Electric charge and Coulomb's law - TCE Physics (Tasmania)

Syllabus dot point

“Apply Coulomb's law to the force between point charges and describe how objects become charged.”

TCEPhysicsUnit 3: Gravity and Electromagnetism9 min readUpdated 2 Jun 2026

Quick answer

There are two kinds of charge, positive and negative; like charges repel and unlike attract. The force between two point charges follows Coulomb's law $F = \dfrac{kQ_1Q_2}{r^2}$ , an inverse-square law just like gravity but able to push as well as pull.

What this dot point is asking

This dot point introduces electrostatics: where charge comes from, how charge is conserved, and how to calculate the force between charged objects. TCE markers expect you to substitute carefully in SI units and to keep direction separate from the numerical magnitude.

Two kinds of charge

Charge is a fundamental property of matter, carried by protons (positive) and electrons (negative). Charge is quantised in units of the elementary charge $e = 1.6 \times 10^{-19}\ \text{C}$ , and it is conserved, meaning it can be transferred but never created or destroyed. An object becomes negatively charged by gaining electrons and positively charged by losing them; the protons in the nucleus do not move.

Like charges repel and unlike charges attract. This single rule, combined with electron transfer, explains all electrostatic effects.

Charging methods

Objects can be charged in three ways:

Friction: rubbing two insulators transfers electrons from one to the other, as when a balloon is rubbed on hair.
Conduction: touching a charged object to a conductor shares charge between them, leaving both with the same sign.
Induction: bringing a charge near a conductor without touching it pushes the conductor's electrons to one side, separating charge; grounding then leaves a net charge of the opposite sign.

Conservation of charge

The total electric charge in an isolated system is constant. When objects are charged by friction or conduction, electrons move from one object to another, so any charge gained by one is exactly balanced by the charge lost by the other.

Coulomb's law

The electrostatic force between two point charges $Q_1$ and $Q_2$ separated by a distance $r$ is:

F = \frac{kQ_1Q_2}{r^2}

where the Coulomb constant $k = 8.99 \times 10^9\ \text{N m}^2\,\text{C}^{-2}$ . The force acts along the line joining the charges: repulsive for like charges, attractive for unlike. The structure is identical to Newton's law of gravitation, an inverse-square dependence on distance and a product of the two source quantities, but the electric force can be either attractive or repulsive and is vastly stronger.

Inverse-square scaling and ratios

Because $F \propto \dfrac{1}{r^2}$ , separation strongly controls the force. Halving the distance quadruples the force; tripling it cuts the force to one ninth. Many TCE questions are ratio problems best solved by writing $\dfrac{F_2}{F_1} = \left(\dfrac{r_1}{r_2}\right)^2$ so that $k$ and the unchanged charges cancel, avoiding repeated substitution. The same logic lets you predict how the force changes if one charge is doubled or if both spheres are touched together to share charge.

Worked example: force between two charges

Two point charges of $+3.0\ \mu\text{C}$ and $-5.0\ \mu\text{C}$ are $0.20\ \text{m}$ apart. Find the magnitude of the electrostatic force between them and state whether it is attractive or repulsive.

Step 1: Convert to SI units before doing anything else

Coulomb's law requires charge in coulombs and distance in metres. Micro-coulombs are a common pitfall: $1\ \mu\text{C} = 10^{-6}\ \text{C}$ . Converting immediately avoids errors later when the numbers are inside the formula:

Q_1 = 3.0 \times 10^{-6}\ \text{C}, \quad Q_2 = 5.0 \times 10^{-6}\ \text{C}, \quad r = 0.20\ \text{m}

Use the magnitudes of the charges here; the sign tells us direction (attractive or repulsive), which we state separately at the end.

Step 2: Substitute the magnitudes into Coulomb's law

The force between two point charges is inversely proportional to the square of the separation, so squaring $r$ before multiplying is essential. With $k = 8.99\times10^9\ \text{N m}^2\,\text{C}^{-2}$ :

F = \frac{kQ_1Q_2}{r^2} = \frac{(8.99\times10^9)(3.0\times10^{-6})(5.0\times10^{-6})}{(0.20)^2}

Step 3: Evaluate and state the direction

Carrying out the arithmetic:

F = \frac{(8.99\times10^9)(1.5\times10^{-11})}{0.040} = 3.4\ \text{N}

Because $Q_1$ is positive and $Q_2$ is negative, the charges are of opposite sign and attract one another.

Final answer: The electrostatic force is $3.4\ \text{N}$ , attractive.

Adding 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 every other charge (the principle of superposition). Calculate each pairwise force separately with magnitudes, draw its direction (toward an attracting charge, away from a repelling one), then add the forces as vectors, resolving into $x$ and $y$ components when they are not along a single line. For charges at the corners of a triangle, symmetry often makes one component cancel, which saves work.

The electric force dwarfs gravity

Between two protons the Coulomb repulsion is about $10^{36}$ times stronger than their gravitational attraction. This is why electromagnetic forces dominate atomic and chemical behaviour while gravity dominates only on astronomical scales where matter is electrically neutral.

Putting the sign of charge into the force equation

Students sometimes substitute negative charge values and get a negative force, then misread it. Use the magnitudes of the charges to find the size of the force, then decide direction separately: attraction for unlike charges, repulsion for like. The sign of $F$ from the formula is not a reliable direction indicator unless your axis convention is fully consistent.

In the exam, sketch the charges and the force directions first, then use magnitudes in Coulomb's law to find the size. For more than two charges, find each pairwise force as a vector and add them, since electrostatic forces superpose. Check that every charge is in coulombs and every distance in metres before pressing the calculator.

Exam-style questions

Questions in the style of TASC 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.

Practice2 marks

A calcium atom loses two of its electrons to form a doubly-charged positive ion with a charge of

+3.2 \times 10^{-19}\ \text{C}

. Calculate the magnitude of the electrostatic force that it exerts on an electron which is

4.0 \times 10^{-10}\ \text{m}

from it.

Show worked answer →

Use Coulomb's law $F = \dfrac{kq_1 q_2}{r^2}$ with $k = 8.99 \times 10^9\ \text{N m}^2\,\text{C}^{-2}$ .

Here $q_1 = 3.2 \times 10^{-19}\ \text{C}$ (the ion), $q_2 = 1.6 \times 10^{-19}\ \text{C}$ (the electron) and $r = 4.0 \times 10^{-10}\ \text{m}$ :

F = \frac{(8.99 \times 10^9)(3.2 \times 10^{-19})(1.6 \times 10^{-19})}{(4.0 \times 10^{-10})^2} = \frac{4.60 \times 10^{-28}}{1.6 \times 10^{-19}} = 2.88 \times 10^{-9}\ \text{N}.

The attractive force is about $2.9 \times 10^{-9}\ \text{N}$ . Markers want the correct charges substituted into Coulomb's law and $r$ squared, in coulombs and metres.

Practice5 marks

Three equal-sized charges A, B and C are placed near each other. A and B are positive while C is negative. Charge A exerts a force of

4.00\ \text{N}

on charge B. C exerts a force of equal magnitude on B from a perpendicular direction. Calculate the magnitude of the total force on charge B and find its direction relative to the line AB.

Show worked answer →

By Coulomb's law each pair of equal-magnitude charges at equal separation gives equal-magnitude forces, so C also exerts $4.00\ \text{N}$ on B. A (positive) repels B along AB; C (negative) attracts B toward C, and the two forces act at $90^\circ$ .

Combine the perpendicular forces as a vector sum:

F = \sqrt{(4.00)^2 + (4.00)^2} = \sqrt{32.0} = 5.66\ \text{N}.

Direction: $\theta = \tan^{-1}\!\left(\dfrac{4.00}{4.00}\right) = 45^\circ$ from the line AB, deflected toward C.

Markers reward recognising both forces are $4.00\ \text{N}$ from Coulomb's law, the perpendicular vector addition, and the $45^\circ$ resultant.

Practice3 marks

A physicist repeats Coulomb's experiment, measuring the force between two identically charged spheres at separations of

1.00

1.50

and

2.00\ \text{mm}

, recording forces of

9.92

4.41

and

2.48 \times 10^{-3}\ \text{N}

. Explain what should be plotted to obtain a straight-line graph and how the result confirms Coulomb's law.

Show worked answer →

Coulomb's law predicts $F = \dfrac{kq^2}{r^2}$ , so $F$ is proportional to $\dfrac{1}{r^2}$ . To get a straight line through the origin, plot $F$ (vertical) against $\dfrac{1}{r^2}$ (horizontal).

Check the data: at $r = 1.00\ \text{mm}$ , $\dfrac{1}{r^2} = 1.00 \times 10^6\ \text{m}^{-2}$ with $F = 9.92 \times 10^{-3}\ \text{N}$ ; at $r = 2.00\ \text{mm}$ , $\dfrac{1}{r^2} = 0.25 \times 10^6\ \text{m}^{-2}$ with $F = 2.48 \times 10^{-3}\ \text{N}$ . Doubling the distance gives one quarter of the force ( $9.92 / 4 = 2.48$ ), confirming the inverse-square relationship.

The straight line through the origin verifies $F \propto \dfrac{1}{r^2}$ ; its gradient equals $kq^2$ , so the charge follows from gradient $= kq^2$ . Markers want the variable $\dfrac{1}{r^2}$ and the inverse-square confirmation.

Practise this

Exam-style questions 3 on this page
Flashcards Physics deck

Sources & how we know this

Physics Level 4 (PHY415115) course document (2015 onwards) — TASC