Specification: AQA A-level Physics (7408) 3.7.3.2 | Paper: 2 | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Electric field strength: the force per unit positive charge acting on a small positive test charge placed at that point, E = F ÷ Q. It is a vector, measured in N C⁻¹ (equivalent to V m⁻¹).
Electric field lines: show the direction of the force on a positive charge, so they point away from positive charges and towards negative charges. A negative charge feels a force opposite to the field direction.
Radial field: the field of a point charge, or outside a uniformly charged sphere, E = Q ÷ (4πε₀r²), where r is measured from the centre. E obeys an inverse square law.
The equation
electric field strength = force ÷ charge E = F ÷ Q and, for a radial field, E = charge ÷ (4π × permittivity of free space × distance²) E = Q ÷ (4πε₀r²)
| Quantity | Symbol | Unit |
|---|---|---|
| electric field strength | E | newtons per coulomb (N C⁻¹), equivalent to V m⁻¹ |
| force on the charge | F | newtons (N) |
| charge (test charge in E = F ÷ Q; source charge in the radial equation) | Q | coulombs (C) |
| distance from the centre of the source charge | r | metres (m) |
| permittivity of free space | ε₀ | farads per metre (F m⁻¹) |
Rearranged: F = EQ | Q = F ÷ E | Q = 4πε₀r²E | r = √(Q ÷ 4πε₀E) | In the radial equation, Q is the charge making the field, not the charge placed in it. Measure r from the centre of a charged sphere.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A small test charge of +2.0 µC experiences a force of 0.050 N to the right at a point in an electric field. Calculate the electric field strength at that point and state its direction.
Show solution
Convert first: 2.0 µC = 2.0 × 10−6 C
F: E = F ÷ Q
I: E = 0.050 ÷ 2.0 × 10−6
F: E is already the subject
A: E = 25 000 N C⁻¹ = 2.5 × 104 N C⁻¹ to the right (the same direction as the force, because the test charge is positive)
Q2 (Core) Calculate the electric field strength 30 cm from a point charge of +8.0 nC in air, to 2 significant figures, and state its direction. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 8.0 nC = 8.0 × 10−9 C; 30 cm = 0.30 m
F: E = Q ÷ (4πε₀r²)
I: E = 8.0 × 10−9 ÷ (4π × 8.85 × 10−12 × 0.30²)
F: E = 8.0 × 10−9 ÷ 1.000… × 10−11
A: E = 799.2… N C⁻¹ = 8.0 × 102 N C⁻¹ (2 s.f.), radially away from the positive charge
Q3 (Core) The electric field between two parallel plates has a strength of 3.0 kV m⁻¹, directed vertically downwards. Calculate the size of the force on an electron in this field and state its direction. (e = 1.60 × 10−19 C)
Show solution
Convert first: 3.0 kV m⁻¹ = 3.0 × 103 V m⁻¹ = 3.0 × 103 N C⁻¹
F: E = F ÷ Q
I: 3.0 × 103 = F ÷ 1.60 × 10−19
F: F = 3.0 × 103 × 1.60 × 10−19
A: 4.8 × 10−16 N, vertically upwards
Watch out: the field direction is the force on a positive charge. An electron is negative, so the force on it is opposite to the field.
Q4 (Core) The electric field strength 12 cm from the centre of a small charged sphere is 45 kN C⁻¹, directed away from the sphere. Calculate the charge on the sphere, to 2 significant figures. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 45 kN C⁻¹ = 4.5 × 104 N C⁻¹; 12 cm = 0.12 m
F: E = Q ÷ (4πε₀r²)
I: 4.5 × 104 = Q ÷ (4π × 8.85 × 10−12 × 0.12²)
F: Q = 4.5 × 104 × 4π × 8.85 × 10−12 × 0.12²
A: Q = 7.20… × 10−8 C = +7.2 × 10−8 C (+72 nC, 2 s.f.); positive because the field points away from the sphere
Q5 (Exam) A metal sphere of radius 5.0 cm carries a charge of +30 nC spread uniformly over its surface. Calculate the electric field strength (a) at the surface of the sphere and (b) at a point 10 cm from the surface, both to 2 significant figures. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 30 nC = 3.0 × 10−8 C; at the surface r = 0.050 m; 10 cm from the surface r = 0.050 + 0.10 = 0.15 m
F: E = Q ÷ (4πε₀r²)
I: (a) E = 3.0 × 10−8 ÷ (4π × 8.85 × 10−12 × 0.050²); (b) E = 3.0 × 10−8 ÷ (4π × 8.85 × 10−12 × 0.15²)
F: E is already the subject
A: (a) 1.07… × 105 N C⁻¹ = 1.1 × 105 N C⁻¹; (b) 1.19… × 104 N C⁻¹ = 1.2 × 104 N C⁻¹ (2 s.f.), both radially outwards
Watch out: using r = 0.10 m (the distance from the surface) gives 2.7 × 104 N C⁻¹. Always measure r from the centre of the sphere.
Q6 (Exam) In a hydrogen atom the electron is 5.3 × 10−11 m from the proton. Calculate the electric field strength due to the proton at the position of the electron, and use it to find the force on the electron, both to 2 significant figures. (e = 1.60 × 10−19 C, ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Step 1 – field from the proton
F: E = Q ÷ (4πε₀r²)
I: E = 1.60 × 10−19 ÷ (4π × 8.85 × 10−12 × (5.3 × 10−11)²)
A: E = 5.12… × 1011 N C⁻¹ = 5.1 × 1011 N C⁻¹ (2 s.f.), away from the proton
Step 2 – force on the electron
F = EQ = 5.12… × 1011 × 1.60 × 10−19 = 8.19… × 10−8 N
A: 8.2 × 10−8 N (2 s.f.), towards the proton (the same answer Coulomb’s law gives)
Q7 (Exam) The electric field strength 0.20 m from a point charge is 2.4 × 104 N C⁻¹. Without calculating the charge, find (a) the field strength 0.50 m from the charge, to 2 significant figures, and (b) the distance from the charge at which the field strength is 6.0 × 103 N C⁻¹.
Show solution
F: E = Q ÷ (4πε₀r²), so E ∝ 1 ÷ r² and E₂ = E₁ × (r₁ ÷ r₂)²
I: (a) E₂ = 2.4 × 104 × (0.20 ÷ 0.50)² = 2.4 × 104 × 0.16
A: (a) 3840 N C⁻¹ = 3.8 × 103 N C⁻¹ (2 s.f.)
F: (b) r₂ = r₁ × √(E₁ ÷ E₂) = 0.20 × √(2.4 × 104 ÷ 6.0 × 103) = 0.20 × √4
A: (b) 0.40 m
Watch out: E is not inversely proportional to r. A quarter of the field strength needs double the distance, not four times the distance.
Q8 (Exam) A proton is released from rest 6.0 cm from a fixed point charge of +4.0 nC in a vacuum. Calculate the electric field strength at the proton’s starting position and the initial acceleration of the proton, both to 2 significant figures. (e = 1.60 × 10−19 C, mₚ = 1.67 × 10−27 kg, ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 4.0 nC = 4.0 × 10−9 C; 6.0 cm = 0.060 m
Step 1 – field strength
F: E = Q ÷ (4πε₀r²)
I: E = 4.0 × 10−9 ÷ (4π × 8.85 × 10−12 × 0.060²)
A: E = 9990.8… N C⁻¹ = 1.0 × 104 N C⁻¹ (2 s.f.)
Step 2 – force and acceleration
F = EQ = 9990.8… × 1.60 × 10−19 = 1.59… × 10−15 N; a = F ÷ mₚ = 1.59… × 10−15 ÷ 1.67 × 10−27
A: a = 9.57… × 1011 m s⁻² = 9.6 × 1011 m s⁻² (2 s.f.), directed away from the charge
Watch out: the acceleration is only the initial value. As the proton moves away, E falls with 1 ÷ r², so you cannot use suvat for its later motion.
Q9 (Stretch) Point charges of +4.0 nC and +9.0 nC are fixed 30 cm apart in air. Find the point on the line between them where the resultant electric field strength is zero, and confirm the field strengths there. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Step 1 – set up the condition
let the point be x from the +4.0 nC charge, so it is (0.30 − x) from the +9.0 nC charge. Between two like charges the fields point in opposite directions, so the resultant is zero where they are equal in size.
F: Q₁ ÷ (4πε₀x²) = Q₂ ÷ (4πε₀(0.30 − x)²); the 4πε₀ cancels
I: 4.0 × 10−9 ÷ x² = 9.0 × 10−9 ÷ (0.30 − x)²
F: square root both sides: 2 ÷ x = 3 ÷ (0.30 − x), so 0.60 − 2x = 3x and x = 0.60 ÷ 5
A: x = 0.12 m (12 cm from the +4.0 nC charge, 18 cm from the +9.0 nC charge)
Step 2 – check
E from +4.0 nC = 4.0 × 10−9 ÷ (4π × 8.85 × 10−12 × 0.12²) = 2497.7… N C⁻¹; E from +9.0 nC = 9.0 × 10−9 ÷ (4π × 8.85 × 10−12 × 0.18²) = 2497.7… N C⁻¹, equal and opposite
Watch out: the zero point is nearer the smaller charge, not at the midpoint. For two like charges it lies between them; for two unlike charges it lies outside, beyond the smaller charge.
Q10 (Stretch) A +3.0 nC point charge and a −2.0 nC point charge are fixed 20 cm apart in air. Point P is midway between them. Calculate (a) the resultant electric field strength at P, to 2 significant figures, giving its direction, and (b) the force on an electron placed at P, to 2 significant figures. (e = 1.60 × 10−19 C, ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 3.0 nC = 3.0 × 10−9 C; 2.0 nC = 2.0 × 10−9 C; P is 0.10 m from each charge
Step 1 – field from each charge
F: E = Q ÷ (4πε₀r²)
I: from +3.0 nC: E = 3.0 × 10−9 ÷ (4π × 8.85 × 10−12 × 0.10²) = 2697.5… N C⁻¹, away from the +3.0 nC charge (towards the −2.0 nC charge)
I: from −2.0 nC: E = 2.0 × 10−9 ÷ (4π × 8.85 × 10−12 × 0.10²) = 1798.3… N C⁻¹, towards the −2.0 nC charge
Step 2 – resultant
both fields point the same way, so they add: E = 2697.5… + 1798.3… = 4495.9… N C⁻¹
A: (a) 4.5 × 103 N C⁻¹ (2 s.f.), towards the −2.0 nC charge
Step 3 – force on the electron
F = EQ = 4495.9… × 1.60 × 10−19 = 7.19… × 10−16 N
A: (b) 7.2 × 10−16 N (2 s.f.), towards the +3.0 nC charge
Watch out: subtracting the two field strengths (giving 9.0 × 102 N C⁻¹) is a common error. Between unlike charges the fields point the same way and add; draw an arrow for each field before combining.
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