Use Coulomb's law for point charges
Apply the inverse-square law to point charges and model the external field of a spherical conductor as a central point charge.
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 18. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use force, charge, work, energy, potential difference, inverse-square relationships and SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What happens to the Coulomb force magnitude if point-charge separation doubles?
What you will learn
- Model the external field of a spherical conductor as a point charge at its centre
- Use Coulomb's law for the force between charges in free space
- Determine whether point charges attract or repel
A spherical conductor acts like a point charge from outside
At a point outside a spherical conductor, its charge can be treated as if it were concentrated at the centre. This point-charge model applies to the external field; do not infer that the field inside a conductor follows the same inverse-square expression.
For two point charges separated by distance r in free space, Coulomb's law gives the force magnitude F = |Q₁Q₂|/(4πε₀r²), where ε₀ is the permittivity of free space. The force acts along the line joining the charges.
Signs set attraction or repulsion
Like signs repel and unlike signs attract. The equation with magnitudes gives a positive force magnitude; state the direction separately from the charge signs. Doubling separation makes the force one quarter as large if the charges remain fixed.
Use coulombs and metres with ε₀ = 8.85 × 10⁻¹² F m⁻¹, or use 1/(4πε₀) ≈ 8.99 × 10⁹ N m² C⁻². Keep powers of ten visible when multiplying microcoulomb or nanocoulomb charges.
Pause and explain
How do two negative point charges interact?
Worked example
Charges +2.0 μC and −3.0 μC are 0.30 m apart in free space. Find the force magnitude and state whether it is attractive.
Show the worked solution
- Use F = k|Q₁Q₂|/r² with k = 8.99 × 10⁹ N m² C⁻².
- Convert charges to coulombs and substitute.
- Opposite signs attract.
Answer F = 8.99 × 10⁹ × (2.0 × 10⁻⁶ × 3.0 × 10⁻⁶)/(0.30²) ≈ 0.60 N; the force is attractive.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Two charges +4.0 μC and +1.0 μC are separated by 0.20 m. Calculate the force magnitude and state its direction for each charge.
Give me a hint
Use the Coulomb constant form and remember that like charges repel.
Compare my reasoning
- F = k|Q₁Q₂|/r².
- F = 8.99 × 10⁹ × (4.0 × 10⁻⁶)(1.0 × 10⁻⁶)/(0.20²) ≈ 0.90 N.
- The forces have equal magnitude and opposite directions along the line; the charges repel.
Each charge experiences a 0.90 N repulsive force, in opposite directions.
Look for these in your work
- Inverse-square law used
- Coulomb conversion applied
- Repulsion and paired directions stated
A spherical conductor carries charge +6.0 nC. Explain how to calculate the field at an external point 0.40 m from its centre.
Give me a hint
Use the stated external point-charge model for a spherical conductor.
Compare my reasoning
- Treat the conductor's charge as concentrated at its centre for an external point.
- Use the point-charge field expression E = k|Q|/r².
- Because Q is positive, the field points radially outward.
Use E = k(6.0 × 10⁻⁹)/(0.40²), directed radially outward; this point-charge model is for the exterior.
Look for these in your work
- External model applied
- Correct field equation identified
- Direction and scope stated
The separation between two fixed point charges increases from r to 3r. By what factor does the force magnitude change?
Give me a hint
Compare 1/(3r)² with 1/r².
Compare my reasoning
- Initial force is proportional to 1/r².
- At 3r, the factor is 1/(3r)² = 1/(9r²).
- The new force is one ninth of the original.
The force magnitude becomes one ninth as large.
Look for these in your work
- Inverse-square dependence used
- Distance factor squared
- Force ratio stated
Common mistakes
- Forgetting to square the separation
- Using charge in microcoulombs without conversion
- Using charge signs to make a negative magnitude instead of stating direction
- Applying the spherical-conductor model to a point inside the conductor
What can you explain now?
What happens to the Coulomb force magnitude if point-charge separation doubles?
Compare with the explanation
It becomes one quarter
Force is proportional to 1/r².
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Make it stick
How do charge signs and separation determine the direction and magnitude of Coulomb force?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.