OCR A Physics · Module 6: Particles and medical physics
Electric fields: practice questions with answers
6 exam-style questions on electric fields. Try each one on paper first, then open the answer to see the full working and the common slips that cost marks.
- Unit prefixes2 marks
Two parallel plates 2.0 cm apart have a potential difference of 3.0 kV. Calculate the force on an electron between the plates.
F = N
Show the answer and working
Answer F = 2.40 × 10⁻¹⁴ N
E = V/d = 3000 ÷ 0.020 = 1.5 × 10⁵ V m⁻¹. F = eE = 1.6 × 10⁻¹⁹ × 1.5 × 10⁵ = 2.40 × 10⁻¹⁴ N.
- Got 2.40 × 10⁻¹⁶ N? You used the separation in centimetres. d = 0.020 m, so E = V ÷ d = 1.5 × 10⁵ V m⁻¹ and F = eE = 2.40 × 10⁻¹⁴ N.
- Got 2.40 × 10⁻¹⁷ N? You used 3.0 V instead of 3000 V. E = 3000 ÷ 0.020 = 1.5 × 10⁵ V m⁻¹ and F = eE = 2.40 × 10⁻¹⁴ N.
- Using the right equation3 marks
A +3.0 nC point charge and a -5.0 nC point charge are 4.0 cm apart in air. Calculate the size of the force between them, in μN.
F = μN
Show the answer and working
Answer F = 84.3 μN
F = Q₁Q₂ ÷ (4πε₀r²) = (3.0 × 10⁻⁹ × 5.0 × 10⁻⁹) ÷ (4π × 8.85 × 10⁻¹² × 0.040²) = 84.3 × 10⁻⁶ N = 84.3 μN.
- Got 3.37 μN? You did not square the separation. F = Q₁Q₂ ÷ (4πε₀r²) with r² = 1.6 × 10⁻³ m², giving 84.3 μN.
- Got 1060 μN? You left out the 4π in the denominator. F = Q₁Q₂ ÷ (4πε₀r²) = 84.3 μN.
- Using the right equation3 marks
Calculate the electric field strength 0.15 m from a point charge of +8.0 nC, in kV m⁻¹.
E = kV m⁻¹
Show the answer and working
Answer E = 3.2 kV m⁻¹
E = Q ÷ (4πε₀r²) = 8.0 × 10⁻⁹ ÷ (4π × 8.85 × 10⁻¹² × 0.15²) = 3.2e+03 V m⁻¹ = 3.2 kV m⁻¹.
- Got 0.48 kV m⁻¹? That is the potential (in kV), not the field. Field strength uses r²: E = Q ÷ (4πε₀r²) = 3.2 kV m⁻¹.
- Got 3200 kV m⁻¹? That is in V m⁻¹. Divide by 1000 to give kV m⁻¹: 3.2 kV m⁻¹.
- eV and joules3 marks
An electron is accelerated from rest through a potential difference of 2.5 kV in a vacuum. Calculate its final speed.
v = m s⁻¹
Show the answer and working
Answer v = 2.96 × 10⁷ m s⁻¹
eV = ½mv², so v = √(2eV/m) = √(2 × 1.6 × 10⁻¹⁹ × 2500 ÷ 9.11 × 10⁻³¹) = 2.96 × 10⁷ m s⁻¹.
- Got 2.10 × 10⁷ m s⁻¹? You left out the 2 in ½mv² = eV. v = √(2eV ÷ m) = 2.96 × 10⁷ m s⁻¹.
- Got 9.37 × 10⁵ m s⁻¹? You used 2.5 V rather than 2500 V. v = √(2 × 1.6 × 10⁻¹⁹ × 2500 ÷ 9.11 × 10⁻³¹) = 2.96 × 10⁷ m s⁻¹.
- Using the right equation1 mark
The potential difference between two parallel plates is kept constant while the plate separation is doubled. What happens to the field strength between them?
- A It doubles.
- B It stays the same.
- C It halves.
Show the answer and working
Answer C: It halves.
E = V/d with V fixed, so doubling d halves the field strength.
- Why not A? E = V/d, so doubling d halves E; it does not double.
- Why not B? E depends on d, so it changes.
- Using the right equation1 mark
Electric field lines around a positive point charge are drawn closer together near the charge. What does this show?
- A The field is stronger there.
- B The field is weaker there.
- C The potential is lower there.
Show the answer and working
Answer A: The field is stronger there.
The density of field lines is proportional to the field strength.
- Why not B? Closer lines mean a stronger field.
- Why not C? Near a positive charge the potential is higher, not lower.