AQA 7408 · Section 3.7.3 · Paper 2
Electric fields, potential and charged particles
Master Coulomb’s law, radial and uniform fields, electric potential, graph methods and charged-particle deflection. Diagnose the sign convention first, test it in the dual-mode simulator, then complete the original assessment.
Content diagnostic
An electron enters between charged plates
Active revision
Use a three-pass electric-field routine
Fix the source and signs
Field direction is the force direction on a positive test charge. Decide the source-charge sign before considering the moving particle.
Choose the field model
Use inverse-square and 1/r equations for a point charge; use E = V/d only between uniform parallel plates.
Separate field from force
E and V belong to the source field. The test charge sets F = qE, U = qV and whether force follows or opposes the field.
Coulomb’s law
Calculate magnitude, then determine direction
Coulomb’s law applies to point charges in a vacuum. For typical A-level calculations, air can be treated as a vacuum and a charged sphere can be modelled as if its charge were concentrated at its centre.
Permittivity
The permittivity of free space ε₀ links source charge to field strength. A larger permittivity reduces the electrostatic interaction for fixed charges and separation.
Force comparison
For a proton–electron pair, both electrostatic and gravitational forces vary as 1/r², so their ratio is independent of separation:
Electric field strength
Distinguish radial fields from uniform fields
Field lines and definition
Electric field strength is force per unit positive charge, measured in N C⁻¹ or V m⁻¹. Lines leave positive charges, enter negative charges, never cross, and are perpendicular to equipotentials.
Radial field
Magnitude decreases as 1/r². Direction is outward from a positive point charge and inward towards a negative point charge.
Uniform field
Between ideal parallel plates away from the edges, field lines are straight, parallel and equally spaced. E is constant and points from the positive plate to the negative plate.
- Moving charge q through plate separation d changes its electric potential energy by qV.
- For force parallel to displacement, work is Fd.
- Set Fd = qV and divide by qd.
- Since E = F/q, obtain E = V/d.
Potential, work and graphs
Track energy per unit charge
Absolute electric potential is work done per unit positive charge in bringing a small test charge from infinity. Zero potential is defined at infinity.
Radial potential varies as 1/r and has the sign of the source charge. Unlike gravitational potential, electric potential can be positive or negative.
Potential difference is the work done per unit charge. A negative test charge reverses the sign of its potential-energy change.
The signed electric field is the negative potential gradient. The signed area under an Eᵣ–r graph is −ΔV.
Every point on an equipotential has the same V. Moving a charge along it gives ΔV = 0, so the electric field does no work even though a force may still exist.
Useful simulation
Connect signs, equations, graphs and trajectories
The PhysicsUK explorer combines point-charge field lines and E–V graphs with uniform plates and animated electron or proton motion. Every visual change is paired with numerical results.
Source sign
Switch from +Q to −Q while q stays positive. Explain what reverses and what remains a magnitude.
Inverse powers
Double r. Verify that |E| and |F| become one quarter while |V| and |U| become one half.
Uniform plates
Double V, then double d. Predict E and the trajectory before comparing the readouts.
Electron versus proton
Use identical plate settings. Compare force direction and acceleration using charge sign and q/m.
Charged-particle motion
Treat the two directions independently
Particle enters at right angles
- The electric force qE is constant in a uniform field.
- Acceleration a = qE/m is constant parallel or antiparallel to the field.
- No horizontal force means horizontal speed u stays constant, so x = ut.
- Vertical displacement is y = ½at².
- Eliminating t gives y = qEx²/(2mu²), a parabola.
Direction and scale
- A positive particle accelerates with the field.
- An electron accelerates opposite to the field.
- Deflection is proportional to E and plate length squared.
- Deflection is inversely proportional to mass and entry speed squared.
Data and graph skills
Linearise inverse-power field data
Coulomb force or radial field
- For fixed charges, plot F vertically against 1/r² horizontally.
- Gradient = |Q₁Q₂|/(4πε₀).
- For a point source, plot E against 1/r²; gradient = |Q|/(4πε₀).
- Use centre-to-centre r in metres and charge in coulombs.
Electric potential
- Plot V vertically against 1/r horizontally.
- Gradient = Q/(4πε₀), including its sign.
- A non-zero intercept suggests a reference offset or another potential contribution.
- Potential is scalar, so contributions add algebraically rather than vectorially.
Misconception repair
Eight ideas that commonly lose marks
- “Field direction follows an electron.” Field direction is defined using a positive test charge; an electron’s force is opposite.
- “A negative source has a negative field strength everywhere.” Field strength is a vector. State direction; use signed components only after defining an axis.
- “E and V are both vectors.” E is a vector; V is a scalar.
- “E = V/d works around a point charge.” It applies to a uniform field. A radial field uses E = Q/(4πε₀r²).
- “E and V both follow an inverse-square law.” E ∝ 1/r², but V ∝ 1/r.
- “Use the gap between charged-sphere surfaces.” Coulomb’s law uses centre-to-centre separation.
- “No work along an equipotential means no force exists.” Force can exist but is perpendicular to the displacement, and ΔV = 0.
- “A proton and electron curve equally because |q| is equal.” Their force magnitudes match in one field, but acceleration depends on q/m.
Original exam-style assessment
41 marks with coded mark schemes
Attempt every question before opening its mark scheme. All constants and data needed for each answer are supplied. No response requires drawing, plotting or annotating; graph and trajectory marks are awarded through calculations and written reasoning.
1. Two small charged spheres in air carry charges +3.2 μC and −1.8 μC. Their centres are 0.240 m apart. Calculate the magnitude of the force between them and state whether it is attractive or repulsive. Treat air as a vacuum and use ε₀ = 8.85 × 10⁻¹² F m⁻¹.[4 marks]
Mark scheme
- (M1) Uses F = |Q₁Q₂|/(4πε₀r²).
- (M1) Substitutes the charges in coulombs and centre-to-centre distance 0.240 m.
- (A1) F = 0.899 N, accept 0.90 N.
- (B1) The force is attractive because the charges have opposite signs.
Exact AQA content: 3.7.3.1(a), 3.7.3.1(b), 3.7.3.1(c)
2. An electron and a proton are separated by distance r. Show that the ratio of the electrostatic force to the gravitational force is independent of r, then calculate the ratio. Use k = 8.99 × 10⁹ N m² C⁻², e = 1.60 × 10⁻¹⁹ C, G = 6.67 × 10⁻¹¹ N m² kg⁻², mₚ = 1.67 × 10⁻²⁷ kg and mₑ = 9.11 × 10⁻³¹ kg.[4 marks]
Mark scheme
- (M1) Writes Fₑ = ke²/r² and Fᵍ = Gmₚmₑ/r².
- (B1) Dividing cancels r², so the ratio is independent of separation.
- (M1) Uses Fₑ/Fᵍ = ke²/(Gmₚmₑ).
- (A1) Fₑ/Fᵍ = 2.27 × 10³⁹, so the electrostatic force is vastly larger.
Exact AQA content: 3.7.3.1(e)
3. A point charge of −6.0 μC is isolated in air. Calculate the electric field strength 0.150 m from the charge, state its direction and calculate the force on a +2.0 nC test charge placed there. Use k = 8.99 × 10⁹ N m² C⁻².[4 marks]
Mark scheme
- (M1) Uses E = k|Q|/r².
- (A1) E = 2.40 × 10⁶ N C⁻¹.
- (B1) The field is directed radially inward, towards the negative source charge.
- (A1) F = qE = 4.80 × 10⁻³ N inward because the test charge is positive.
Exact AQA content: 3.7.3.2(a), 3.7.3.2(b), 3.7.3.2(f)
4. Two parallel plates have a potential difference of 2.40 kV and separation 32.0 mm. Starting from Fd = QΔV, derive E = V/d. Then calculate the field strength and the magnitude and direction of the force on an electron. Use e = 1.60 × 10⁻¹⁹ C.[4 marks]
Mark scheme
- (M1) Divides Fd = QΔV by Qd and uses E = F/Q.
- (A1) Obtains E = ΔV/d for a uniform field.
- (A1) E = 2400/0.0320 = 7.50 × 10⁴ N C⁻¹.
- (A1) |F| = eE = 1.20 × 10⁻¹⁴ N, opposite to the electric-field direction because the electron is negative.
Exact AQA content: 3.7.3.2(b), 3.7.3.2(c), 3.7.3.2(d)
5. An electron enters a uniform electric field at right angles to the field with horizontal speed 3.00 × 10⁷ m s⁻¹. The field strength is 1.50 × 10⁴ N C⁻¹ and the plate length is 0.0800 m. Describe the trajectory, calculate the vertical deflection while between the plates, and state the deflection if the entry speed is doubled. Use e = 1.60 × 10⁻¹⁹ C and mₑ = 9.11 × 10⁻³¹ kg.[4 marks]
Mark scheme
- (B1) Horizontal velocity remains constant while the electron has constant acceleration opposite to the field.
- (B1) The resulting trajectory is parabolic and bends towards the positive plate.
- (M1) Uses t = L/v and y = ½(eE/mₑ)t² to obtain y = 9.37 × 10⁻³ m.
- (A1) Doubling speed halves transit time, so deflection becomes one quarter: 2.34 × 10⁻³ m.
Exact AQA content: 3.7.3.2(b), 3.7.3.2(c), 3.7.3.2(e)
6. Define absolute electric potential. Explain why the potential around an isolated negative point charge is negative, and state the work done by the electric field when a test charge moves along one equipotential surface.[4 marks]
Mark scheme
- (B1) Electric potential is work done per unit positive charge in bringing a small test charge from infinity to the point.
- (B1) Absolute potential is defined as zero at infinity.
- (B1) V = Q/(4πε₀r), so the potential has the sign of the negative source charge.
- (B1) No work is done along an equipotential because ΔV = 0, equivalently the displacement is perpendicular to the field.
Exact AQA content: 3.7.3.3(a), 3.7.3.3(d), 3.7.3.3(e)
7. A −3.0 nC particle is moved slowly from 0.120 m to 0.300 m from a +5.0 μC point charge. Calculate the change in electric potential and the change in potential energy. State whether the electric field does positive or negative work. Use k = 8.99 × 10⁹ N m² C⁻².[4 marks]
Mark scheme
- (M1) Uses V = kQ/r at each position.
- (A1) ΔV = V₂ − V₁ = −2.25 × 10⁵ V.
- (A1) ΔU = qΔV = +6.74 × 10⁻⁴ J.
- (B1) The electric field does negative work; +6.74 × 10⁻⁴ J of external work is required for the slow move.
Exact AQA content: 3.7.3.3(b), 3.7.3.3(c), 3.7.3.3(e)
8. Graphs of signed radial field Eᵣ and potential V against radius are drawn for a positive point charge. At radius A, Eᵣ = 360 kN C⁻¹ and V = 180 kV. Radius B is twice radius A. Determine Eᵣ and V at B, the potential change Vᴮ − Vᴬ, and the signed area under the Eᵣ–r graph from A to B.[4 marks]
Mark scheme
- (A1) Eᵣ at B = 360/2² = 90 kN C⁻¹.
- (A1) V at B = 180/2 = 90 kV.
- (A1) Vᴮ − Vᴬ = −90 kV.
- (A1) Using ∫Eᵣdr = −ΔV, the signed area is +90 kV, equivalent to +90 kJ C⁻¹.
Exact AQA content: 3.7.3.3(e), 3.7.3.3(f), 3.7.3.3(g)
9. An isolated conducting sphere has radius 0.040 m and charge −2.2 μC. Treat its charge as concentrated at the centre. Calculate the electric potential and electric field strength at its surface, including field direction, and calculate the potential energy of a proton at the surface. Use k = 8.99 × 10⁹ N m² C⁻² and e = 1.60 × 10⁻¹⁹ C.[4 marks]
Mark scheme
- (A1) V = kQ/r = −4.94 × 10⁵ V.
- (M1) Uses E = k|Q|/r².
- (A1) E = 1.24 × 10⁷ N C⁻¹, directed towards the sphere.
- (A1) U = qV = −7.91 × 10⁻¹⁴ J for the proton.
Exact AQA content: 3.7.3.1(d), 3.7.3.2(f), 3.7.3.3(e)
10. A negatively charged oil droplet of mass 4.59 × 10⁻¹⁵ kg is stationary between horizontal parallel plates separated by 8.00 mm. The upper plate is positive and the potential difference is 450 V. Calculate the magnitude and sign of the droplet charge and determine the number of excess electrons. Use g = 9.81 m s⁻² and e = 1.60 × 10⁻¹⁹ C.[5 marks]
Mark scheme
- (A1) E = V/d = 450/0.00800 = 5.625 × 10⁴ N C⁻¹ downward.
- (B1) For equilibrium, the upward electric force equals the downward weight: |q|E = mg.
- (M1) Uses |q| = mg/E.
- (A1) |q| = 8.00 × 10⁻¹⁹ C and q is negative so the electric force is upward.
- (A1) Number of excess electrons = |q|/e = 5.
Exact AQA content: 3.7.3.2(b), 3.7.3.2(c)
Specification coverage
AQA 7408 section 3.7.3
- 3.7.3.1(a) Use Coulomb's law for point charges in a vacuum: F = (1/(4πε0))(Q1Q2/r²).
- 3.7.3.1(b) Use permittivity of free space ε0.
- 3.7.3.1(c) Treat air as a vacuum for force calculations between charges.
- 3.7.3.1(d) Model a charged sphere as having charge at its centre.
- 3.7.3.1(e) Compare magnitudes of gravitational and electrostatic forces between subatomic particles.
- 3.7.3.2(a) Represent electric fields using electric field lines.
- 3.7.3.2(b) Define electric field strength E = F/Q.
- 3.7.3.2(c) Use E = V/d for a uniform electric field.
- 3.7.3.2(d) Derive E = V/d from Fd = QΔV.
- 3.7.3.2(e) Describe the trajectory of a charged particle entering a uniform electric field initially at right angles.
- 3.7.3.2(f) Use E = (1/(4πε0))(Q/r²) for a radial electric field.
- 3.7.3.3(a) Define absolute electric potential with zero at infinity.
- 3.7.3.3(b) Define and use electric potential difference.
- 3.7.3.3(c) Use ΔW = QΔV for work done moving charge in an electric field.
- 3.7.3.3(d) Use equipotential surfaces and state that no work is done moving charge along an equipotential.
- 3.7.3.3(e) Use V = (1/(4πε0))(Q/r) for electric potential in a radial field.
- 3.7.3.3(f) Interpret graphs of electric field strength E and potential V against radius r.
- 3.7.3.3(g) Use E = ΔV/Δr and area under an E-r graph to find ΔV.
Continue through AQA Fields and Paper 2
Return to the AQA 3.7 fields module hub. Compare inverse-square and potential ideas with gravitational fields, potential and orbits, then connect electric potential energy with capacitors, charge and energy. Revisit particles and radiation for electron and proton properties. Continue through the AQA Paper 2 revision hub, problem-solving practice and MCQ practice.
Written against all 18 AQA Physics 7408 points in section 3.7.3. Questions, values and contexts are original. The parallel-plate stimulus is an original generated PNG; the interactive model is the original PhysicsUK electric-field explorer.