Mini Physics
Part of Mini EducationSwitch subject
Search

Coulomb's Law

Key idea: Use Coulomb’s law to calculate the electric force between point charges, including direction (attraction/repulsion) and inverse-square scaling (A Level Physics).

  • GCE A-Level H2 Physics 2027
On this page

Learning objectives

  • Apply Coulomb's law to the force between point charges.

1. Definitions (Must Know)

A. Coulomb’s law (magnitude)

For two point charges Q₁ and Q₂ separated by a distance r in free space / air, the magnitude of the electrostatic force between them is:

F = (1/4πε₀)|Q₁Q₂|/r²

where ε₀ is the permittivity of free space.

B. Constants (often given)

  • Permittivity of free space, ε₀ ≈ 8.85 × 10⁻¹² F m⁻¹.
  • Coulomb constant, k = 1/4πε₀ ≈ 8.99 × 10⁹ N m² C⁻².

C. Direction: attraction vs repulsion

  • If Q₁Q₂ > 0 (same sign), the force is repulsive.
  • If Q₁Q₂ < 0 (opposite signs), the force is attractive.

The forces act along the line joining the charges, and the forces on the two charges are equal in magnitude and opposite in direction.

2. Key Ideas (What Earns Marks)

  • F ∝ |Q₁Q₂| (double one charge ⇒ double F).
  • F ∝ 1/r² (double r ⇒ F becomes 1/4).
  • Use magnitude + direction. Don’t say “F is negative” unless you have defined a sign convention.
  • You can write Coulomb’s law as F = k|Q₁Q₂|/r² where k = 1/4πε₀.
High-frequency trap

r is the separation between the two charges. Convert cm and mm to metres before substitution.

Exam pitfall: scalar answer with missing direction

Coulomb’s law gives magnitude; exam answers still need direction (attractive or repulsive, and axis direction if relevant). Treat force as a vector quantity.

3. Detailed Explanations

A. What “inverse-square” means (ratio method)

If Q₁ and Q₂ stay the same:

F ∝ 1/r²

So:

F₂/F₁ = (r₁/r₂)²

This is the fastest way to answer “how does the force change?” questions.

B. How to handle direction (vector thinking without heavy notation)

  1. Decide which charge you are finding the force on.
  2. Draw the other charge and decide attraction/repulsion from the signs.
  3. The force on your chosen charge points:
    • away from the other charge (repulsion),
    • towards the other charge (attraction).

If you are working on a line (e.g. x-axis), choose a positive direction and assign signs at the very end.

The electric field strength due to a point charge Q is:

E = (1/4πε₀)Q/r²

and the force on a charge q in an electric field is:

F = qE

Continue with Electric Field Strength of a Point Charge for field direction and vector superposition.

4. Common Mistakes

  • Writing F as negative/positive without defining a direction first.
  • Forgetting the square on r.
  • Using r in cm or mm instead of metres.
  • Using Coulomb’s law for non-point charges without justification (use it when charges can be modelled as point charges).

5. Exam Tips

  • Write the equation before substitution, then show units, then final answer.
  • For “compare” questions, use the ratio method instead of full calculation.
  • State direction using words (“towards”, “away”, “to the left/right”) based on your diagram/sign convention.

6. Worked Examples

Modelled example 1

Force between two charges (magnitude)

Core

Problem

Charges Q₁ = +3.0 μC and Q₂ = -5.0 μC are separated by 0.20 m. Find the force magnitude.
Study the worked solution

  1. Use magnitudes in Coulomb's law

    Method

    Convert microcoulombs and square the separation.

    Reason

    The scalar formula gives force magnitude; charge signs determine direction separately.

    Working

    F = (8.99 × 10⁹)((3.0 × 10⁻⁶)(5.0 × 10⁻⁶))/(0.20)²

  2. Evaluate

    Method

    Obtain approximately 3.4 N.

    Reason

    All substituted quantities are in SI units.

    Working

    F = 3.37 N ≈ 3.4 N

Guided practice 2

Direction of the force

About 3 min

Problem

For Q₁ > 0 and Q₂ < 0, state the direction of the force on Q₁.

Try this before viewing the solution

Force on Q₁

Hints

Hint 1: compare signs
Unlike charges attract; like charges repel.
View solution step by step

  1. Classify the interaction

    Method

    Identify attraction.

    Reason

    The charges have opposite signs.

    Working

    Q₁Q₂ < 0: attractive pair.

  2. State the vector direction

    Method

    Point the force on Q₁ towards Q₂.

    Reason

    Coulomb force lies along the line joining the charges.

    Working

    vector F₁ is directed from Q₁ towards Q₂.

Common misconception 3

Inverse-square scaling

Find and correct the mistake

Learner claim

The separation is tripled while both charges stay fixed. A learner says force becomes one-third because distance is in the denominator. Diagnose the claim.

Try this before viewing the solution

New force factor

View solution step by step

  1. Form a ratio

    Method

    Apply the inverse square to both separations.

    Reason

    With charges fixed, only r⁻² changes.

    Working

    F₂/F₁ = (r₁/r₂)²

  2. Square the factor

    Method

    Obtain one-ninth.

    Reason

    r₂ = 3r₁.

    Working

    F₂/F₁ = (1/3)² = 1/9

Examiner practice 4

Find separation for a required force

3 marks

Examination question

Two + 2.0 μC charges repel with force 0.50 N. Find their separation. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Rearrange

    1 mark

    Method

    Make r the subject and take a square root.

    Reason

    Separation appears as r² in the denominator.

    Working

    r = square root of (k|Q₁Q₂|/F)

  2. Convert and substitute

    1 mark

    Method

    Use charges in coulombs.

    Reason

    The Coulomb constant is in SI units.

    Working

    r = square root of (((8.99 × 10⁹)(2.0 × 10⁻⁶)²)/0.50)

  3. Evaluate

    1 mark

    Method

    Obtain 0.268 m.

    Reason

    Use the positive separation magnitude.

    Working

    r = 0.268 m.

Challenge 5

Net force on a charge on a line

Minimal support

Independent transfer

Charges + 4.0 μC and + 1.0 μC are at x = 0 and 0.30 m. Find the net force on + 2.0 μC at x = 0.10 m.

Try this before viewing the solution

Hints

Hint 1: draw both force arrows
The left charge repels right; the right charge repels left. Use their different separations before subtracting.
View solution step by step

  1. Resolve the geometry

    Method

    Use separations 0.10 m and 0.20 m.

    Reason

    Force depends on source-to-test distance, not coordinate alone.

    Working

    r_L = 0.10 m; r_R = 0.20 m.

  2. Calculate opposing forces

    Method

    Find each magnitude separately.

    Reason

    Both interactions are repulsive but point oppositely.

    Working

    F_L = 7.19 N right, F_R = 0.449 N left

  3. Superpose vectors

    Method

    Subtract magnitudes and retain the stronger direction.

    Reason

    The forces are collinear and opposite.

    Working

    Fₙₑₜ = 7.19-0.449 = 6.74 N right

7. Mind Stretchers

Mind stretcher 1: Symmetry cancelationExtension

Three charges lie on a straight line: + Q at x = 0, + Q at x = 2d, and -Q at x = d. Find the direction of the net force on the middle charge (-Q).

Show Answer

The middle charge is negative, so it is attracted to both positive charges.

  • The + Q at x = 0 attracts -Q to the left.
  • The + Q at x = 2d attracts -Q to the right.

The distances are equal (d), so the forces are equal in magnitude and opposite. Net force is zero.

Mind stretcher 2: Zero-force point for two like chargesExtension

Two identical positive charges are fixed. You place a third positive charge at a point where the net force on it is zero. Explain what you can deduce about that point.

Show Answer

For the net force to be zero, the electric field contributions from the two fixed charges must cancel (equal magnitude, opposite directions). For two identical positive charges on a line, this happens at the midpoint between them.

Mind stretcher 3: Optional (Enrichment)Extension

A. Vector form (for compact working)

Sometimes Coulomb’s law is written in a direction-aware form using a unit vector along the line joining the charges. This is not required for most A Level questions, but it can make multi-charge problems shorter.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027