AP Physics C: Electricity and Magnetism

Electric Forces and Fields

Coulomb’s Law

Learning Objectives

By the end of this lesson, students should be able to:

• State and apply Coulomb’s Law.

• Calculate the electric force between point charges.

• Determine the direction of electric forces.

• Apply the principle of superposition to systems of charges.

• Analyze electric interactions using vector methods.

• Solve AP Physics C free-response and multiple-choice problems involving electrostatic forces.

Introduction

The Fundamental Law of Electrostatics

Coulomb’s Law describes the electric force between two point charges.

It is the electrostatic equivalent of Newton’s Law of Universal Gravitation.

Just as gravity explains the interaction between masses, Coulomb’s Law explains how charged particles interact.

It was experimentally established by Charles-Augustin de Coulomb in the eighteenth century.

Electric Charge

Types of Charge

There are two kinds of electric charge:

• Positive charge

• Negative charge

The basic rule of electric interaction is:

• Like charges repel.

• Opposite charges attract.

Unit of Charge

The SI unit of charge is the coulomb.

$$
1,C
$$

The elementary charge is:

$$
e = 1.60 \times 10^{-19},C
$$

Examples include:

• Proton:

$$
+e
$$

• Electron:

$$
-e
$$

Coulomb’s Law

Fundamental Equation

The magnitude of the electric force between two point charges is:

$$
F=k_e\frac{|q_1q_2|}{r^2}
$$

where:

• \(F\) = magnitude of electric force

• \(q_1,q_2\) = point charges

• \(r\) = distance between the charges

• \(k_e\) = Coulomb’s constant

Coulomb’s Constant

The electrostatic constant is:

$$
k_e=8.99\times10^9
;N\cdot m^2/C^2
$$

It can also be written as:

$$
k_e=\frac{1}{4\pi\varepsilon_0}
$$

where

$$
\varepsilon_0=8.85\times10^{-12};C^2/(N\cdot m^2)
$$

is the permittivity of free space.

Vector Form of Coulomb’s Law

Direction of the Force

The vector form is:

$$
\vec{F}_{12}=k_e\frac{q_1q_2}{r^2}\hat{r}
$$

where:

• (\hat{r}) is the unit vector pointing from one charge toward the other.

The sign of:

$$
q_1q_2
$$

determines whether the force is attractive or repulsive.

Newton’s Third Law

The forces between two charges satisfy:

$$
\vec{F}_{12}=-\vec{F}_{21}
$$

The forces are equal in magnitude and opposite in direction.

Attractive and Repulsive Forces

Like Charges

If:

$$
q_1q_2>0
$$

then the force is repulsive.

Examples:

• Positive-positive

• Negative-negative

Opposite Charges

If:

$$
q_1q_2<0
$$

then the force is attractive.

Examples:

• Positive-negative

Dependence on Distance

Inverse-Square Law

Coulomb’s Law follows an inverse-square relationship:

$$
F\propto\frac{1}{r^2}
$$

This means:

• Doubling the distance reduces the force to one-fourth.

• Tripling the distance reduces the force to one-ninth.

Comparison with Gravity

Newton’s Law of Gravitation:

$$
F_G=G\frac{m_1m_2}{r^2}
$$

Coulomb’s Law:

$$
F_E=k_e\frac{|q_1q_2|}{r^2}
$$

Both are inverse-square laws.

However, unlike gravity:

• Electric forces can attract or repel.

• Electric forces are much stronger than gravitational forces at atomic scales.

Principle of Superposition

Multiple Charges

When more than two charges are present, the net force is the vector sum of individual forces.

Mathematically:

$$
\vec{F}_{net}=\sum \vec{F}
$$

Each pairwise interaction is calculated separately.

Vector Addition

When forces act in different directions:

• Resolve into components.

• Add the (x)-components.

• Add the (y)-components.

Then calculate:

$$
F_{net}=\sqrt{F_x^2+F_y^2}
$$

Example 1

Two Positive Charges

Two charges are separated by:

$$
r=0.50,m
$$

with values:

$$
q_1=2.0\times10^{-6},C
$$

and

$$
q_2=3.0\times10^{-6},C
$$

Find the magnitude of the force.

Solution

Use Coulomb’s Law:

$$
F=k_e\frac{|q_1q_2|}{r^2}
$$

Substitute:

$$
F=(8.99\times10^9)\frac{(2.0\times10^{-6})(3.0\times10^{-6})}{(0.50)^2}
$$

$$
F=0.216,N
$$

Answer

$$
F=0.216,N
$$

The force is repulsive because both charges are positive.

Example 2

Positive and Negative Charges

Two charges are separated by:

$$
r=0.20,m
$$

with values:

$$
q_1=4.0\times10^{-6},C
$$

$$
q_2=-2.0\times10^{-6},C
$$

Find the force magnitude.

Solution

Use:

$$
F=k_e\frac{|q_1q_2|}{r^2}
$$

Substitute:

$$
F=(8.99\times10^9)\frac{(4.0\times10^{-6})(2.0\times10^{-6})}{(0.20)^2}
$$

$$
F=1.80,N
$$

Answer

$$
F=1.80,N
$$

The force is attractive because the charges have opposite signs.

Example 3

Distance Change

If the distance between two charges doubles:

$$
r\rightarrow2r
$$

then:

$$
F’=k_e\frac{|q_1q_2|}{(2r)^2}
$$

Therefore:

$$
F’=\frac{F}{4}
$$

Answer

Doubling the distance reduces the force to one-fourth its original value.

Common AP Exam Mistakes

Mistake 1

Ignoring the direction of the force.

Always determine whether the force is attractive or repulsive.

Mistake 2

Using charge signs incorrectly.

The magnitude equation uses:

$$
|q_1q_2|
$$

Determine the direction separately.

Mistake 3

Forgetting vector addition.

In multi-charge systems:

$$
\vec{F}_{net}=\sum\vec{F}
$$

Forces must be added as vectors.

AP Free-Response Strategy

Draw Force Diagrams

Before calculating:

• Sketch the charges.

• Identify attraction or repulsion.

• Draw force vectors.

This reduces sign errors.

Solve Pair by Pair

For multiple charges:

1. Calculate each individual force.

2. Determine directions.

3. Add vectors carefully.

Check Units

The final force should be reported in:

$$
N
$$

Always verify unit consistency.

Summary

Key Takeaways

• Coulomb’s Law gives the electric force between point charges:

$$
F=k_e\frac{|q_1q_2|}{r^2}
$$

• Coulomb’s constant is:

$$
k_e=8.99\times10^9;N\cdot m^2/C^2
$$

• Like charges repel and opposite charges attract.

• Electric forces obey the inverse-square law:

$$
F\propto\frac{1}{r^2}
$$

• Forces satisfy Newton’s Third Law:

$$
\vec{F}_{12}=-\vec{F}_{21}
$$

• Multiple-charge problems use superposition:

$$
\vec{F}_{net}=\sum\vec{F}
$$

• Coulomb’s Law serves as the foundation for understanding electric fields, electric potential, and much of electrostatics in AP Physics C E&M.