How does Coulomb’s law describe the force between two point charges?
Quick answer
For two point charges, Coulomb’s law gives the force magnitude as F = k|q₁q₂|/r². Larger charge magnitudes mean a stronger force; greater separation weakens it according to the square of the distance. The force acts along the line joining the charges: like signs repel, while opposite signs attract. This model is most directly applicable when charged objects are small compared with their separation, and it describes stationary charges.
The question, the details, the next step.
What does Coulomb’s law say, in plain language?
It describes the electric force between two point charges. Its magnitude is F = k|q₁q₂|/r²: multiply the magnitudes of the charges, then divide by the square of the distance between them, with k as the proportionality constant. The force acts along the line connecting the charges. Like signs push apart; opposite signs pull together. [S1]
The absolute-value bars give the magnitude, not the direction. Use the signs of the charges separately to decide whether the force is attractive or repulsive. The model is most directly suited to particles or objects small compared with their separation, and the cited explanation concerns stationary charges. [S1]
How can I tell what changing the distance does?
Distance is squared in the denominator. If the separation doubles while the charges stay the same, the force magnitude becomes one quarter as large, because the denominator becomes four times larger. If the separation is halved, the magnitude becomes four times as large, because the denominator becomes one quarter as large. These comparisons follow from the inverse-square relationship in the law. [S1]
That tells you about the magnitude only. Whether the charges attract or repel still depends on their signs.
Can you show a calculation and check the units?
Assume an idealized stationary pair of point charges: q₁ = +2.0 μC, q₂ = −3.0 μC, and r = 0.50 m. Keep k symbolic, since the supplied reference gives the formula but does not specify a numerical value for k.
Convert the charges first: 2.0 μC = 2.0 × 10⁻⁶ C and 3.0 μC = 3.0 × 10⁻⁶ C. Then:
F = k × |(2.0 × 10⁻⁶ C)(−3.0 × 10⁻⁶ C)| / (0.50 m)² = k × (6.0 × 10⁻¹² C² / 0.25 m²) = k × 2.4 × 10⁻¹¹ C²/m².
So the magnitude is 2.4 × 10⁻¹¹ k C²/m², with k still to be supplied in consistent units. For the result to be in newtons, k must have units N·m²/C². The signs are opposite, so the force is attractive. This is an idealized calculation, not a claim about an actual measurement.
What should I check before using this result?
Check that you used charge in coulombs and separation in metres if k is in SI units; keep the distance squared; and report magnitude separately from direction. Also consider whether the point-charge, stationary-charge model suits the situation. The reference does not establish how to calculate forces for arbitrary extended objects, moving charges, or other more complex situations. [S1]
A quick reusable setup is: write the charge values with signs, convert units, identify r, substitute magnitudes into the formula, check units, then use charge signs to state attraction or repulsion.
A practical checklist
Write each charge with its sign and convert its magnitude to coulombs.
Measure or identify the separation r and express it in metres for SI units.
Square r in the denominator and use absolute charge magnitudes to find force magnitude.
Check that k has compatible units so the result is in newtons.
Use the charge signs to label the interaction attractive or repulsive.
Confirm that the stationary point-charge model is appropriate for the situation.
A few more useful answers
Does the minus sign on a charge make the force magnitude negative?
No. The magnitude uses |q₁q₂| and is nonnegative; the charge signs determine whether the interaction attracts or repels.
If the distance triples, what happens to the force magnitude?
With the charges unchanged, the force becomes one ninth as large because the separation is squared in the denominator.
Can I apply this calculation directly to any charged object?
Not automatically. The point-charge model is most directly applicable when objects are small compared with their separation, and the cited treatment concerns stationary charges; it does not establish a general method for every situation. [S1]
Sources
References used for this guide.
- [S1] OpenStax Physics: Coulomb’s Law openstax.org · Checked