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

A focused answer to the WACE Year 12 Physics Unit 3 content point on Coulomb's law. The inverse-square electrostatic force, attraction and repulsion of point charges, comparison with gravitation, and adding forces from several charges as vectors.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

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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 $q_B$ .

Force from $q_A$ on $q_B$ (both positive, so repulsive, pushing $q_B$ to the right):

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

Force from $q_C$ on $q_B$ (opposite signs, so attractive, pulling $q_B$ to the right toward $q_C$ ):

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

Both point right, so they add:

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

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.

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