OCR A Level Physics A: Electric fields, Coulomb's law, potential and uniform fields (H556)

This guide covers section 6.2 (Electric fields) of the OCR A Level Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017: 6.2.1 point and spherical charges, 6.2.2 Coulomb’s law, 6.2.3 the uniform electric field and 6.2.4 electric potential and energy. H556 is the full A level. Also see the revision notes and the practice questions. Links: OCR A Level Physics hub, printable checklist, free diagnostics and Foundations of Physics.

What you must be able to do

Spec	You must be able to
6.2.1(a)-(d)	Say that fields are due to charges; treat a uniformly charged sphere as a point charge at its centre; draw field lines; use E = F/q
6.2.2(a)-(d)	Use Coulomb’s law and E = Q/(4πε₀r²); compare electric and gravitational fields; see fields as forms of field giving rise to a force
6.2.3(a)-(c)	Use E = V/d; use C = ε₀A/d, C = εA/d and ε = εᵣε₀; analyse charged particles in a uniform field
6.2.4(a)-(e)	Define electric potential; use V = Q/(4πε₀r); derive C = 4πε₀R for an isolated sphere; use the area under a force-distance graph; use potential energy Vq = Qq/(4πε₀r)

Data used below (treat as given in each question): G = 6.67 × 10⁻¹¹ N m² kg⁻², ε₀ = 8.85 × 10⁻¹² F m⁻¹, proton mass 1.67 × 10⁻²⁷ kg, electron mass 9.11 × 10⁻³¹ kg, elementary charge 1.60 × 10⁻¹⁹ C.

6.2.1 Point and spherical charges

Electric fields are due to charges. A charge sets up a field, and any other charge placed in it feels a force.

Modelling a charged sphere (b). A uniformly charged sphere acts, for points outside it, as though all its charge Q sat at its centre. Distances are measured from the centre, not the surface. This is a model (HSW1) and applies only outside the sphere.

Field lines (c). A field line is drawn so that its tangent gives the direction in which a small positive charge would be pushed.

• Lines leave positive charges and end on negative charges (or go to infinity).
• The closer the lines, the stronger the field.
• Lines never cross.
• An isolated point or sphere gives radial lines; two opposite charges give curved lines joining them; two parallel plates give evenly spaced parallel lines in the gap, bending outwards at the edges.

Electric field strength (d). E is the force per unit positive charge:

E = F/q, in N C⁻¹ (equivalently V m⁻¹).

It is a vector, in the direction of the force on a positive charge; a negative charge feels the opposite force.

6.2.2 Coulomb’s law

Coulomb’s law (a). For two point charges Q and q a distance r apart, the force on each is

F = Qq/(4πε₀r²)

The force acts along the line joining them. By Newton’s third law the two forces are equal and opposite. Like charges repel; unlike charges attract. With signs substituted, a negative F means attraction; otherwise give the size and state the direction in words.

Field of a point charge (b). Put q in the field of Q and divide the force by q:

E = Q/(4πε₀r²)

Doubling r divides E by 4.

Worked example 1: force between two small spheres

Spheres carry +7.2 nC and −1.8 nC, with centres 4.5 cm apart in air. Treat them as point charges.

F = Qq/(4πε₀r²) = (7.2 × 10⁻⁹ × 1.8 × 10⁻⁹) / (4π × 8.85 × 10⁻¹² × 0.045²)

= 1.296 × 10⁻¹⁷ / 2.252 × 10⁻¹³ = 5.8 × 10⁻⁵ N, attractive (unlike charges).

Worked example 2: field of a charged sphere

A sphere of radius 2.0 cm carries −45 nC. Find E at a point 6.0 cm from its centre.

The point is outside the sphere, so r = 0.060 m.

E = 45 × 10⁻⁹ / (4π × 8.85 × 10⁻¹² × 0.060²) = 1.1 × 10⁵ N C⁻¹, directed towards the centre (the charge is negative).

Worked example 3: a point where the field is zero

Charges +12 nC and +3.0 nC are 0.30 m apart. Where on the line between them is the resultant field zero?

At the point, the two fields are equal and opposite. Let x be the distance from the 12 nC charge:

12/x² = 3.0/(0.30 − x)²

Take square roots: √12/x = √3/(0.30 − x), so 2(0.30 − x) = x. Then x = 0.20 m.

The point is 0.20 m from the +12 nC charge (0.10 m from the +3.0 nC charge). Check: 12/0.20² = 300 and 3.0/0.10² = 300.

Gravitational and electric fields compared (c)

	Gravitational field of a point mass	Electric field of a point charge
Force law	F = Gm₁m₂/r²	F = Qq/(4πε₀r²)
Field strength	force per unit mass	force per unit charge
Variation	inverse square of r	inverse square of r
Direction	always attractive	attractive or repulsive
Strength	very weak	much stronger for typical charges

Both are radial around a point source and take potential as zero at infinity.

Worked example 4: how much stronger is the electric force?

Compare the electric and gravitational forces between two protons.

Electric force e²/(4πε₀r²) over gravitational Gm_p²/r²: the r² cancels, so separation does not matter.

Ratio = e² / (4πε₀Gm_p²) = 1.24 × 10³⁶.

Fields as forms of field (d)

Gravitational, electric and magnetic fields are all fields in which a body experiences a force at a distance: mass in a gravitational field, charge in an electric field, and a moving charge in a magnetic field (section 6.3). Link this to section 5.4: in each case field strength is a force per unit of the relevant property.

6.2.3 Uniform electric field

Field strength between plates (a). Between two large parallel plates with p.d. V and separation d, the field is uniform (away from the edges):

E = V/d

Work done moving charge q across the gap gives qV = Fd, so F/q = V/d. V m⁻¹ and N C⁻¹ are the same unit.

Example: plates 3.0 cm apart at 1500 V give E = 1500 / 0.030 = 5.0 × 10⁴ V m⁻¹; halving the separation at the same p.d. doubles E.

Parallel-plate capacitor and permittivity (b). For plates of area A separated by d in a vacuum (air is close enough):

C = ε₀A/d

With an insulating material between the plates, C = εA/d, where ε is the permittivity of the material and ε = εᵣε₀. The relative permittivity εᵣ is at least 1; you do not need to explain why.

Worked example 5

Square plates of side 12 cm are 0.50 mm apart.

A = 0.12² = 1.44 × 10⁻² m². In air: C = 8.85 × 10⁻¹² × 1.44 × 10⁻² / 5.0 × 10⁻⁴ = 2.5 × 10⁻¹⁰ F.

A material with εᵣ = 4.5 fills the gap: ε = 4.5 × 8.85 × 10⁻¹² = 3.98 × 10⁻¹¹ F m⁻¹, and C = εA/d = 1.1 × 10⁻⁹ F. At 24 V the charge stored is Q = VC = 2.8 × 10⁻⁸ C.

Charged particles in a uniform field (c) (HSW2). The force on a charge q is F = qE = qV/d. Newton’s second law gives a = qE/m. Use the kinematics of section 3.

Along the field (acceleration). Work done = qV = change in kinetic energy.

Worked example 6: an electron accelerated across a gap

An electron starts from rest at one plate. The plates are 2.4 cm apart with a p.d. of 900 V.

E = 900/0.024 = 3.75 × 10⁴ V m⁻¹, F = eE = 6.0 × 10⁻¹⁵ N and a = F/m = 6.59 × 10¹⁵ m s⁻². From energy: ½mv² = eV, so v = √(2eV/m) = √(2 × 1.60 × 10⁻¹⁹ × 900 / 9.11 × 10⁻³¹) = 1.8 × 10⁷ m s⁻¹.

Across the field (projectile-like motion). The velocity component parallel to the plates is constant. The component along the field changes uniformly. The path in the field is a parabola, then a straight line.

Worked example 7: deflection

An electron enters midway between plates at 1.5 × 10⁷ m s⁻¹, parallel to them. The plates are 4.0 cm long, 1.2 cm apart, with a p.d. of 80 V.

t = 0.040 / 1.5 × 10⁷ = 2.67 × 10⁻⁹ s. a = eV/(md) = 1.17 × 10¹⁵ m s⁻². Sideways displacement y = ½at² = 4.2 mm towards the positive plate.

This is less than 6.0 mm, so the electron emerges. Its sideways speed on leaving is at = 3.1 × 10⁶ m s⁻¹, so it leaves at tan⁻¹(3.12 × 10⁶ / 1.5 × 10⁷) = 11.8° to its original path.

6.2.4 Electric potential and energy

Electric potential (a). The electric potential at a point is the work done in bringing unit positive charge from infinity to that point. Potential is zero at infinity. It is a scalar, measured in volts (J C⁻¹).

Potential of a point charge (b).

V = Q/(4πε₀r)

V has the sign of Q: positive near a positive charge, negative near a negative charge. A change in potential ΔV is the work done per unit charge between two points, so moving charge q through ΔV involves work qΔV.

Worked example 8: work done between two radii

A small sphere carries +8.0 nC. Find the work done in moving +0.60 nC from 0.40 m to 0.10 m from the sphere’s centre.

V at 0.40 m = 8.0 × 10⁻⁹ / (4π × 8.85 × 10⁻¹² × 0.40) = 180 V. V at 0.10 m = 719 V. ΔV = 539 V.

W = qΔV = 0.60 × 10⁻⁹ × 539 = 3.2 × 10⁻⁷ J. The work is done on the charge because like charges repel.

Capacitance of an isolated sphere (c). Derivation, as the specification expects: for a sphere of radius R carrying charge Q, the potential at its surface (outside points treated as a point charge at the centre) is V = Q/(4πε₀R). Capacitance is C = Q/V, so

C = Q / [Q/(4πε₀R)] = 4πε₀R.

For R = 0.25 m, C = 4π × 8.85 × 10⁻¹² × 0.25 = 2.8 × 10⁻¹¹ F.

Force-distance graph (d) (HSW5). Plot F against r for a point or spherical charge. The curve falls steeply and flattens, since F ∝ 1/r². The area under the graph between two radii is the work done moving the charge between them. Counting squares or trapezia estimate it; trapezia over-estimate here because the curve is concave up.

Electric potential energy (e). The potential energy of charge q at distance r from a point charge Q is

E_p = Vq = Qq/(4πε₀r)

It is positive for like charges, negative for unlike.

Worked example 9: closest approach

A particle of charge +2e has 5.0 MeV of kinetic energy and heads straight for a fixed nucleus of charge +29e. How close does it get? (1 MeV = 1.60 × 10⁻¹³ J.)

At closest approach all the kinetic energy has become potential energy: 5.0 × 1.60 × 10⁻¹³ = 8.0 × 10⁻¹³ J = (2e)(29e)/(4πε₀r).

r = 58 × (1.60 × 10⁻¹⁹)² / (4π × 8.85 × 10⁻¹² × 8.0 × 10⁻¹³) = 1.7 × 10⁻¹⁴ m.

Common errors

• Using the sphere’s surface instead of its centre for r.
• Using E = V/d for a point charge, or E = Q/(4πε₀r²) between plates.
• Forgetting to square r or convert units.
• Adding potentials as vectors, or fields as scalars, from two charges.
• Treating ε as ε₀ in a capacitor with a dielectric.

Official syllabus

OCR Level 3 Advanced GCE in Physics A (H556), Version 3.0 (March 2026), published by OCR: section 6.2 (Electric fields), learning outcomes 6.2.1(a)-(d), 6.2.2(a)-(d), 6.2.3(a)-(c) and 6.2.4(a)-(e).