Calculate the field of a point charge
Use the inverse-square field equation and infer direction from the sign of a source charge.
Section 18 · Electric fields
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
How does point-charge field magnitude depend on distance?
What you will learn
- Use E = Q/(4πε₀r²) for a point charge in free space
- Determine radial field direction from source-charge sign
- Connect point-charge field strength to force on a test charge
Field strength decreases with the square of distance
For a point charge Q in free space, the electric-field strength magnitude at distance r is E = |Q|/(4πε₀r²). The field is radial and has unit N C⁻¹. Doubling r reduces E to one quarter for a fixed source charge.
The signed source charge determines direction: field lines point outward for Q > 0 and inward for Q < 0. The magnitude expression uses |Q|; apply direction separately. A spherical conductor has the same point-charge form for points outside it.
A test charge feels a force in the field
Once E is known, a test charge q experiences F = qE. This separates the field created by source charge Q from the force on a chosen test charge. A negative test charge's force is opposite to the radial field direction.
Do not substitute the test charge q into the source-field equation unless q itself is the charge creating the field being evaluated. Identify which charge creates the field and where the field is measured.
Pause and explain
What is the field direction around a negative point source?
Worked example
Find the field magnitude 0.30 m from a +4.0 nC point charge. State its direction.
Show the worked solution
- Use E = k|Q|/r².
- Substitute 8.99 × 10⁹, 4.0 × 10⁻⁹ C and 0.30 m.
- A positive source charge produces an outward field.
Answer E = 8.99 × 10⁹ × 4.0 × 10⁻⁹/0.30² ≈ 4.0 × 10² N C⁻¹, radially outward.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
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.
Calculate E at 0.20 m from a +2.0 nC point charge. Use k = 8.99 × 10⁹ N m² C⁻².
Give me a hint
Use E = k|Q|/r² and convert nanocoulombs to coulombs.
Compare my reasoning
- Q = 2.0 × 10⁻⁹ C.
- E = (8.99 × 10⁹)(2.0 × 10⁻⁹)/(0.20²).
- Evaluate the magnitude and give the outward direction.
E ≈ 4.5 × 10² N C⁻¹, radially outward.
Look for these in your work
- Charge converted
- Distance squared used
- Magnitude and direction stated
At the same distance from two point charges, one +Q and one −Q, compare field magnitudes and directions.
Give me a hint
Use the magnitude equation and then apply the source signs.
Compare my reasoning
- The charges have equal magnitude, so |E| is equal at equal r.
- The +Q field points away from its source.
- The −Q field points towards its source.
Field magnitudes are equal; directions are radially outward for +Q and inward for −Q.
Look for these in your work
- Magnitude comparison correct
- Positive-source direction correct
- Negative-source direction correct
A +3.0 nC source charge creates a field of 600 N C⁻¹ at a point. Find the force on a −2.0 μC test charge there and give its direction relative to E.
Give me a hint
First use F = qE; the negative sign reverses the field direction.
Compare my reasoning
- |F| = |q|E = 2.0 × 10⁻⁶ × 600.
- |F| = 1.2 × 10⁻³ N.
- Because q is negative, the force is opposite to E.
The force magnitude is 1.2 × 10⁻³ N, opposite the field direction.
Look for these in your work
- Test charge used in force relation
- Magnitude calculated
- Negative-charge direction applied
Common mistakes
- Using r instead of r²
- Confusing the source charge with a test charge
- Reporting field magnitude without direction when direction is asked
- Using a negative field magnitude rather than a direction convention
What can you explain now?
How does point-charge field magnitude depend on distance?
Compare with the explanation
It is proportional to 1/r²
The field follows an inverse-square relationship.
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Make it stick
How do you keep the source charge and test charge roles separate in an electric-field calculation?
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.