Charge and Electric Force

AP Physics C: Electricity and Magnetism· AP Physics C: E&M CED — Electrostatics· 14 min read

1. Fundamental Properties of Electric Charge★☆☆☆☆⏱ 3 min

Charge is the fundamental property of matter that causes electrostatic interactions, one of the four fundamental forces of nature. Charge can be positive (deficit of electrons) or negative (excess of electrons), with SI unit the coulomb ( $C$ ). Electric force can be attractive (opposite charges) or repulsive (like charges), unlike gravity which is only attractive.

📘 Definition

Quantization of Charge

q = ne

All free net charge in nature exists as an integer multiple of the elementary charge $e$ , where $e = 1.602 \times 1 0^{- 19} C$ . Electrons have charge $- e$ , protons have charge $+ e$ .

Example:

Where $n$ is any positive or negative integer, representing net excess or missing electrons.

📘 Definition

Conservation of Charge

Total net charge in an isolated system never changes. Charge can transfer between objects, but cannot be created or destroyed.

Example:

When identical conducting spheres touch, total charge splits equally between them after separation.

📐 Worked Example

Three identical conducting spheres initially have charges $q_{A} = + 6 μ C$ , $q_{B} = - 8 μ C$ , and $q_{C} = 0 μ C$ . First A touches B, then they separate. Next A touches C, then separate. What is the final charge on C, and how many electrons are transferred to C?

1
When identical conductors touch, total charge is conserved and splits equally. Total charge after A and B touch:
2
$q_{A,B,\text{total}} = +6\ \mu\text{C} - 8\ \mu\text{C} = -2\ \mu\text{C}$
3
After separation, $q_{A} = q_{B} = - 1 μ C$ .
4
Next, A ( $- 1 μ C$ ) touches neutral C. Total charge is $- 1 μ C$ , which splits equally:
5
$q_C = -0.5\ \mu\text{C} = -5 \times 10^{-7}\ \text{C}$
6
Number of electrons transferred is $n = ∣Δ q_{C} ∣/ e$ :
7
$n = \frac{5 \times 10^{-7}\ \text{C}}{1.602 \times 10^{-19}\ \text{C/electron}} \approx 3.1 \times 10^{12}\ \text{electrons}$

Exam tip:

Only assume equal charge division after contact for identical conducting spheres. If spheres are non-identical or insulating, never assume equal division without explicit problem statement.

2. Coulomb's Law★★☆☆☆⏱ 4 min

Coulomb's Law describes the electrostatic force between two stationary point charges. A point charge is any charged object much smaller than the distance between it and other charges, so it can be treated as a single point in space.

📘 Definition

Coulomb's Law

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

Gives the force between two stationary point charges. Magnitude depends on the product of charges and inverse square of distance between them.

Example:

Where $k \approx 9 \times 1 0^{9} N \cdotp m^{2} / C^{2}$ for most AP calculations.

\vec{F}_{12} = k \frac{q_1 q_2}{r_{12}^2} \hat{r}_{12}

Where $\overset{r}{^}_{12}$ is the unit vector pointing from charge 1 to charge 2. Direction is attractive for opposite charges, repulsive for like charges.

📐 Worked Example

Two point charges are placed on the x-axis: $q_{1} = + 2 nC$ at $x = 0$ , and $q_{2} = - 8 nC$ at $x = 3 m$ . What is the magnitude and direction of the force that $q_{1}$ exerts on $q_{2}$ ?

1
Convert all values to SI units: $q_{1} = 2 \times 1 0^{- 9} C$ , $q_{2} = - 8 \times 1 0^{- 9} C$ , $r = 3 m$ .
2
Calculate magnitude using Coulomb's Law:
3
$F = k \frac{|q_1 q_2|}{r^2} = \frac{(9 \times 10^9)(2 \times 10^{-9})(8 \times 10^{-9})}{3^2} = 1.6 \times 10^{-8}\ \text{N}$
4
$q_{1}$ and $q_{2}$ have opposite signs, so force is attractive. $q_{1}$ pulls $q_{2}$ toward itself, in the negative x-direction.
5
Vector form confirms direction: $\overset{r}{^}_{12}$ points in +x direction, $q_{1} q_{2}$ is negative so $F_{12}$ points opposite to $\overset{r}{^}_{12}$ .

Exam tip:

Always use absolute values to calculate force magnitude, then assign direction separately based on charge signs. This avoids sign errors from unit vector misinterpretation.

3. Superposition of Electric Forces★★☆☆☆⏱ 3 min

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. Pairwise interactions are independent of the presence of other charges, so we calculate each force separately with Coulomb's Law then add them as vectors.

\vec{F}_{\text{total}} = \sum_{i=1}^n \vec{F}_{i0}

The most common student error is adding force magnitudes directly instead of breaking into x and y components first. Always use vector component addition for forces acting at different angles.

📐 Worked Example

Three charges are at the vertices of a right triangle: $q_{1} = + 1 μ C$ at $(0, 0)$ , $q_{2} = + 1 μ C$ at $(3 m, 0)$ , $q_{3} = + 2 μ C$ at $(0, 4 m)$ . What is the total force on $q_{1}$ ?

1
Calculate force from $q_{2}$ on $q_{1}$ : repulsive, pushes $q_{1}$ in negative x-direction. Magnitude:
2
$F_{21} = \frac{k q_1 q_2}{r_{21}^2} = \frac{(9 \times 10^9)(1 \times 10^{-6})(1 \times 10^{-6})}{3^2} = 1 \times 10^{-3}\ \text{N}$
3
Components: $F_{21} = (- 1 \times 1 0^{- 3} N) \hat{i} + 0 \hat{j}$
4
Calculate force from $q_{3}$ on $q_{1}$ : repulsive, pushes $q_{1}$ in negative y-direction. Magnitude:
5
$F_{31} = \frac{(9 \times 10^9)(1 \times 10^{-6})(2 \times 10^{-6})}{4^2} = 1.125 \times 10^{-3}\ \text{N}$
6
Components: $F_{31} = 0 \hat{i} + (- 1.125 \times 1 0^{- 3} N) \hat{j}$
7
Add components, then calculate total magnitude:
8
$|\vec{F}_{\text{total}}| = \sqrt{(-1 \times 10^{-3})^2 + (-1.125 \times 10^{-3})^2} \approx 1.5 \times 10^{-3}\ \text{N}$

Exam tip:

Always draw a free-body diagram for the charge of interest before adding forces, to confirm force directions before calculating components.

4. Equilibrium of Charged Systems★★★☆☆⏱ 4 min

Equilibrium occurs when the net force on a charge is zero. A common AP exam problem asks you to find the position where a third charge experiences zero net force from two fixed charges, and classify the equilibrium as stable or unstable.

📐 Worked Example

Two identical positive point charges $+ Q$ are fixed at $x = - a$ and $x = + a$ . A third charge $+ q$ is placed at the origin, free to move along the x-axis. What is the net force on $+ q$ , and is the equilibrium stable or unstable?

1
By superposition: left $+ Q$ exerts a repulsive force of magnitude $k Qq / a^{2}$ pushing $+ q$ right. Right $+ Q$ exerts an equal force pushing $+ q$ left. Forces cancel, so net force is zero.
2
To check stability: displace $+ q$ a small distance right. The force from the right $+ Q$ becomes larger than the force from the left $+ Q$ , pushing $+ q$ further right.
3
Any displacement away from the origin results in a net force pushing further away, so equilibrium is unstable.

5. Common Pitfalls

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. Quick Reference Cheatsheet

Category	Formula	Notes
Elementary Charge	$e = 1.602 \times 1 0^{- 19} C$	Electrons: $q = - e$ , Protons: $q = + e$
Charge Quantization	$q = n e$	All free net charge is integer multiple of $e$ ; $n$ is positive/negative integer
Conservation of Charge	$Q_{total, initial} = Q_{total, final}$	Holds for all isolated systems; charge transfer does not change total net charge
Coulomb's Law (Magnitude)	$F = k \frac{∣ q _{1} q _{2} ∣}{r ^{2}}$	For stationary point charges; $k \approx 9 \times 1 0^{9} N \cdotp m^{2} / C^{2}$
Coulomb's Law (Vector)	$F_{12} = k \frac{q _{1} q _{2}}{r _{12}^{2}} \overset{r}{^}_{12}$	$\overset{r}{^}_{12}$ points from $q_{1}$ to $q_{2}$
Superposition Principle	$F_{total} = \sum_{i = 1}^{n} F_{i 0}$	Total force is vector sum of individual pairwise forces
Equal Charge Division	$q_{1} = q_{2} = \frac{q _{1 i} + q _{2 i}}{2}$	Only applies to identical conducting spheres after contact

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Charge conservation identical spheres
2022 · FRQ
Equilibrium of three point charges

Going deeper

Official SyllabusAP Physics C: E&M Course and Exam DescriptionOfficial College Board syllabus reference

What's Next

Charge and electric force is the foundation of all electrostatics, which makes up 25-30% of your total AP Physics C: E&M exam score. The concepts you learned here directly extend to electric fields, where we replace the force on a test charge with the field per unit charge, and all the same rules of quantization, conservation, and superposition apply. Next, you will use Coulomb's Law and superposition to calculate electric fields from continuous charge distributions, a core skill for free-response questions. Understanding vector addition of forces will also prepare you to solve Gauss's Law problems later in the unit.