Coulomb's Law
Two point charges with adjustable signs and magnitudes — the electrostatic force vector follows F = kq₁q₂/r² and flips direction when charge signs differ.
Objective
Verify that the electrostatic force between two point charges obeys Coulomb's inverse-square law: F = k·|q₁|·|q₂|/r². Set the charge magnitudes, signs, and separation, and read the static Force F before launching. Then press Start to release the charges and watch them accelerate under their mutual force — opposite signs draw together until their edges touch, like signs fly apart to the frame edge — while a marker rides the force-versus-separation curve at the current r and the velocity-versus-separation r graph fills in below. The simulation treats both charges as ideal point charges in vacuum with no dielectric medium.
Setup
- Set q₁ to +5 μC and q₂ to −5 μC (defaults) and separation r to 2.0 m. Read the static Force F before launching — it should sit near 0.0562 N. Press Start: the opposite-sign charges accelerate toward each other and stop when their edges touch (attractive configuration).
- Press Reset. Change q₂ to +5 μC so both charges are positive. Note the static Force F is unchanged because |q₁|·|q₂| is the same. Press Start: the like-sign pair now accelerates apart and the charges stop at the frame edge (repulsive configuration).
- Press Reset. Keep q₁ = +5 μC and q₂ = +5 μC and set r to 1.0 m. Read the static Force F — it should be near 0.225 N, roughly four times the r = 2.0 m value, confirming the inverse-square law before any motion begins.
- Press Reset. Set r back to 2.0 m and change q₁ to +10 μC. The static Force F should read about 0.112 N, roughly double the q₁ = +5 μC value, confirming F ∝ |q₁|. Press Start to watch the heavier-force pair separate faster on the velocity graph.
Analytical Prediction
Coulomb's law gives F = k·|q₁|·|q₂|/r² with k = 8.99 × 10⁹ N·m²/C². With |q₁| = |q₂| = 5 × 10⁻⁶ C and r = 2.0 m:
At r = 1.0 m the inverse-square law predicts:
Doubling q₁ to 10 μC at r = 2.0 m:
Sign of q₁·q₂ determines direction only — magnitude is the same for opposite-sign and same-sign pairs of equal absolute value.
Results Analysis
With defaults (q₁ = +5 μC, q₂ = −5 μC, r = 2.0 m) the static Force F reads near 0.056 N and the arrows point inward (attractive). Switching both charges to +5 μC keeps F ≈ 0.056 N but flips the arrows outward. Setting r = 1.0 m with equal charges raises the static F to ≈ 0.225 N — a factor of 4 rise consistent with (2.0/1.0)² = 4. Pressing Start releases the charges along the reference axis: the attractive pair accelerates inward and halts when its edges touch, while a like-sign pair accelerates outward and halts at the frame edge. The right panel stacks two graphs with separation r on the x-axis. The top graph is the analytical force-versus-separation curve with a marker that rides the curve at the current separation r, climbing toward smaller r as the gap closes (or sliding outward as the charges separate). The bottom graph plots each particle's velocity versus separation r as two mirrored lines, one per charge, showing the equal-and-opposite accelerations demanded by Newton's third law.
Source of Error
Both charges are modelled as ideal point charges — no finite size, no charge distribution, no self-energy, and no relativistic corrections. The medium is assumed to be vacuum (permittivity ε₀ only); any real dielectric would reduce the force by a factor of εᵣ. No gravitational, magnetic, or inductive effects are included. The analytical prediction in the Prediction section shares all of these idealizations, so they cancel in the comparison. The r slider minimum of 0.5 m avoids the unphysical divergence that would occur for macroscopic objects at separations below their own radii. The residual gap between the predicted 0.056 N and the Force F readout is therefore purely numerical.
Further Exploration
- Sweep q₁ from −10 μC to +10 μC while keeping q₂ fixed at +5 μC. At q₁ = 0 the force drops to zero — does the direction flip continuously as q₁ crosses zero, or is there a discontinuity in the arrow?
- With q₁ = q₂ = 5 μC, drag r from 0.5 m to 5.0 m and watch the F-vs-r plot. At what separation does the force drop below 0.010 N? Does the ratio F(r)/F(2r) stay close to 4 throughout the full range?
- Set q₁ = +10 μC and q₂ = −10 μC, then compare Force F to the q₁ = +5 μC, q₂ = −5 μC case at the same r. Does the force scale as |q₁|·|q₂| — a factor of 4 — as the law predicts?
- Try the extreme case: q₁ = +10 μC, q₂ = +10 μC, r = 0.5 m. What is the predicted force from F = k·(10e−6)²/(0.5)²? Does the readout agree within rounding?