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Electrostatics · Coulomb's Law

Coulomb's Law and the Inverse Square Law

Place two charges, read the electrostatic force between them, and move one charge to different distances to discover how the force falls off with separation. Test the inverse square law, compare measured forces with Coulomb's Law, calculate percent error, and plot force against 1/r² to recover a straight line.

Theory — Coulomb's Law

The Electrostatic Force

Two electric charges exert a force on each other along the line joining them. Like charges (both positive or both negative) repel; opposite charges attract. The size of the force grows with the product of the charges and shrinks rapidly as the charges move apart.

Coulomb's Law F = k · |q₁ · q₂| / r²

F = electrostatic force (N)
q₁, q₂ = charges (C), r = separation (m)
k = 8.99 × 10⁹ N·m²/C²
Force is proportional to 1/r² — the inverse square law

The Inverse Square Law

The most striking feature of Coulomb's Law is the in the denominator. Because the force depends on 1/r², the way it changes with distance is dramatic: if you double the separation, the force drops to one quarter; triple it and the force drops to one ninth. This same inverse-square pattern appears in gravity and in the brightness of light.

Effect of Changing Distance r → 2r  gives  F → F/4
r → 3r  gives  F → F/9
r → r/2  gives  F → 4F
A plot of F against 1/r² is a straight line through the origin

Linearising the Data

A graph of force against distance is a curve, which is hard to read. But if you plot force against 1/r² instead, the inverse-square relationship becomes a straight line through the origin. The slope of that line equals k·|q₁·q₂| — a neat way to test the law and even recover the constant k.

Slope of the F vs 1/r² Graph F = (k·|q₁·q₂|) × (1/r²)
slope = k · |q₁ · q₂|
Straight line ⇒ the inverse square law holds

Percent Error

To judge how well the measured (simulated) force agrees with the value predicted by Coulomb's Law, compute the percent error.

Percent Error % Error = |(Theoretical − Simulated) / Theoretical| × 100

Bigger charges

Force grows with the product q₁·q₂ — double either charge and the force doubles.

Greater distance

Force falls as 1/r². Doubling r cuts the force to a quarter.

Like vs opposite

Like charges repel, opposite charges attract; the magnitude follows the same formula.

QuantityRelationshipNotes
Electrostatic forceF = k|q₁q₂|/r²k = 8.99×10⁹ N·m²/C²
Distance dependenceF ∝ 1/r²inverse square law
Linear plotF vs 1/r²slope = k|q₁q₂|
Percent error|(T−S)/T|×100compares sim to theory

Instructions — Running the Virtual Experiment

The Force tab lets you set the charges and move q₂ to read the force at each distance; the Inverse Square tab records your readings and plots force against 1/r². Record every reading in your lab report and include a labelled graph.

Part A — Force vs Distance (Force tab)
1
Open Simulation → Force. Set q₁ = +4 μC and q₂ = +8 μC (both positive, so the force is repulsive). q₁ sits at the 0 cm mark.
2
Move q₂ with the distance slider to each separation (for example 2, 4, 6, 8, 10 cm) and read the force from the readout. Convert each distance from cm to metres and record the force with the digits shown.
3
For each distance, use F = k|q₁q₂|/r² to compute the theoretical force, then the percent error between simulated and theoretical values.
Part B — Inverse Square Plot (Inverse Square tab)
1
Open Inverse Square. Click each distance to record its (1/r², force) point, then Plot & fit. The points should lie on a straight line through the origin.
2
Confirm the slope equals k·|q₁·q₂|, and check the inverse-square prediction: doubling the distance should cut the force to one quarter.

Simulation — Two Charges

Coulomb's Law Virtual LabSet the charges and move q₂ to read the force
q₁ (fixed at 0 cm)
q₂ (movable)
Like charges repel, unlike charges attract. k = 8.99×10⁹ N·m²/C².

Charges & distance

Force readout
q₁+4 μC
q₂+8 μC
Distance r0.02 m
1 / r²2500 m⁻²
Force F719 N
Repulsive force (both charges positive).
Each point is (1/r², force) at the current charges. Slope = k·|q₁q₂|.

Record a distance (q₁=+4 μC, q₂=+8 μC)

Fit result
Slope (k|q₁q₂|)— N·m²
Expected k|q₁q₂|0.2877 N·m²
r (m)1/r² (m⁻²)Force (N)
Click a distance to record it.

Team Questions

Question 1. Two charges, +4 μC and +8 μC, are 0.02 m apart. Calculate the force using F = k|q₁q₂|/r² with k = 8.99×10⁹. (Type just the number in N — e.g. 719)
Question 2. For the same charges at 0.04 m (double the distance), what is the force? (Type just the number in N)
Question 3. When the distance doubled from 0.02 m to 0.04 m, by what factor did the force change? (Type the factor — e.g. 0.25)
Question 4. When you plot force (y) against 1/r² (x), what shape is the graph? (One word — e.g. "line", "curve")
Question 5. If a simulated force reads 78.5 N where the theoretical value is 79.9 N, what is the percent error? Use |(T−S)/T|×100. (Type just the number — e.g. 1.8)
Question 6. The slope of the F-vs-1/r² line equals k·|q₁·q₂|. For +4 μC and +8 μC, what is this slope, in N·m²? (Type just the number — e.g. 0.288)
Question 7 — Challenge. The simulation is highly precise, yet you may not get exactly 0% error. Name one limiting factor that could cause a small error. (A few words)

Example Lab Report

Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission, and include a clearly labelled graph of force against 1/r².

Electrostatic Force and the Inverse Square Law

Physics | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To measure the electrostatic force between two positive charges as a function of their separation, to verify the inverse square law by comparing measured forces with values calculated from Coulomb's Law, to compute the percent error at each distance, and to confirm that a graph of force against 1/r² is a straight line through the origin.

Theory

Coulomb's Law gives the force between two point charges as F = k|q₁q₂|/r², with k = 8.99×10⁹ N·m²/C². Because F depends on 1/r², doubling the separation reduces the force to one quarter. Plotting F against 1/r² gives a straight line through the origin whose slope is k|q₁q₂|.

F = k|q₁q₂|/r² · slope of F vs 1/r² = k|q₁q₂|
% Error = |(T − S)/T| × 100

Calculations — Sample: r = 0.02 m

Theoretical force: F = (8.99×10⁹)(4×10⁻⁶)(8×10⁻⁶)/(0.02)² = (8.99×10⁹)(3.2×10⁻¹¹)/(4×10⁻⁴) = 719 N

Percent error (if simulated = 719 N): |(719 − 719)/719| × 100 = 0%

Results Table (q₁ = +4 μC, q₂ = +8 μC)

Trialr (m)1/r² (m⁻²)Simulated F (N)Theoretical F (N)% Error
10.0225007197190.0
20.046251801800.0
30.0627879.979.90.0
40.0815645.045.00.0
50.1010028.828.80.0

A plot of force against 1/r² was a straight line through the origin with slope ≈ 0.288 N·m² = k|q₁q₂|.

Discussion

The measured forces matched the values from Coulomb's Law, and the graph of force against 1/r² was a straight line through the origin, confirming that the force follows the inverse square law. The slope of that line, about 0.288 N·m², equalled k|q₁q₂| = (8.99×10⁹)(4×10⁻⁶)(8×10⁻⁶), providing an independent check of the law. Doubling the distance from 0.02 m to 0.04 m reduced the force from 719 N to 180 N — a factor of about one quarter — exactly as the inverse square law predicts.

The simulation is very precise, so the main limiting factors on accuracy are small ones: aligning the charge exactly on the ruler mark, and rounding in the simulation's force display. These can introduce a percent error of a fraction of a percent even when the physics is exact.

Conclusion

The electrostatic force obeyed Coulomb's Law and the inverse square law: force was proportional to 1/r², the F-vs-1/r² graph was linear through the origin with slope k|q₁q₂|, and doubling the distance quartered the force. Percent errors were negligible, limited only by ruler alignment and display rounding.

Practice Questions

Show all work and include units. Use k = 8.99×10⁹ N·m²/C².

Question 1
Two charges of +5 μC and +3 μC are 0.05 m apart. Calculate the electrostatic force between them and state whether it is attractive or repulsive.
Hint: F = k|q₁q₂|/r²; both positive ⇒ repulsive.
Question 2
A force of 90 N acts between two charges separated by 0.03 m. What force would act if the separation were increased to 0.09 m (three times as far)?
Hint: tripling r divides the force by 9.
Question 3
A student measures forces at five distances and plots F against 1/r², getting a straight line of slope 0.36 N·m². If one charge is +6 μC, what is the other charge?
Hint: slope = k|q₁q₂|, so |q₂| = slope/(k·q₁).
Question 4
A simulated force reads 44.2 N where Coulomb's Law predicts 45.0 N. Compute the percent error.
Hint: |(45.0 − 44.2)/45.0| × 100.
Question 5
Explain why plotting force against 1/r² (rather than against r) is useful when testing the inverse square law.
Hint: it turns a curve into a straight line, which is easy to test and whose slope has physical meaning.
Question 6 — Challenge
Two equal charges repel with a force of 0.50 N when 0.10 m apart. Find the size of each charge.
Hint: F = kq²/r², so q = √(F·r²/k).