OxfordAQA International AS and A-level Physics: Coulomb's law and electric field strength (9630)

This guide follows the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, for exams from May/June 2020 onwards. It covers sections 3.8.1 (Coulomb’s law) and 3.8.2 (Electric field strength). Section 3.8 is marked International A-level only, so you meet this content on the A-level route and not in the International AS. Use it with the course hub and the printable checklist.

Companion pages: the revision notes and the practice questions. You will use ideas from motion and Newton’s laws. The gravitational version is in Newton’s law of gravitation. To find out what to revise first, take one of the free diagnostics.

Data used below: G = 6.67 × 10⁻¹¹ N m² kg⁻²; g = 9.81 N kg⁻¹; ε₀ = 8.85 × 10⁻¹² F m⁻¹; elementary charge 1.60 × 10⁻¹⁹ C; masses: electron 9.11 × 10⁻³¹ kg, proton 1.67 × 10⁻²⁷ kg.

What 3.8.1 and 3.8.2 ask you to do

Specification statement	What you must be able to do
3.8.1 Force between point charges in a vacuum, F = Q₁Q₂/(4πε₀r²)	State the law, use it, and handle ratio and scaling problems
3.8.1 Air can be treated as a vacuum when calculating force between charges	Say why you may use ε₀ for charges in air
3.8.1 For a charged sphere, charge may be considered to be at the centre	Measure r from the centre of the sphere, not its surface
3.8.1 Comparison of the gravitational and electrostatic forces between subatomic particles	Calculate both forces and compare their magnitudes
3.8.2 Representation of electric fields by field lines	Draw and interpret field lines
3.8.2 E as force per unit charge, E = F/Q	Define E, give its units, treat it as a vector
3.8.2 Uniform field, E = V/d, and the derivation from Fd = QΔV	Use the equation and reproduce the derivation
3.8.2 Trajectory of a charged particle entering a uniform field at right angles	Calculate the deflection and describe the path
3.8.2 Radial field, E = Q/(4πε₀r²)	Use it, including how E changes with distance

Coulomb’s law (3.8.1)

In a vacuum, point charges Q₁ and Q₂ separated by r push or pull on each other with forces of equal size and opposite direction:

F = Q₁Q₂ / (4πε₀r²)

It has the same inverse-square form as Newton’s law of gravitation. The force acts along the line joining the charges. Like charges repel and unlike charges attract. The equation gives magnitude only: decide direction from the signs of the charges.

ε₀ is the permittivity of free space. The factor 1/(4πε₀) is about 8.99 × 10⁹ N m² C⁻². Air can be treated as a vacuum: it changes the force by a negligible amount, so you use ε₀ for charges in air.

For a charged sphere, the charge may be considered to be at the centre. So r is the distance from the centre of the sphere to the other charge.

Worked example 1: force between two charges

Two small spheres carry +4.8 nC and −3.2 nC. Their centres are 6.0 cm apart in air. Find the force on each.

F = (4.8 × 10⁻⁹ × 3.2 × 10⁻⁹) / (4π × 8.85 × 10⁻¹² × 0.060²)

Numerator: 1.536 × 10⁻¹⁷ C². Denominator: 4π × 8.85 × 10⁻¹² × 3.6 × 10⁻³ = 4.00 × 10⁻¹³.

F = 3.8 × 10⁻⁵ N. The charges are unlike, so each sphere is pulled towards the other. By Newton’s third law both forces have this magnitude.

Worked example 2: scaling

The force between two charges is 0.80 mN. One charge is doubled and the separation is tripled. Find the new force.

Doubling a charge doubles F. Tripling r divides F by 3² = 9. New force = 0.80 × 2/9 = 0.18 mN.

Worked example 3: a charged sphere

A conducting sphere of radius 2.5 cm carries +60 nC. A point charge of −5.0 nC is held 10 cm from the surface of the sphere. Find the force on the point charge.

The charge on the sphere acts as if at its centre, so r = 2.5 + 10 = 12.5 cm = 0.125 m.

F = (60 × 10⁻⁹ × 5.0 × 10⁻⁹) / (4π × 8.85 × 10⁻¹² × 0.125²) = 1.7 × 10⁻⁴ N, towards the sphere (unlike charges).

Using 0.10 m by mistake gives a force 1.6 times too large.

Gravitational and electrostatic forces between subatomic particles

For an electron and a proton, the same r appears in both forces, so it cancels in the ratio:

Fₑ / F_g = [e² / (4πε₀r²)] ÷ [G mₑ mₚ / r²] = e² / (4πε₀ G mₑ mₚ)

Substituting: (1.60 × 10⁻¹⁹)² = 2.56 × 10⁻³⁸ in the numerator. The denominator is 4π × 8.85 × 10⁻¹² × 6.67 × 10⁻¹¹ × 9.11 × 10⁻³¹ × 1.67 × 10⁻²⁷ = 1.13 × 10⁻⁷⁷.

Ratio = 2.3 × 10³⁹.

For a separation of 5.3 × 10⁻¹¹ m, the electrostatic force is 8.2 × 10⁻⁸ N and the gravitational force is 3.6 × 10⁻⁴⁷ N. Both forces fall as 1/r², so the ratio is the same at any separation. Gravity is negligible between subatomic particles.

Electric field lines (3.8.2)

An electric field is a region where a charge experiences a force. You show it with field lines. The rules are:

• An arrow on each line shows which way a positive test charge would be pushed.
• Lines leave positive charges and arrive at negative ones (an isolated charge has lines running to or from infinity).
• The lines never cross.
• Where the lines are closer together, the field is stronger.

Three patterns to know:

Situation	Field line pattern
Isolated positive point charge	Straight lines pointing radially outwards, spreading out so the field weakens with distance
Isolated negative point charge	The same lines pointing radially inwards
Two oppositely charged parallel plates	Parallel, equally spaced lines from the positive plate to the negative plate, running at right angles to the plates. They curve outwards only near the edges

Equally spaced parallel lines show a uniform field (ignoring the edges).

Electric field strength E = F/Q (3.8.2)

The electric field strength at a point is the force per unit positive charge placed at that point:

E = F / Q

E is a vector that points the way a positive charge would be pushed. Its unit is N C⁻¹, which is equivalent to V m⁻¹. For a negative charge the force is opposite to E, so the force on −Q is −QE.

Example: a +2.0 nC test charge feels 6.0 × 10⁻⁵ N, so E = 6.0 × 10⁻⁵ / 2.0 × 10⁻⁹ = 3.0 × 10⁴ N C⁻¹.

Uniform field: E = V/d

Between two parallel plates with potential difference V and separation d, the field is uniform and its magnitude is

E = V / d

Derivation

Move a charge Q from one plate to the other through a field of strength E. The force on it is F = QE, and it moves a distance d along the field.

Work done = Fd. The work done per unit charge is the potential difference, so the work done is also QΔV.

So Fd = QΔV. Substituting F = QE gives QEd = QΔV, and cancelling Q:

E = ΔV / d

Write out each step: work = force × distance, work = QΔV, and F = QE.

Worked example 4: balancing a droplet

A tiny oil droplet of mass 1.6 × 10⁻¹⁵ kg hangs stationary between horizontal plates 1.5 cm apart with a p.d. of 480 V. Find the charge on the droplet.

E = V/d = 480 / 0.015 = 3.2 × 10⁴ V m⁻¹.

For equilibrium, the electric force equals the weight: QE = mg.

Q = mg/E = (1.6 × 10⁻¹⁵ × 9.81) / 3.2 × 10⁴ = 4.9 × 10⁻¹⁹ C.

This is about 3.1 times e, so the droplet carries about three excess electrons. The force acts upwards, so the upper plate is positive and the droplet is negative.

Particle crossing a uniform field

A particle enters the plates with speed v at right angles to the field. There is no force along its original direction, so that speed stays constant. A constant force acts across the field, so it accelerates in that direction. This matches a projectile in a uniform gravitational field: the path is a parabola between the plates, then a straight line after it leaves.

The steps are:

1. Field: E = V/d.
2. Force: F = QE.
3. Acceleration across the field: a = F/m.
4. Time between the plates: t = L/v (L is the plate length).
5. Deflection: y = ½at² (starting from zero sideways speed).
6. Sideways speed on leaving: v_y = at, and the angle to the original direction is tan⁻¹(v_y/v).

Worked example 5: a proton in a field

A proton enters the gap between two horizontal plates at 6.0 × 10⁵ m s⁻¹, parallel to the plates. The plates are 4.5 cm long and 2.4 cm apart, with a p.d. of 450 V.

1. E = 450 / 0.024 = 1.875 × 10⁴ V m⁻¹.
2. F = eE = 1.60 × 10⁻¹⁹ × 1.875 × 10⁴ = 3.0 × 10⁻¹⁵ N, towards the negative plate.
3. a = F/m = 3.0 × 10⁻¹⁵ / 1.67 × 10⁻²⁷ = 1.8 × 10¹² m s⁻².
4. t = 0.045 / 6.0 × 10⁵ = 7.5 × 10⁻⁸ s.
5. y = ½ × 1.796 × 10¹² × (7.5 × 10⁻⁸)² = 5.1 mm towards the negative plate.
6. v_y = at = 1.3 × 10⁵ m s⁻¹, so the angle is tan⁻¹(1.347 × 10⁵ / 6.0 × 10⁵) = 13° to the original direction.

The proton’s weight is 1.6 × 10⁻²⁶ N, about 10¹¹ times smaller than the electric force, so weight is ignored. Check that the deflection is below 12 mm (half the gap), or the particle hits a plate. A lighter particle with the same charge is deflected further, because a = QE/m.

Radial field: E = Q/(4πε₀r²)

Put a small test charge q a distance r from a source charge Q. Coulomb’s law gives F = Qq/(4πε₀r²). Dividing by q:

E = Q / (4πε₀r²)

• E is inversely proportional to r². Doubling the distance cuts the field to a quarter.
• For a charged sphere, r is measured from the centre and the equation applies outside the sphere.
• The direction is away from a positive Q and towards a negative Q.

Worked example 6

A sphere carries −35 nC. Find the field strength 0.30 m from its centre, and the force on a +4.0 nC charge placed there.

E = 35 × 10⁻⁹ / (4π × 8.85 × 10⁻¹² × 0.30²) = 3.5 × 10³ N C⁻¹, directed towards the sphere.

F = qE = 4.0 × 10⁻⁹ × 3.497 × 10³ = 1.4 × 10⁻⁵ N, towards the sphere.

Common errors

• Taking r to the surface of a sphere rather than to its centre.
• Forgetting to square r.
• Using E = V/d for a radial field, or E = Q/(4πε₀r²) between parallel plates.
• Leaving the separation in centimetres or the charge in nC.
• Giving the force on an electron in the direction of E. The electron is negative, so the force is opposite to E.
• Drawing the path between the plates as a circle. It is a parabola.
• Saying field lines show a moving charge’s path. They show the direction of the force.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Sections 3.8.1 Coulomb’s law and 3.8.2 Electric field strength (International A-level only).