Coulomb's law and point charges: WACE Year 12 Physics Unit 3

Syllabus dot point

“Apply Coulomb's law to calculate the electrostatic force between point charges”

8 min readUpdated 2 Jun 2026

Quick answer

Two point charges exert equal and opposite forces on each other given by $F=\dfrac{kq_1 q_2}{r^2}$ , where $k=8.99\times10^9\ \text{N m}^2\text{C}^{-2}$ . Like charges repel and unlike charges attract, both along the line joining them. Like gravity it is an inverse-square law, but it can push as well as pull and is enormously stronger.

What this dot point is asking

WACE wants you to calculate electrostatic forces between point charges and compare the law structurally with Newton's law of gravitation. Treating charges as points (or uniform spheres) lets you use the centre-to-centre distance directly.

The law

F=\frac{kq_1 q_2}{r^2},\qquad k=\frac{1}{4\pi\varepsilon_0}=8.99\times10^9\ \text{N m}^2\text{C}^{-2}.

The magnitude depends on the product of the charges and the inverse square of their separation. The direction follows the sign rule: positive product means repulsion, negative product means attraction. The forces form a Newton's third law pair, equal in size and opposite in direction regardless of the relative size of the charges.

Comparing with gravitation

Coulomb's law and Newton's law of gravitation share the same inverse-square form, but with two key differences. Gravity is always attractive, whereas the electrostatic force can be attractive or repulsive depending on sign. The electrostatic force is also vastly stronger: between a proton and an electron it exceeds their gravitational attraction by roughly $10^{39}$ times, which is why gravity is irrelevant at the atomic scale.

Inverse-square scaling

Because $F\propto 1/r^2$ , separation strongly controls the force. Halving the distance quadruples the force; tripling it reduces the force to one ninth. Many questions are ratio problems, solved by writing $F_2/F_1$ so that $k$ and unchanged charges cancel.

Forces from several charges

When more than two charges are present, the net force on one charge is the vector sum of the individual Coulomb forces from each other charge (the principle of superposition). Calculate each pairwise force separately, then add them as vectors, resolving into components if they are not along the same line.

Net force on a charge from two others

Three charges in a line

Charge $q_A=+3.0\ \mu\text{C}$ sits $0.20\ \text{m}$ to the left of $q_B=+2.0\ \mu\text{C}$ , and $q_C=-4.0\ \mu\text{C}$ sits $0.10\ \text{m}$ to the right of $q_B$ . Find the net force on $q_B$ .

Step 1: Find the force from $q_A$ on $q_B$ .

Both charges are positive, so the force is repulsive. By Coulomb's law, $q_A$ pushes $q_B$ to the right, away from itself. The separation is $r_{AB}=0.20\ \text{m}$ .

F_{AB}=\frac{(8.99\times10^9)(3.0\times10^{-6})(2.0\times10^{-6})}{(0.20)^2}=1.35\ \text{N}.

Step 2: Find the force from $q_C$ on $q_B$ .

$q_C$ is negative and $q_B$ is positive, so the force is attractive: $q_C$ pulls $q_B$ to the right toward itself. The separation is $r_{CB}=0.10\ \text{m}$ .

F_{CB}=\frac{(8.99\times10^9)(2.0\times10^{-6})(4.0\times10^{-6})}{(0.10)^2}=7.19\ \text{N}.

Step 3: Add the forces as vectors.

Both $F_{AB}$ and $F_{CB}$ point to the right, so they reinforce each other. The net force is their sum, directed to the right.

F_{\text{net}}=1.35+7.19=8.5\ \text{N to the right.}

Charges off the line: resolving into components

When the charges do not lie on one straight line, the superposition is genuinely two-dimensional. The method is to compute each pairwise magnitude with $F=kq_1 q_2/r^2$ , draw the direction of each force as an arrow (toward an attracting charge, away from a repelling one), then resolve every force into $x$ and $y$ components. Add the components separately to get $F_x$ and $F_y$ , and recombine with $F=\sqrt{F_x^2+F_y^2}$ and $\theta=\tan^{-1}(F_y/F_x)$ . A common configuration is three charges at the corners of a triangle, where symmetry can make one component cancel and save work.

A worked feel for the numbers

To build intuition, two charges of $1\ \mu\text{C}$ each held $1\ \text{m}$ apart repel with $F=(8.99\times10^9)(10^{-6})(10^{-6})/(1)^2=8.99\times10^{-3}\ \text{N}$ , about the weight of a small paperclip. Bring them to $1\ \text{cm}$ apart and the inverse square multiplies this by $10^4$ to nearly $90\ \text{N}$ , the force needed to lift a $9\ \text{kg}$ mass. The steepness of the inverse-square law is why electrostatic forces feel negligible at arm's length yet become overwhelming at atomic separations.

Signs versus directions

Use the signs of the charges to decide attraction or repulsion, then assign directions on your diagram. Do not carry the negative sign of a charge straight into a vector sum; let the physics of attract or repel set the arrow, and treat magnitudes as positive when adding.

Forgetting to convert microcoulombs

Charges are usually quoted in microcoulombs ( $\mu\text{C}$ ) or nanocoulombs, and forgetting to convert to coulombs throws the answer off by huge powers of ten. Always write $q=3.0\ \mu\text{C}=3.0\times10^{-6}\ \text{C}$ before substituting, and check that $r$ is in metres, not centimetres.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style7 marks

Three point charges lie on a straight line. Charge

q_1=+5.0\ \mu\text{C}

is at the origin,

q_2=-3.0\ \mu\text{C}

is at

x=0.30\ \text{m}

, and

q_3=+2.0\ \mu\text{C}

is at

x=0.50\ \text{m}

. Calculate the magnitude and direction of the net electrostatic force on

q_3

Show worked answer →

A 7 mark calculation rewards two correct pairwise forces with directions and a valid vector sum.

Force from $q_1$ on $q_3$: Separation $r_{13}=0.50\ \text{m}$ , like signs so repulsive (pushes $q_3$ in the $+x$ direction):
$F_{13}=\frac{(8.99\times10^9)(5.0\times10^{-6})(2.0\times10^{-6})}{(0.50)^2}=0.36\ \text{N}.$
Force from $q_2$ on $q_3$: Separation $r_{23}=0.20\ \text{m}$ , unlike signs so attractive (pulls $q_3$ toward $q_2$ , in the $-x$ direction):
$F_{23}=\frac{(8.99\times10^9)(3.0\times10^{-6})(2.0\times10^{-6})}{(0.20)^2}=1.35\ \text{N}.$
Net force: Taking $+x$ as positive: $F_{\text{net}}=0.36-1.35=-0.99\ \text{N}$ , so $0.99\ \text{N}$ directed in the $-x$ direction (toward $q_2$ ).

Markers reward both magnitudes, correct attract/repel directions, the sign bookkeeping and the final $0.99\ \text{N}$ toward $q_2$ .

WACE 2023-style4 marks

Coulomb's law and Newton's law of universal gravitation have the same mathematical form. Explain two physical differences between the electrostatic and gravitational interactions, referring to the relevant equations.

Show worked answer →

A 4 mark explain answer needs two clearly distinct differences tied to the equations.

Direction of the interaction. Gravitation $F=\dfrac{Gm_1 m_2}{r^2}$ uses masses, which are always positive, so the force is always attractive. Coulomb's law $F=\dfrac{kq_1 q_2}{r^2}$ uses charges that can be positive or negative, so the product $q_1 q_2$ can give attraction (unlike charges) or repulsion (like charges).

Relative strength. The constant $k=8.99\times10^9$ is enormous compared with $G=6.67\times10^{-11}$ , so for fundamental particles the electrostatic force exceeds the gravitational force by around $10^{39}$ times. This is why gravity is negligible inside atoms while electrostatic forces dominate.

Markers reward the attractive-only versus attract-or-repel distinction and the vast difference in strength via the constants.

Practise this

Exam-style questions 2 on this page
Flashcards Physics deck

Sources & how we know this

Physics ATAR course: syllabus and support materials — SCSA
Physics: past ATAR course examinations — SCSA