675 \text{ N}$ Direction: C is positive, B is negative →

Net force on B: $F_{\text{net}} = -0.60 + 0.675 = +0.075 \text{ N (rightward)}$

The forces partially cancel because they pull in opposite directions!

2D Superposition — Component Method

When charges aren't in a line, you must use vector components:

The Workflow

Find each force magnitude using Coulomb's Law
Find the angle of each force
Break each force into $x$ $x$ and $y$ $y$ components:
- $F_x = F\cos\theta$
- $F_y = F\sin\theta$
Add all $x$ -components, add all $y$ -components
Find the magnitude: $F_{\text{net}} = \sqrt{F_x^2 + F_y^2}$
Find the angle: $\theta = \tan^{-1}(F_y / F_x)$

Common AP Geometry

The AP exam loves equilateral triangle and right-angle charge arrangements because the geometry is clean.

For equilateral triangles: angles are 60° For squares: diagonal distance is $r\sqrt{2}$

Equilibrium Problems

A charge is in equilibrium when the net force on it is zero:

$\vec{F}_{\text{net}} = 0$

Classic Problem Type

"Where can a third charge be placed so that it's in equilibrium?"

For two charges of the same sign:

The equilibrium point is between the charges
Closer to the smaller charge (it needs less distance to match the larger charge's pull)

For two charges of opposite sign:

The equilibrium point is outside the pair
On the side of the smaller charge

Setting Up the Equation

At equilibrium: $F_1 = F_2$

$k\frac{|q_1 q_3|}{r_1^2} = k\frac{|q_2 q_3|}{r_2^2}$

The $k$ and $q_3$ cancel, leaving:

$\frac{|q_1|}{r_1^2} = \frac{|q_2|}{r_2^2}$

This is a clean equation to solve for position!

1D Superposition Drill

Two charges on the x-axis:

$q_1 = +4\ \mu\text{C}$ at origin
$q_2 = +4\ \mu\text{C}$ at $x = 0.4$ m
A test charge $q_3 = +1\ \mu\text{C}$ at $x = 0.2$ m (the midpoint)

Use $k = 9 \times 10^9$ .

Enter:

Force from $q_1$ on $q_3$ (magnitude in N, round to 1 decimal)
Force from $q_2$ on (magnitude in N, round to 1 decimal)

Round all answers to 3 significant figures.

Superposition & Equilibrium Quiz

Before You Move On — Superposition traps.

B has charge

-4\ \mu\text{C}

They are 0.30 m apart

(a) What is the force between them? (b) They are touched together and separated back to 0.30 m. What is the new force?

(a) Before contact:

$F = k\frac{|q_A q_B|}{r^2} = (9 \times 10^9)\frac{(12 \times 10^{-6})(4 \times 10^{-6})}{(0.30)^2}$

$F = (9 \times 10^9)\frac{48 \times 10^{-12}}{0.09} = (9 \times 10^9)(5.33 \times 10^{-10}) = 4.8 \text{ N}$

Direction: Attractive (opposite charges)

(b) After contact:

Charge sharing: $q_{\text{each}} = \frac{+12 + (-4)}{2} = +4\ \mu\text{C}$

$F = (9 \times 10^9)\frac{(4 \times 10^{-6})^2}{(0.30)^2} = (9 \times 10^9)\frac{16 \times 10^{-12}}{0.09} = 1.6 \text{ N}$

Direction: Repulsive (both positive now)

The force decreased from 4.8 N attractive to 1.6 N repulsive — and changed direction!

Synthesis Problem 2 — Scaling + Superposition

Problem: Two charges $+Q$ and $+4Q$ are separated by distance $d$ .

(a) Where along the line between them is the net force on a test charge zero?

Solution

Let the test charge $+q$ be at distance $x$ from $+Q$ (so distance $d - x$ from $+4Q$ ).

At equilibrium: $F_1 = F_2$

$k\frac{Qq}{x^2} = k\frac{4Qq}{(d-x)^2}$

Cancel $k$ and $q$ :

$\frac{Q}{x^2} = \frac{4Q}{(d-x)^2}$

$\frac{1}{x^2} = \frac{4}{(d-x)^2}$

Cross-multiply:

$(d-x)^2 = 4x^2$

$d - x = 2x \quad (\text{taking positive root})$

$d = 3x \quad \Rightarrow \quad x = \frac{d}{3}$

The equilibrium point is at d/3 from the smaller charge — closer to the smaller charge, as expected!

Mixed Concepts Drill

Two identical spheres, each with charge $+6\ \mu\text{C}$ , are 0.20 m apart.

Force between them (in N, use $k = 9 \times 10^9$ )
One sphere touches a neutral identical sphere, then is returned. Its new charge? (in μC)
New force between the original pair (in N, round to 2 decimals)

Round all answers to 3 significant figures.

Final Mastery Quiz — All concepts combined.

Final Check — The two mistakes that cost the most points on the AP exam.

F \times \frac{1}{9}

(0.20)2∣(3.0×10−6)(−5.0×10−6)∣

(0.30)2(4.0×10−6)2

109)(0.3)2(2×10−6)(3×10−6)=

attractive → toward A → leftward (−x)

109)(0.2)2(1×10−6)(3×10−6)=

attractive → toward C → rightward (+x)

Net force on

q_3

(in N)

Electric Charge and Coulomb's Law

Quantization of Charge

What This Equation Tells You

Direction Rules

Comparison with Gravity

Proportional Reasoning — The AP Shortcut

The Key Relationships

Quick Scaling Examples

Worked Example 1

Worked Example 2

Three Methods of Charging

1. Charging by Friction (Triboelectric)

2. Charging by Conduction (Contact)

3. Charging by Induction

Summary Table

Polarization — Why Neutral Objects Are Attracted

Sharing Charge Between Conductors

Quantization Problems

Finding Number of Electrons

Finding Charge from Electron Count

Checking if a Charge is Possible

What This Means in Practice

Critical Rule

1D Superposition — Three Charges in a Line

Solution

2D Superposition — Component Method

The Workflow

Common AP Geometry

Equilibrium Problems

Classic Problem Type

Setting Up the Equation

Solution

Synthesis Problem 2 — Scaling + Superposition

Solution