Coulomb's law is the electric counterpart of Newton's law of gravitation — and it is 10³⁹ times stronger! Understand the force between charges, how it compares to gravity, and how charges move in uniform electric fields.
⚡State and apply Coulomb's law: F = kQ₁Q₂/r²
⚖️Identify the similarities and differences between Coulomb and gravitational forces
🔢Determine the superposition of forces from multiple charges
📐Analyse the motion of a charge in a uniform electric field using kinematics
🏹Solve projectile-style problems for charged particles in electric fields
🔬Discuss where Coulomb's law applies and breaks down
Coulomb's Law
The electrostatic force between two point charges Q₁ and Q₂ separated by distance r is given by Coulomb's Law:
F = kQ₁Q₂ / r² where k = 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻²
Key features:
Attractive if charges have opposite signs (F is negative in vector form)
Repulsive if charges have the same sign (F is positive)
Inverse-square law: F ∝ 1/r². Double distance → quarter force.
Acts along the line joining the two charges
Newton's 3rd law: force on Q₁ due to Q₂ = force on Q₂ due to Q₁ (equal magnitude, opposite direction)
Permittivity of free space: ε₀ = 8.85 × 10⁻¹² F m⁻¹. In a material with relative permittivity ε_r (dielectric constant), the force becomes F = kQ₁Q₂/(ε_r r²) — reduced by factor ε_r. Water has ε_r ≈ 80, greatly reducing electrostatic forces between ions in solution.
The superposition principle: when multiple charges are present, the net force on any charge is the vector sum of the individual Coulomb forces from all other charges.
⚠️ Coulomb's law applies to point charges (or uniformly charged spheres, treated as point charges at their centres). It does not directly apply to extended charge distributions without integration.
Comparing Coulomb and Gravitational Forces
Feature
Coulomb
Newton (Gravity)
Formula
F = kQ₁Q₂/r²
F = Gm₁m₂/r²
Distance dependence
Inverse-square (1/r²)
Inverse-square (1/r²)
Source
Electric charge Q
Mass m
Constant
k = 8.99×10⁹
G = 6.67×10⁻¹¹
Direction
Repulsive OR attractive
Always attractive
Superposition
Yes (vector sum)
Yes (vector sum)
Shielding
Yes (Faraday cage)
No known shielding
Relative strength
~10³⁶ to 10³⁹ times stronger
Weakest fundamental force
Both forces are "action at a distance" — they act through fields that permeate space. Modern physics explains both as particle exchanges: electromagnetism via photons, gravity via (hypothetical) gravitons. Both are long-range forces with infinite theoretical reach.
Despite electric force being so much stronger, gravity dominates at large scales because: (1) there are no negative masses, so gravitational force always accumulates; (2) large objects are electrically neutral — positive and negative charges cancel — so large-scale electric forces are negligible.
Motion of Charges in Uniform Electric Fields
A charge Q in a uniform electric field E experiences a constant force F = QE. This is directly analogous to a mass in a uniform gravitational field experiencing F = mg. The motion analysis follows identical kinematic equations.
Key scenarios:
1. Charge launched parallel to field: Uniformly accelerated/decelerated motion. a = QE/m along field direction.
2. Charge launched perpendicular to field (electric projectile):
Along field: uniform acceleration a = QE/m
Perpendicular to field: uniform velocity (no force component)
Trajectory is a parabola — identical in form to gravitational projectile with g replaced by QE/m
For a positive charge entering a field pointing upward: the charge is deflected upward. For a negative charge (e.g. electron): it is deflected in the opposite direction to the field lines.
For charge entering field perpendicular to E: y = ½at² = ½(QE/m)(x/v₀)² — parabolic path
This is the principle behind cathode ray tubes (CRT displays), oscilloscopes, mass spectrometers, and inkjet printers.
Applications of Coulomb's Law
The Millikan oil drop experiment (1909): A charged oil drop is balanced between two plates. At equilibrium: electric force = gravitational force:
QE = mg → Q = mg/E = mgd/V
By measuring the mass (from terminal velocity in air) and voltage needed to suspend the drop, Millikan determined the charge on an electron. He found all measured charges were integer multiples of a smallest unit e = 1.6×10⁻¹⁹ C — demonstrating charge is quantised.
Rutherford scattering: Alpha particles (charge +2e) are deflected by gold nuclei (charge +Ze) due to Coulomb repulsion. The closest approach distance d is found by equating initial KE to Coulomb PE:
KE = k(2e)(Ze)/d → d = 2kZe²/KE
The fact that some alpha particles bounced nearly straight back proved the nucleus is tiny and massive.
Quantisation of charge: All observable charges are integer multiples of the elementary charge e = 1.6×10⁻¹⁹ C. Quarks have fractional charges (e/3, 2e/3) but are never found isolated (confinement). In macroscopic physics, charge is always a multiple of e.
Calculate the Coulomb force between two electrons separated by 1.0 nm (1.0×10⁻⁹ m). Compare this to the gravitational force between them.
F_E = 2.3×10⁻¹⁰ N (repulsive); F_G = 5.5×10⁻⁵³ N (attractive). Electric force is 4×10⁴² times stronger!
Three charges are placed on the x-axis: Q₁ = +6.0 μC at x = 0, Q₂ = −4.0 μC at x = 0.30 m, Q₃ = +3.0 μC at x = 0.60 m. Find the net force on Q₂.
1Force on Q₂ from Q₁: F₁₂ = k|Q₁||Q₂|/r₁₂² = (8.99×10⁹ × 6.0×10⁻⁶ × 4.0×10⁻⁶)/(0.30)² = (8.99×10⁹ × 2.4×10⁻¹¹)/0.09 = 0.2158/0.09 = 2.397 N. Q₁ is + and Q₂ is −, so force is attractive: F₁₂ points toward Q₁ (in −x direction). F₁₂ = −2.40 N
2Force on Q₂ from Q₃: F₂₃ = k|Q₂||Q₃|/(0.30)² = (8.99×10⁹ × 4.0×10⁻⁶ × 3.0×10⁻⁶)/0.09 = (8.99×10⁹ × 1.2×10⁻¹¹)/0.09 = 1.199 N. Q₂ is − and Q₃ is +, so force on Q₂ is attractive: points toward Q₃ (in +x direction). F₂₃ = +1.20 N
3Net force: F_net = −2.40 + 1.20 = −1.20 N (in −x direction, toward Q₁)
Net force on Q₂ = 1.20 N in the −x direction (toward Q₁). Q₂ is pulled toward the larger charge.
In a Millikan experiment, an oil drop of mass 1.2×10⁻¹⁴ kg is suspended stationary between plates 5.0 mm apart with 1800 V across them. (a) Find the charge on the drop. (b) How many excess electrons does this represent?
1Electric field: E = V/d = 1800/(5.0×10⁻³) = 3.6×10⁵ V m⁻¹
2Equilibrium: QE = mg → Q = mg/E = (1.2×10⁻¹⁴ × 9.81)/(3.6×10⁵) = 1.177×10⁻¹³/3.6×10⁵ = 3.27×10⁻¹⁹ C
3Number of electrons: n = Q/e = 3.27×10⁻¹⁹/1.6×10⁻¹⁹ = 2.04 ≈ 2
Q = 3.27×10⁻¹⁹ C ≈ 2 × elementary charges. Drop has 2 excess electrons. (Demonstrates charge quantisation.)
Q1. If the distance between two charges is halved, the Coulomb force between them:
Q2. In Millikan's oil drop experiment, a drop is stationary when the electric field acts upward. This means the drop is:
Q3. An alpha particle (charge +2e) approaches a gold nucleus (charge +79e). As the alpha particle gets closer, the Coulomb force:
Q4. A proton enters a uniform electric field perpendicular to the field lines. Its path is:
Q5. Why does gravity dominate at astronomical scales even though electric force is ~10³⁶ times stronger?
Challenge 1. In Rutherford scattering, an alpha particle (m = 6.64×10⁻²⁷ kg, charge = +2e) is fired directly at a gold nucleus (Z = 79) with initial KE = 7.7 MeV. Find the distance of closest approach. Comment on its significance.
✓ KE = 7.7×10⁶ × 1.6×10⁻¹⁹ = 1.232×10⁻¹² J. At closest approach, all KE converts to Coulomb PE: KE = kQ_α Q_gold/d → d = kQ_α Q_gold/KE = (8.99×10⁹ × 2×1.6×10⁻¹⁹ × 79×1.6×10⁻¹⁹)/1.232×10⁻¹² = (8.99×10⁹ × 3.2×10⁻¹⁹ × 1.264×10⁻¹⁷)/1.232×10⁻¹². Numerator: 8.99×10⁹ × 4.045×10⁻³⁶ = 3.636×10⁻²⁶. d = 3.636×10⁻²⁶/1.232×10⁻¹² = 2.95×10⁻¹⁴ m ≈ 30 fm. Significance: This is an upper bound on the nuclear size — the nucleus must be smaller than 30 fm (the gold nucleus radius is ~7 fm). Rutherford's results showed most alpha particles passed through undeflected (the atom is mostly empty space) but some bounced almost straight back — only possible if all the positive charge and most of the mass is concentrated in a tiny nucleus.
Challenge 2. An electron (m_e = 9.11×10⁻³¹ kg, e = 1.6×10⁻¹⁹ C) moves in a circular orbit around a proton at radius r = 5.3×10⁻¹¹ m (Bohr radius). (a) Find the orbital speed using Coulomb force as centripetal force. (b) Calculate the total energy (KE + PE). (c) Convert to eV.
✓ (a) Coulomb force = centripetal force: ke²/r² = m_e v²/r → v² = ke²/(m_e r) = (8.99×10⁹ × (1.6×10⁻¹⁹)²)/(9.11×10⁻³¹ × 5.3×10⁻¹¹) = (8.99×10⁹ × 2.56×10⁻³⁸)/(4.828×10⁻⁴¹) = (2.301×10⁻²⁸)/(4.828×10⁻⁴¹) = 4.766×10¹² m² s⁻². v = 2.18×10⁶ m s⁻¹ ≈ 0.73% of speed of light. (b) KE = ½m_e v² = ½ × 9.11×10⁻³¹ × 4.766×10¹² = 2.171×10⁻¹⁸ J. PE = −ke²/r = −(8.99×10⁹ × 2.56×10⁻³⁸)/(5.3×10⁻¹¹) = −2.301×10⁻²⁸/5.3×10⁻¹¹ = −4.342×10⁻¹⁸ J. Total E = 2.171×10⁻¹⁸ − 4.342×10⁻¹⁸ = −2.171×10⁻¹⁸ J = −KE. (c) E = −2.171×10⁻¹⁸ / 1.6×10⁻¹⁹ = −13.6 eV. This is the ionisation energy of hydrogen — precisely the Bohr model result! Note E_total = −GMm/2r analogy: both systems have E_total = −KE = PE/2.
Challenge 3 (Synoptic). A proton is accelerated through 25 kV and then enters a uniform magnetic field B = 0.50 T perpendicular to its velocity. (a) Find the proton's speed after acceleration. (b) Find the radius of the circular path in the magnetic field. (c) How does this differ if an alpha particle (charge +2e, mass 4m_p) is used instead?
✓ (a) Work-energy: eV = ½m_p v² → v = √(2eV/m_p) = √(2×1.6×10⁻¹⁹×25000/1.67×10⁻²⁷) = √(8.0×10⁻¹⁵/1.67×10⁻²⁷) = √(4.79×10¹²) = 2.19×10⁶ m s⁻¹. (b) Magnetic force provides centripetal: evB = m_p v²/r → r = m_p v/(eB) = (1.67×10⁻²⁷ × 2.19×10⁶)/(1.6×10⁻¹⁹ × 0.50) = 3.657×10⁻²¹/8.0×10⁻²⁰ = 0.0457 m ≈ 4.6 cm. (c) Alpha particle: charge = 2e, mass = 4m_p. Speed after acceleration through 25 kV: 2eV = ½(4m_p)v_α² → v_α = √(eV/m_p) = √(4.79×10¹²/2) = √(2.395×10¹²) = 1.548×10⁶ m s⁻¹ (slower because heavier). Radius: r_α = m_α v_α/(qB) = (4m_p × 1.548×10⁶)/(2e × 0.50) = (4 × 1.67×10⁻²⁷ × 1.548×10⁶)/(2 × 1.6×10⁻¹⁹ × 0.50) = (1.034×10⁻²⁰)/(1.6×10⁻¹⁹) = 0.0646 m ≈ 6.5 cm. Alpha particle has larger radius — used in cyclotrons and mass spectrometers to separate particles by mass-to-charge ratio.