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

Condensed notes for section 6.2 (Electric fields) of OCR A Level Physics A (H556), Version 3.0 (March 2026), first assessment 2017. Use the full study guide for explanations and worked examples, then try the practice questions. Links: OCR A Level Physics hub, printable checklist and free diagnostics.

Definitions

• Electric field strength, E: force per unit positive charge, E = F/q. Unit N C⁻¹ (= V m⁻¹). A vector; direction of the force on a positive charge. (6.2.1(d))
• Field line: a line whose tangent gives the direction of the force on a small positive charge. Lines start on positive, end on negative; closer lines mean a stronger field; lines never cross. (6.2.1(c))
• Electric potential at a point: work done bringing unit positive charge from infinity to that point. Zero at infinity. Scalar. Unit V = J C⁻¹. (6.2.4(a))
• Permittivity, ε: property of the material between plates, ε = εᵣε₀. Relative permittivity εᵣ is at least 1. (6.2.3(b))
• Point-charge model: a uniformly charged sphere acts, for points outside, as if its charge were at its centre. (6.2.1(b))

Formulae

Result	Equation	Spec
Field strength	E = F/q	6.2.1(d)
Coulomb’s law	F = Qq/(4πε₀r²)	6.2.2(a)
Field of a point charge	E = Q/(4πε₀r²)	6.2.2(b)
Uniform field	E = V/d	6.2.3(a)
Capacitor, vacuum or air	C = ε₀A/d	6.2.3(b)
Capacitor with material	C = εA/d, ε = εᵣε₀	6.2.3(b)
Potential of a point charge	V = Q/(4πε₀r)	6.2.4(b)
Isolated sphere	C = 4πε₀R	6.2.4(c)
Potential energy	E_p = Vq = Qq/(4πε₀r)	6.2.4(e)

Data used in the examples: ε₀ = 8.85 × 10⁻¹² F m⁻¹; elementary charge 1.60 × 10⁻¹⁹ C; masses of electron (9.11 × 10⁻³¹ kg) and proton (1.67 × 10⁻²⁷ kg).

Methods in steps

Force between two charges

1. Convert units (nC to 10⁻⁹ C, cm to m).
2. Use the sizes of Q and q in F = Qq/(4πε₀r²).
3. State direction in words: repel (like), attract (unlike).

Field from two charges at a point

1. Find the size and direction of each field separately.
2. Add as vectors: along a line, subtract if opposite directions; at right angles, use Pythagoras.
3. Zero-field point between like charges: equate Q₁/x² and Q₂/(d − x)².

Charged particle in a uniform field

1. E = V/d, F = qE, a = F/m.
2. Along the field from rest: qV = ½mv².
3. Entering across the field: horizontal speed constant, time t = length/speed, sideways displacement y = ½at².

Work done and potential energy

1. V = Q/(4πε₀r) at each radius, keeping the sign of Q.
2. W = qΔV (or the change in Qq/(4πε₀r)).
3. Positive work done on the charge when like charges are pushed together.

Must-know distinctions

Compare	Point or spherical charge	Parallel plates
Field pattern	radial lines	parallel, evenly spaced lines
Field strength	E ∝ 1/r², varies with r	E = V/d, same at every point in the gap
Equation	E = Q/(4πε₀r²)	E = V/d

Compare	Gravitational	Electric
Force	F = Gm₁m₂/r², always attractive	F = Qq/(4πε₀r²), attractive or repulsive
Field strength	force per unit mass	force per unit charge
Common ground	inverse square of distance, radial field, potential zero at infinity	same

Field E and potential V: E is a vector, V is a scalar. Add fields with direction; add potentials as numbers with signs.

Fields as forms of field (6.2.2(d)): gravitational, electric and magnetic fields each give rise to a force on the right kind of object. Link to section 5.4 and, for magnetic, section 6.3.

Area under a force-distance graph (6.2.4(d)): work done moving the charge between the two distances. Estimate by trapezia (over-estimates on this curve) or squares.

Worked reminders

Isolated sphere derivation: V = Q/(4πε₀R) at the surface, C = Q/V, so C = 4πε₀R. For R = 0.10 m, C = 1.1 × 10⁻¹¹ F.

Dielectric: inserting material with εᵣ = 2.5 raises C by a factor 2.5. Plates of area 5.0 × 10⁻³ m² and gap 1.0 mm: C = 4.4 × 10⁻¹¹ F (air), 1.1 × 10⁻¹⁰ F with material.

Potential energy sign: two protons 1.0 × 10⁻¹⁴ m apart store e²/(4πε₀r) = 2.3 × 10⁻¹⁴ J, positive because they repel.

Fields from two opposite charges: +4.0 nC and −4.0 nC, 0.10 m apart. At the midpoint each charge is 0.050 m away and each gives E = 1.44 × 10⁴ N C⁻¹. Both point towards the negative charge, so they add: 2.9 × 10⁴ N C⁻¹. Between like charges at the midpoint they would cancel.

Uniform field and an electron: plates 1.8 cm apart with 450 V. E = 450/0.018 = 2.5 × 10⁴ V m⁻¹, F = eE = 4.0 × 10⁻¹⁵ N, and a = F/m = 4.4 × 10¹⁵ m s⁻², directed towards the positive plate (opposite to E).

V against r: for a +6.0 nC point charge V = 270 V at 0.20 m and 135 V at 0.40 m. Doubling r halves V (V ∝ 1/r) but quarters E (E ∝ 1/r²).

Graph and diagram reminders

• Radial field of a positive point charge: straight lines directed outwards, equally spaced around the charge; lines get further apart as r increases.
• Two opposite charges: curved lines from the positive to the negative charge; none cross.
• Two charged plates: parallel lines at right angles to the plates in the middle; draw them bending outwards near the edges.
• F against r (point charge): a curve that falls steeply then flattens, never touching the axis. Doubling r quarters F.
• V against r (positive charge): a similar but less steep curve, V ∝ 1/r. For a negative charge V is negative and rises towards zero at infinity.

Links to other sections

• Section 3.1 to 3.3 (motion, forces, Newton’s laws): a charged particle in a uniform field has constant force, so use constant-acceleration equations and resolve motion into two perpendicular components.
• Section 5.4 (gravitational fields): the same structure of ideas; compare inverse square laws, field strength as force per unit property, and potential zero at infinity.
• Section 6.1 (capacitors): Q = VC links charge, p.d. and capacitance; C = εA/d tells you how the plate geometry and material set C.
• Section 6.3 (electromagnetism): moving charges in magnetic fields complete the set of fields giving rise to forces.

Quick self-test

1. Define electric field strength and give its unit.
2. Find the force between two point charges of 1.0 µC, 0.30 m apart.
3. Find E at 0.20 m from a point charge of +2.0 nC.
4. By what factor does E change if r is tripled?
5. Plates 0.80 cm apart have a p.d. of 240 V. Find E.
6. Plates of area 5.0 × 10⁻³ m² are 1.0 mm apart in air. Find C.
7. Find the potential 0.30 m from a point charge of +5.0 nC.
8. Find C for an isolated sphere of radius 0.10 m.
9. Find the potential energy of two protons 1.0 × 10⁻¹⁴ m apart.
10. An electron is accelerated from rest through 500 V. Find its final speed.
11. State what the area under a force-distance graph represents.
12. Give one way in which gravitational and electric fields are alike and one way they differ.

Answers

1. Force per unit positive charge, E = F/q; N C⁻¹ (or V m⁻¹).
2. 0.10 N, repulsive.
3. 4.5 × 10² N C⁻¹, directed away from the charge.
4. Divided by 9 (E ∝ 1/r²).
5. 240/0.0080 = 3.0 × 10⁴ V m⁻¹.
6. 8.85 × 10⁻¹² × 5.0 × 10⁻³ / 1.0 × 10⁻³ = 4.4 × 10⁻¹¹ F.
7. 1.5 × 10² V.
8. 1.1 × 10⁻¹¹ F.
9. 2.3 × 10⁻¹⁴ J.
10. v = √(2eV/m) = 1.3 × 10⁷ m s⁻¹.
11. Work done moving the charge between the two distances.
12. Alike: both follow an inverse square law for a point source. Different: gravity only attracts, but electric forces can attract or repel.

Where marks are usually lost

• Measuring r from the surface of a sphere instead of from its centre.
• Leaving nC, µC, cm or mm unconverted.
• Quoting a force or field without a direction when the question asks for one.
• Adding fields from two charges as plain numbers when they point in different directions.
• Using the sign of a charge inconsistently in V = Q/(4πε₀r) so that the change in potential has the wrong sign.
• Using E = V/d for a radial field, or the inverse square law between plates.
• Writing C = ε₀A/d when a material with εᵣ is present.
• Deriving C = 4πε₀R without stating that V = Q/(4πε₀R) applies at the surface, or without starting from C = Q/V.
• Ignoring the sideways displacement limit: checking whether a charged particle actually leaves the plates.

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).