What force acts between electric charges?
Apply Coulomb's law to the force between point charges and describe how objects become charged.
Positive and negative charge, charging by friction, conduction and induction, conservation of charge, and Coulomb's law for the force between point charges.
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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.
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 , 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.
Coulomb's law
The electrostatic force between two point charges and separated by a distance is:
where the Coulomb constant . 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.
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 forces superpose.
Exam-style practice questions
Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2024 TASC2 marksA calcium atom loses two of its electrons to form a doubly-charged positive ion with a charge of +3.2 x 10^-19 C. Calculate the magnitude of the electrostatic force that it exerts on an electron which is 4.0 x 10^-10 m from it.Show worked answer →
Use Coulomb's law: F = k q1 q2 / r^2, with k = 8.99 x 10^9 N m^2 C-2.
q1 = 3.2 x 10^-19 C (ion), q2 = 1.6 x 10^-19 C (electron), r = 4.0 x 10^-10 m.
F = 8.99 x 10^9 x (3.2 x 10^-19)(1.6 x 10^-19) / (4.0 x 10^-10)^2
= 8.99 x 10^9 x 5.12 x 10^-38 / 1.6 x 10^-19
= 4.60 x 10^-28 / 1.6 x 10^-19
= 2.88 x 10^-9 N.
The attractive force is about 2.9 x 10^-9 N. Markers want the correct charges substituted into Coulomb's law and r squared.
2019 TASC5 marksThree 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 N on charge B. Given C exerts a force of equal magnitude on B from a different direction, calculate the magnitude of the total force on charge B and find its direction relative to the line AB (charges arranged so the two forces act at 90 degrees).Show worked answer →
By Coulomb's law each pair of equal-magnitude charges at equal separation gives equal-magnitude forces, so C exerts 4.00 N on B as well. A (positive) repels B along AB; C (negative) attracts B toward C.
With the two forces perpendicular, combine them as a vector sum:
total = sqrt(4.00^2 + 4.00^2) = sqrt(32.0) = 5.66 N.
Direction: theta = arctan(4.00 / 4.00) = 45 degrees from the line AB, deflected toward C.
Markers reward recognising both forces are 4.00 N from Coulomb's law, the perpendicular vector addition, and the 45 degree resultant direction.
2022 TASC3 marksA physicist repeats Coulomb's experiment, measuring the force between two identically charged spheres at separations of 1.00, 1.50 and 2.00 mm, recording forces of 9.92, 4.41 and 2.48 x 10^-3 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 = k q^2 / r^2, so F is proportional to 1 / r^2. To get a straight line through the origin, plot F (vertical) against 1 / r^2 (horizontal).
Check the data: at r = 1.00 mm, 1/r^2 = 1.00 x 10^6 m-2 and F = 9.92 x 10^-3 N; at r = 2.00 mm, 1/r^2 = 0.25 x 10^6 m-2 and F = 2.48 x 10^-3 N. Halving the distance quarters nothing but doubling the distance to 2 mm gives one quarter of 1/r^2 and one quarter of the force (9.92 / 4 = 2.48), confirming the inverse-square relationship.
The straight line through the origin verifies F is proportional to 1 / r^2. Its gradient equals k q^2, so the charge can be found from gradient = k q^2. Markers want the variable 1 / r^2 and the confirmation that F drops as the square of distance.