Two charges of +5.0\,μC and -5.0\,μC are 0.10 m apart. Calculate the force between them (k = 8.9910^9). [2 points]

A charge carries -3.210^-19 C. How many excess electrons does it have? [1 point]

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How does Coulomb's law quantify the force between charges, and how do you combine these forces as vectors?

Topic 8.1 Electric Charge and Coulomb's Law: model the electrostatic force between point charges with Coulomb's law and add the forces from several charges as vectors.

A calculus-based answer to AP Physics C E&M Topic 8.1, covering electric charge, Coulomb's law for point charges, the inverse-square form, and combining Coulomb forces by superposition, with worked vector problems.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Electric charge
Coulomb's law
Superposition of forces
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What this topic is asking

The College Board (Topic 8.1) wants you to model the electrostatic force between point charges using Coulomb's law, recognize that it is an inverse-square law analogous to gravitation, and combine the forces from several charges by vector superposition. This is the foundation of all of electrostatics: every field, potential and flux result later in the course traces back to this force law.

Electric charge

Like charges repel and unlike charges attract. A neutral object has equal positive and negative charge; charging moves electrons, never creating or destroying net charge. The magnitude of any charge is an integer multiple of $e$ , so $q = ne$ for some integer $n$ .

Coulomb's law

The force between two stationary point charges $q_1$ and $q_2$ separated by a distance $r$ has magnitude

F = \frac{1}{4\pi\varepsilon_0}\frac{|q_1 q_2|}{r^2} = k\frac{|q_1 q_2|}{r^2}

where $\varepsilon_0 = 8.85\times10^{-12}$ C squared per N m squared is the permittivity of free space and $k = \dfrac{1}{4\pi\varepsilon_0} \approx 8.99\times10^9$ N m squared per C squared. In vector form, the force on charge $2$ due to charge $1$ is

\vec{F}_{12} = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}\,\hat{r}_{12}

where $\hat{r}_{12}$ points from $1$ toward $2$ . The sign of the product $q_1 q_2$ then sets the direction automatically: positive product gives repulsion (force away), negative product gives attraction (force toward).

Superposition of forces

The electric force obeys the principle of superposition: the net force on a charge from several others is the vector sum of the separate Coulomb forces, each computed as if the others were absent.

\vec{F}_{net} = \sum_i \vec{F}_i = \sum_i \frac{1}{4\pi\varepsilon_0}\frac{q\,q_i}{r_i^2}\,\hat{r}_i

For charges not on a single line, resolve each force into $x$ and $y$ components, sum the components, then recombine. The discipline is identical to adding any forces in mechanics: directions matter, so you cannot add magnitudes unless they happen to be collinear.

Net force from two charges at right angles

A charge $q = +2.0\,\mu\text{C}$ sits at the origin. A charge $q_1 = +3.0\,\mu\text{C}$ is at $(0.30\text{ m}, 0)$ and a charge $q_2 = +3.0\,\mu\text{C}$ is at $(0, 0.30\text{ m})$ . Find the net force on $q$ . Use $k = 8.99\times10^9$ N m squared per C squared.

step 1 Find each force magnitude

Both source charges are $0.30$ m away. $F_1 = k\dfrac{|q\,q_1|}{r^2} = (8.99\times10^9)\dfrac{(2.0\times10^{-6})(3.0\times10^{-6})}{(0.30)^2} = 0.599$ N. By symmetry $F_2 = 0.599$ N.

step 2 Assign directions from the signs

All charges are positive, so each force on $q$ is repulsive, pushing $q$ away from the source. $q_1$ is in $+x$ , so it pushes $q$ in $-x$ ; $q_2$ is in $+y$ , so it pushes $q$ in $-y$ .

step 3 Add as vectors

$\vec{F}_{net} = (-0.599)\hat{x} + (-0.599)\hat{y}$ N. Magnitude $F_{net} = \sqrt{0.599^2 + 0.599^2} = 0.847$ N.

step 4 Find the direction

The components are equal, so the resultant points at $45^\circ$ below the $-x$ axis, that is into the third quadrant, directly away from the corner where the two source charges sit.

Try this

Q1. Two charges of $+5.0\,\mu\text{C}$ and $-5.0\,\mu\text{C}$ are $0.10$ m apart. Calculate the force between them ( $k = 8.99\times10^9$ ). [2 points]

Cue. $F = k\dfrac{|q_1 q_2|}{r^2} = (8.99\times10^9)\dfrac{(5.0\times10^{-6})^2}{(0.10)^2} = 22.5$ N, attractive.

Q2. A charge carries $-3.2\times10^{-19}$ C. How many excess electrons does it have? [1 point]

Cue. $n = \dfrac{|q|}{e} = \dfrac{3.2\times10^{-19}}{1.60\times10^{-19}} = 2$ electrons.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)1 marksSection I (multiple choice). Two point charges separated by a distance

r

exert a force of magnitude

F

on each other. If both charges are doubled and the separation is also doubled, the new force is (A)

F

(B)

2F

(C)

4F

(D)

F/2

. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the structure of Coulomb's law. The answer is (A).

Coulomb's law is $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}$ . Doubling each charge multiplies $q_1 q_2$ by $4$ ; doubling $r$ multiplies $r^2$ by $4$ . The factors cancel, $\dfrac{4}{4} = 1$ , so the force is unchanged at $F$ . The trap is treating the $r$ dependence as linear instead of inverse-square.

AP 2024 (style)5 marksSection II (FRQ, quantitative). Three point charges lie on the x-axis:

q_1 = +3.0\,\mu\text{C}

at the origin,

q_2 = -2.0\,\mu\text{C}

x = 0.20

m, and

q_3 = +4.0\,\mu\text{C}

x = 0.40

m. (a) Calculate the magnitude and direction of the net force on

q_2

. (b) Explain why the two contributing forces must be added as vectors. Use

k = 8.99 \times 10^9

N m squared per C squared.

Show worked answer →

A 5-point FRQ on Coulomb's law with superposition.

(a) Force from $q_1$ on $q_2$ (2 points): separation $0.20$ m, opposite signs so attractive (toward $q_1$ , in the $-x$ direction). $F_{12} = k\dfrac{|q_1 q_2|}{r^2} = (8.99\times10^9)\dfrac{(3.0\times10^{-6})(2.0\times10^{-6})}{(0.20)^2} = 1.35$ N in $-x$ .
Force from $q_3$ on $q_2$ (1 point): separation $0.20$ m, opposite signs so attractive (toward $q_3$ , in the $+x$ direction). $F_{32} = (8.99\times10^9)\dfrac{(4.0\times10^{-6})(2.0\times10^{-6})}{(0.20)^2} = 1.80$ N in $+x$ .
Net (1 point): $F_{net} = 1.80 - 1.35 = 0.45$ N in the $+x$ direction (toward $q_3$ ).
(b) Vector reasoning (1 point): each Coulomb force has a direction set by the line joining the charges and the signs; forces in different directions cannot be added as plain numbers, so you add components.

Markers reward correct magnitudes, correct attractive directions from the signs, and a vector sum.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)