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Electric Charge & Coulomb’s Law

You’ll be able to

Charge is a conserved, quantized property of matter

Electric charge comes in two signs — positive and negative — and like signs repel while opposites attract. Two rules govern it everywhere. Conservation: the net charge of an isolated system never changes; charge is only transferred, never created or destroyed. Quantization: every observable charge is an integer multiple of the elementary charge e = 1.602 × 10⁻¹⁹ C. A charged object simply has an excess or deficit of electrons.

Coulomb’s law (magnitude)
F = k·q₁q₂ / r², with k = 1/(4πε₀) ≈ 8.99 × 10⁹ N·m²/C²
ε₀ = 8.85 × 10⁻¹² C²/(N·m²) is the permittivity of free space. The force falls off as 1/r², not 1/r.

The force is a vector, and forces superpose

Coulomb’s law gives the magnitude of the force between two point charges; its direction lies along the line joining them — away for like charges, toward for opposite. When several charges act on one target, the total force is the vector sum of the individual Coulomb forces: Fₜₒₜₐₗ = Σ Fᵢ. This principle of superposition is the key that unlocks every later problem, including continuous distributions.

Worked example

A +3.0 μC charge and a −2.0 μC charge are separated by 0.20 m. Find the magnitude of the force between them and state whether it is attractive or repulsive.

  1. 1.Convert to SI: q₁ = 3.0 × 10⁻⁶ C, q₂ = 2.0 × 10⁻⁶ C, r = 0.20 m.
  2. 2.Use magnitudes in Coulomb’s law: F = k·q₁q₂ / r².
  3. 3.Numerator: k·q₁q₂ = (8.99 × 10⁹)(3.0 × 10⁻⁶)(2.0 × 10⁻⁶) = 5.39 × 10⁻² N·m².
  4. 4.Denominator: r² = (0.20)² = 0.040 m².
  5. 5.F = 5.39 × 10⁻² / 0.040 = 1.35 N. The signs are opposite, so the force is attractive.
Answer: ≈ 1.3 N, attractive
Watch out

The distance is squared in the denominator. Doubling the separation cuts the force to one quarter, not one half. Confusing 1/r with 1/r² is the single most common Coulomb’s-law error.

Checkpoint

Two point charges attract each other with a force F. If the distance between them is tripled while the charges are unchanged, the new force is:

Worked example

Three charges sit on the x-axis: q₁ = +4.0 μC at x = 0, q₂ = +1.0 μC at x = 1.0 m, and q₃ = −2.0 μC at x = 2.0 m. Find the net force on q₂.

  1. 1.Force from q₁ on q₂: r = 1.0 m, both positive so repulsive, pushing q₂ in the +x direction. F₁ = k(4.0 × 10⁻⁶)(1.0 × 10⁻⁶)/(1.0)² = 3.6 × 10⁻² N, +x.
  2. 2.Force from q₃ on q₂: r = 1.0 m, opposite signs so attractive, pulling q₂ toward q₃ in the +x direction. F₃ = k(1.0 × 10⁻⁶)(2.0 × 10⁻⁶)/(1.0)² = 1.8 × 10⁻² N, +x.
  3. 3.Both forces point in +x, so add magnitudes: Fₙₑₜ = 3.6 × 10⁻² + 1.8 × 10⁻² = 5.4 × 10⁻² N.
Answer: 5.4 × 10⁻² N in the +x direction
Checkpoint

A neutral metal sphere is touched by a negatively charged rod and shares 5.0 × 10⁹ excess electrons. Approximately what is the sphere’s resulting charge? (e = 1.6 × 10⁻¹⁹ C)

On the exam

On free-response problems, always resolve each Coulomb force into components before summing. Only add magnitudes directly when every force lies along the same line, as it did above. Otherwise sum Fₓ and Fᵧ separately, then recombine.

Answer the 2 checkpoints as you read.

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