Coulomb force between point charges
Key idea: H2 Physics lessons on point-charge interactions, uniform-field motion and capacitor energy.
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The core idea
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Big question: How do two point charges push or pull across empty space?
Coulomb's law gives F = |Qq|/(4πε₀r²) for point charges in vacuum. Like signs repel and unlike signs attract along the line joining the charges. For several sources, calculate each force vector on the chosen charge and add components.
Treat Coulomb's law as a vector interaction
For point charges in vacuum, force magnitude is F = |Qq|/(4πε₀r²). It acts along the line joining the charges: like signs repel and unlike signs attract. Use centre-to-centre separation and convert micro- or nanocoulombs before squaring distance.
The forces on the two charges are equal and opposite by Newton's third law even when their charge magnitudes differ. Coulomb's formula gives a magnitude; signs decide the arrow rather than being inserted carelessly into a scalar result.
Check your understanding: If charge Q doubles and separation triples, what is the force factor?
2/3² = 2/9.
Superpose forces on one chosen charge
With several source charges, draw each force on the same target charge, calculate magnitudes separately and add vectors. Collinear forces add algebraically after choosing a direction; non-collinear forces require components.
Do not add distances or source charges before applying Coulomb's law unless symmetry genuinely makes an equivalent source valid. The inverse-square relationship makes geometry essential.
Check your understanding: At a point, two equal forces act at right angles. What is the resultant magnitude?
√(F² + F²) = √2F, directed along the angle bisector.
Key ideas to keep
- Use centre-to-centre separation and SI charge units.
- The equation gives a magnitude; charge signs determine direction.
- Forces on the two charges are an equal and opposite interaction pair.
See the reasoning
Worked example
Recover a separation from an inverse-square force
Question: Two identical +5.0 nC charges repel with 2.25 × 10⁻⁵ N. Find their separation.
Step 1: Use the magnitude equation
Why: Charge signs determine direction, not the positive force magnitude.
Working: F = kQ²/r² for the two identical charges.
Step 2: Rearrange before substituting
Why: The unknown distance is squared.
Working: r = √(kQ²/F).
Step 3: Substitute SI charge
Why: Nanocoulombs must be converted to coulombs.
Working: r = √[(8.99×10⁹)(5.0×10⁻⁹)²/(2.25×10⁻⁵)] = 0.100 m.
Answer: From F = kQ²/r², r = √(kQ²/F) = √[(8.99 × 10⁹)(5.0 × 10⁻⁹)²/(2.25 × 10⁻⁵)] = 0.100 m. The positive sign determines repulsion, not the magnitude calculation.
Check: Identical positive charges repel, so the two force arrows point apart along their joining line.
Use a hint if needed
Practise with support
Try this
A separation triples while both point charges stay fixed. State the force factor.
Hint: Apply the inverse square to the separation factor, not just its reciprocal.
Check your answer
F ∝ 1/r², so the force becomes 1/3² = 1/9 of its original magnitude.
Now work without the hint
Practise independently
Your turn
Explain all conditions in the syllabus form of Coulomb’s law and solve for the force between +1.5 μC and +4.0 μC separated by 0.25 m.
Check your answer
The charges are treated as points in free space or air, with separation measured centre-to-centre. F = (8.99 × 10⁹)(1.5 × 10⁻⁶)(4.0 × 10⁻⁶)/(0.25)² = 0.863 N, repulsive.
Avoid these traps
Common mistakes
Common mistake
Coulomb force is inversely proportional to separation.
What is wrong with this reasoning?
Show better thinking
For point charges it is inversely proportional to separation squared.
Common mistake
The sign of the formula is enough to describe both force directions.
What is wrong with this reasoning?
Show better thinking
Calculate the magnitude, then use charge signs and the joining line to state attraction or repulsion and each vector direction.
Write for the examiner
Exam guidance
Sketch every force arrow on the chosen charge before resolving components.
Exam-style practice [5 marks]
Two point charges −2.0 nC and −8.0 nC are 0.60 m apart in air. Calculate and describe the force.
Plan before you answer
- Convert nC to C.
- Calculate the positive magnitude.
- Use the signs to state direction.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
F = (8.99 × 10⁹)(2.0 × 10⁻⁹)(8.0 × 10⁻⁹)/(0.60)² = 4.00 × 10⁻⁷ N. Like charges repel along their joining line.
Come back in three days
Check what stayed with you
Recall question
If one charge doubles and separation halves, state the Coulomb-force factor.
Check the answer
F ∝ Q₁Q₂/r², so doubling one charge gives ×2 and halving r gives ×4: the total factor is ×8.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 14 states no explicit exclusions. Coulomb-force, point-field and point-potential equations are used for point charges in free space or air. Electric field and force are vectors; potential and potential energy are scalars. E = ΔV/d is restricted to a uniform field, while E = −dV/dr is the local negative potential gradient.
- GCE A-Level H2 PhysicsTopic 14(a) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027