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In 1785, French physicist Charles-Augustin de Coulomb used a torsion balance to measure the force between charged objects and established one of the fundamental laws of electrostatics. His discovery revealed that the force between two point charges follows an inverse-square law — remarkably similar in structure to Newton’s law of gravitation.
The force between two point charges is:
F = kQ₁Q₂ / r²
where:
The constant k is related to the permittivity of free space (ε₀) by:
k = 1 / (4πε₀)
where ε₀ = 8.85 × 10⁻¹² F m⁻¹ (farads per metre).
So Coulomb’s law can also be written as:
F = Q₁Q₂ / (4πε₀r²)
The permittivity of free space (ε₀) is a fundamental constant that describes how easily an electric field can permeate a vacuum. Its value is:
ε₀ = 8.85 × 10⁻¹² F m⁻¹
A larger permittivity means the medium reduces the electric force between charges. In a dielectric material, the permittivity is greater than ε₀, which is why capacitors use dielectrics to store more charge.
When charges are placed in a medium other than a vacuum, the force is reduced by the relative permittivity εᵣ of the medium:
F = Q₁Q₂ / (4πε₀εᵣr²)
| Medium | Relative Permittivity εᵣ |
|---|---|
| Vacuum | 1.000 (exact) |
| Air | 1.0006 (≈ 1) |
| Paper | 3–4 |
| Glass | 5–10 |
| Water | 80 |
| Barium titanate | ~1200 |
Water has a very high relative permittivity, which is why ionic compounds dissolve in water — the electric force holding the ions together is reduced by a factor of 80, allowing thermal energy to separate them.
By combining Coulomb’s law with the definition E = F / Q, the electric field strength at a distance r from a point charge Q is:
E = kQ / r² = Q / (4πε₀r²)
This is the field strength that a small positive test charge would experience at that distance. The field decreases with the square of the distance — move twice as far away and the field drops to a quarter of its value.
Calculate the electric field strength at a distance of 0.30 m from a charge of +5.0 μC.
E = kQ / r² = (8.99 × 10⁹ × 5.0 × 10⁻⁶) / (0.30)²
E = 44 950 / 0.09 = 4.99 × 10⁵ ≈ 5.0 × 10⁵ N C⁻¹
Two small conducting spheres carry charges of +3.0 μC and −4.0 μC respectively and are separated by 0.15 m in air. Calculate the magnitude and direction of the force between them.
F = kQ₁Q₂ / r² = (8.99 × 10⁹ × 3.0 × 10⁻⁶ × 4.0 × 10⁻⁶) / (0.15)²
F = (8.99 × 10⁹ × 1.2 × 10⁻¹¹) / 0.0225
F = 1.079 / 0.0225 = 4.8 N
The force is attractive because the charges have opposite signs. Each sphere experiences a 4.8 N force directed towards the other sphere.
Two identical small spheres are placed 0.25 m apart and experience a repulsive force of 0.036 N. Calculate the charge on each sphere.
Since the spheres are identical: Q₁ = Q₂ = Q
F = kQ² / r²
Q² = Fr² / k = (0.036 × 0.0625) / 8.99 × 10⁹ = 2.25 × 10⁻³ / 8.99 × 10⁹ = 2.50 × 10⁻¹³
Q = 5.0 × 10⁻⁷ C = 0.50 μC
The mathematical structures are strikingly similar:
| Coulomb’s Law | Newton’s Gravitation | |
|---|---|---|
| Formula | F = kQ₁Q₂ / r² | F = Gm₁m₂ / r² |
| Constant | k = 8.99 × 10⁹ N m² C⁻² | G = 6.67 × 10⁻¹¹ N m² kg⁻² |
| Property | Charge | Mass |
| Direction | Attractive or repulsive | Attractive only |
| Dependence on distance | Inverse-square | Inverse-square |
| Field strength | E = kQ / r² | g = GM / r² |
| Potential | V = kQ / r | V = −GM / r |
Both are inverse-square laws, both act along the line joining the two objects, and both obey Newton’s third law. The crucial difference is that gravity is always attractive (mass is always positive), while the electric force can be attractive or repulsive (charge can be positive or negative).
The ratio k / G ≈ 10²⁰, which tells us that the electric force is vastly stronger than gravity for objects with comparable charge and mass. This is why gravity is negligible at the atomic scale — the electric force between a proton and electron is about 10³⁹ times stronger than the gravitational force between them.
When multiple charges are present, the resultant electric field at any point is the vector sum of the individual fields due to each charge. This is the principle of superposition.
Two charges of +4.0 μC and −4.0 μC are placed 0.20 m apart. Find the electric field at the midpoint between them.
At the midpoint (0.10 m from each charge):
Field due to the +4.0 μC charge: E₁ = kQ / r² = (8.99 × 10⁹ × 4.0 × 10⁻⁶) / (0.10)² = 3.6 × 10⁶ N C⁻¹ (pointing away from the positive charge, towards the midpoint and beyond)
Field due to the −4.0 μC charge: E₂ = kQ / r² = (8.99 × 10⁹ × 4.0 × 10⁻⁶) / (0.10)² = 3.6 × 10⁶ N C⁻¹ (pointing towards the negative charge, in the same direction as E₁)
Since both fields point in the same direction (from the positive charge towards the negative charge), the resultant field is:
E_total = E₁ + E₂ = 7.2 × 10⁶ N C⁻¹
If the charges were both positive, the fields at the midpoint would point in opposite directions and cancel to zero.
Two point charges Q₁ = +9.0 μC and Q₂ = +1.0 μC are placed 0.40 m apart. Find the position along the line joining them where the resultant electric field is zero.
Let the null point be at distance x from Q₁. The fields must be equal in magnitude and opposite in direction:
kQ₁ / x² = kQ₂ / (0.40 − x)²
Q₁ / x² = Q₂ / (0.40 − x)²
Taking square roots: √Q₁ / x = √Q₂ / (0.40 − x)
3 / x = 1 / (0.40 − x)
3(0.40 − x) = x
1.20 − 3x = x
1.20 = 4x
x = 0.30 m from Q₁
The null point is closer to the smaller charge, which makes physical sense — you need to be further from the larger charge for the two fields to balance.
Lightning: During a thunderstorm, charge separation in clouds creates enormous electric fields. When the field strength exceeds the breakdown voltage of air (~3 × 10⁶ V m⁻¹), a conducting channel forms and a lightning bolt discharges the accumulated charge. A typical lightning bolt transfers about 5 C of charge.
Xerography (photocopying): A drum is given a uniform positive charge. Light reflected from the document discharges areas corresponding to white parts of the page. Negatively charged toner particles are attracted to the remaining positively charged areas by the Coulomb force, forming the image.
Particle physics: In Rutherford’s gold foil experiment, alpha particles were deflected by the Coulomb force between the positive alpha particle and the positive gold nucleus. The closest approach distance can be calculated by equating kinetic energy to Coulomb potential energy.
Edexcel 9PH0 specification Topic 7 — Electric and Magnetic Fields covers Coulomb's law for the force between two point charges in the form F=kQ1Q2/r2, with k=1/(4πε0), together with the resulting radial field E=kQ/r2 and the principle of superposition (refer to the official Pearson Edexcel specification document for exact wording). Coulomb's law is the foundation on which the rest of Topic 7 is built — radial fields, electric potential V=kQ/r, capacitance, and the analogy with gravitational fields in Topic 5 all rest on it. It is examined predominantly in 9PH0 Paper 2, with synoptic appearances in Paper 3 through electrostatics demonstrations. The Edexcel formula booklet lists F=kQ1Q2/r2 and E=kQ/r2 and quotes ε0=8.85×10−12 F m⁻¹ on the data sheet; candidates are expected to recall k≈8.99×109 N m² C⁻².
Question (8 marks):
(a) Two small spheres are held at rest 0.080 m apart in a vacuum. Sphere X carries a charge of +2.5×10−8 C and sphere Y carries a charge of +5.0×10−8 C. Calculate the magnitude of the electric force between the spheres and state its direction. (3)
(b) Calculate the magnitude of the gravitational force between the spheres if each has a mass of 1.2×10−3 kg, and hence find the ratio of the electric to the gravitational force. Comment on the result. (5)
(G=6.67×10−11 N m² kg⁻²)
Solution with mark scheme:
(a) Step 1 — apply Coulomb's law with magnitudes.
FE=r2kQ1Q2=(0.080)2(8.99×109)(2.5×10−8)(5.0×10−8)
M1 — correct substitution with magnitudes only. r must be in metres; using r=8.0 would scramble the answer by 104.
Step 2 — evaluate.
FE=6.4×10−3(8.99×109)(1.25×10−15)≈1.8×10−3 N
A1 — magnitude correct to 2 s.f. with units.
Step 3 — direction. The force is repulsive (same signs); X is pushed away from Y along the line joining their centres, with an equal and opposite force on Y (Newton's third law).
B1 — explicit "repulsive" with same-sign justification.
(b) Step 1 — Newton's law of gravitation.
FG=r2Gm1m2=(0.080)2(6.67×10−11)(1.2×10−3)2≈1.5×10−14 N
M1 A1 — correct substitution and evaluation.
Step 2 — ratio.
FGFE=1.5×10−141.8×10−3≈1.2×1011
M1 — correct ratio.
Step 4 — comment.
The electric repulsion is ∼1011 times larger than gravitational attraction here, which is why gravity is ignored in laboratory electrostatics. Between a proton and an electron the ratio is ∼1039, explaining why atomic structure is set by electromagnetic, not gravitational, forces.
A1 (AO3) — comparative comment linking the ratio to a physical principle. A bare number loses this mark.
Total: 8 marks (M3 A3 B1 + 1 evaluative A1).
Question (6 marks): Three small charges sit on a straight line in a vacuum. Charge A (+3.0×10−9 C) is at x=0; charge B (−2.0×10−9 C) is at x=0.040 m; charge C (+4.0×10−9 C) is at x=0.100 m. The charges may be treated as point charges.
(a) Calculate the magnitude and direction of the resultant electric force on charge B due to A and C. (4)
(b) Without further calculation, state and justify whether the resultant force on B would increase, decrease or stay the same if all three charges had their signs reversed. (2)
Mark scheme decomposition by AO:
(a)
(b)
Total: 6 marks split AO1 = 1, AO2 = 3, AO3 = 2. Paper 2 problems on Coulomb's law often combine routine substitution with a vector-superposition step and a "without further calculation" symmetry argument — exactly the synoptic structure rewarded here.
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