AQA A-Level Physics 7408
Measurements and Their Errors
Particles and Radiation
Waves
Mechanics and Materials
Electricity
Further Mechanics and Thermal Physics
Fields and Their Consequences
Nuclear Physics
AstrophysicsOptional topic
Medical PhysicsOptional topic
Engineering PhysicsOptional topic
Turning Points in PhysicsOptional topic
ElectronicsOptional topic
Fields and Their Consequences
Coulomb's law
- 3 things to know
- Papers 2 and 3
Overview
The force between point charges, permittivity of free space, and comparison with gravitation.
What you need to know
Answer each one before you reveal it.
How does the sign of the charges determine whether the force attracts or repels?
What is the value and unit of 1/4πε₀, and what information is needed for an estimate?
Questions from this subtopic
Each one with the model answer underneath.
- 1
State Coulomb's law as an equation for the force between two point charges in a vacuum.
Model answer
F = (1/4πε₀) × (Q₁Q₂/r²), where F is the electrostatic force, Q₁ and Q₂ are the charges, r is the separation, and ε₀ is the permittivity of free space.
- 2
In Coulomb's law, what does each symbol represent in the equation F = (1/4πε₀) × (Q₁Q₂/r²)?
Model answer
F is the electrostatic force between the charges, Q₁ and Q₂ are the magnitudes of the two point charges, r is the distance between the centres of the charges, and ε₀ is the permittivity of free space.
- 3
Calculate the electrostatic force when charge 1 = 2.0 × 10⁻⁶ C, charge 2 = 3.0 × 10⁻⁶ C, separation = 0.44 m.
Model answer
Electrostatic force = (2.0 × 10⁻⁶ × 3.0 × 10⁻⁶) ÷ (4π × 8.85 × 10⁻¹² × 0.44²) = 0.279 N.
- 4
What is the permittivity of free space (ε₀) and what is its approximate value?
Model answer
The permittivity of free space is a constant that appears in Coulomb's law, with a value of approximately 8.85 × 10⁻¹² F m⁻¹.
- 5
What are the units of the permittivity of free space?
Model answer
F m⁻¹ (farads per metre).
- 6
Why can air be treated as a vacuum when applying Coulomb's law to calculate electrostatic forces?
Model answer
Because the permittivity of air is very close to the permittivity of free space, so the difference in calculated force is negligible.
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