Specification: AQA A-level Physics (7408) 3.7.3.2 | Paper: 2 | Data and formulae booklet: given in the exam
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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).
Uniform electric field: a field where the field strength has the same magnitude and direction at every point. Between parallel plates the field lines are parallel, equally spaced and point from the positive plate to the negative plate.
Work done on a charge: when a charge Q moves through a potential difference V, the work done is W = QV. An electron accelerated from rest through V gains kinetic energy eV.
The equation
electric field strength = potential difference ÷ plate separation E = V ÷ d and force = field strength × charge F = EQ
| Quantity | Symbol | Unit |
|---|---|---|
| electric field strength | E | volts per metre (V m⁻¹), equivalent to N C⁻¹ |
| potential difference between the plates | V | volts (V) |
| separation of the plates | d | metres (m) |
| force on the charge | F | newtons (N) |
| charge | Q | coulombs (C) |
Rearranged: V = Ed | d = V ÷ E | Q = F ÷ E | combined: F = QV ÷ d | Convert mm or cm to m and kV to V before you substitute. The field is the same everywhere between the plates, so the force does not depend on where the charge is.
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) Two parallel plates are 0.050 m apart and have a potential difference of 2000 V across them. Calculate the electric field strength between the plates.
Show solution
F: E = V ÷ d
I: E = 2000 ÷ 0.050
F: E is already the subject
A: 40 000 V m⁻¹ = 4.0 × 104 V m⁻¹
Q2 (Core) An alpha particle has a charge of +3.2 × 10−19 C. It is in a uniform electric field of strength 5.0 × 104 V m⁻¹. Calculate the force on it.
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F: F = EQ
I: F = 5.0 × 104 × 3.2 × 10−19
F: F is already the subject
A: 1.6 × 10−14 N, in the direction of the field
Q3 (Core) A uniform field of strength 2.5 × 105 V m⁻¹ is needed between two parallel plates 8.0 mm apart. Calculate the potential difference that must be applied across the plates.
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Convert first: 8.0 mm = 8.0 × 10−3 m
F: E = V ÷ d
I: 2.5 × 105 = V ÷ 8.0 × 10−3
F: V = Ed = 2.5 × 105 × 8.0 × 10−3
A: 2000 V (2.0 kV)
Q4 (Core) Two parallel plates 12 mm apart have a potential difference of 1.5 kV across them. Calculate the force on an electron between the plates. (e = 1.60 × 10−19 C)
Show solution
Convert first: 12 mm = 0.012 m; 1.5 kV = 1500 V
Step 1 – field strength
E = V ÷ d = 1500 ÷ 0.012 = 125 000 V m⁻¹
Step 2 – force
F: F = EQ
I: F = 125 000 × 1.60 × 10−19
F: F is already the subject
A: 2.0 × 10−14 N, towards the positive plate
Watch out: leaving d in mm gives a field 1000 times too small. Always convert the separation to metres.
Q5 (Exam) A proton is released from rest next to the positive plate of a pair of parallel plates 2.0 cm apart with a potential difference of 500 V. Calculate (a) the acceleration of the proton and (b) the time it takes to reach the negative plate. Give both answers to 2 significant figures and ignore gravity. (e = 1.60 × 10−19 C, mass of proton = 1.67 × 10−27 kg)
Show solution
Convert first: 2.0 cm = 0.020 m
Step 1 – field and force
E = V ÷ d = 500 ÷ 0.020 = 25 000 V m⁻¹; F = EQ = 25 000 × 1.60 × 10−19 = 4.0 × 10−15 N
Step 2 – (a) acceleration
F: F = ma
I: 4.0 × 10−15 = 1.67 × 10−27 × a
F: a = 4.0 × 10−15 ÷ 1.67 × 10−27 = 2.395… × 1012 m s⁻²
A: 2.4 × 1012 m s⁻² (2 s.f.)
Step 3 – (b) time, using s = ut + ½at² with u = 0
t = √(2s ÷ a) = √(2 × 0.020 ÷ 2.395… × 1012) = 1.29… × 10−7 s
A: 1.3 × 10−7 s (2 s.f.)
Q6 (Exam) An electron is accelerated from rest through a potential difference of 3000 V. Calculate (a) the work done on the electron and (b) its final speed, to 2 significant figures. (e = 1.60 × 10−19 C, mass of electron = 9.11 × 10−31 kg)
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Step 1 – (a) work done
F: W = QV
I: W = 1.60 × 10−19 × 3000
A: 4.8 × 10−16 J
Step 2 – (b) speed
all the work done becomes kinetic energy: ½mv² = 4.8 × 10−16
F: v = √(2W ÷ m) = √(2 × 4.8 × 10−16 ÷ 9.11 × 10−31)
A: 3.246… × 107 m s⁻¹ = 3.2 × 107 m s⁻¹ (2 s.f.)
Watch out: W = QV depends only on the potential difference, not on the plate separation. Do not calculate E unless the question needs a force.
Q7 (Exam) In a Millikan-style experiment, a negatively charged oil drop of mass 4.9 × 10−15 kg is held stationary between two horizontal plates 15 mm apart when the potential difference across them is 1500 V. (a) State which plate is positive. (b) Calculate the charge on the drop, to 2 significant figures. (c) Find how many excess electrons are on the drop. (g = 9.81 N kg⁻¹, e = 1.60 × 10−19 C)
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Step 1 – (a) direction
the electric force must act upwards to balance the weight; the drop is negative, so it is attracted to the top plate, which is positive
Convert first: 15 mm = 0.015 m
Step 2 – (b) balance the forces
F: mg = EQ = QV ÷ d
I: 4.9 × 10−15 × 9.81 = Q × 1500 ÷ 0.015
F: Q = mgd ÷ V = 4.8069 × 10−14 × 0.015 ÷ 1500
A: 4.8069 × 10−19 C = 4.8 × 10−19 C (2 s.f.), negative
Step 3 – (c) number of electrons
4.8069 × 10−19 ÷ 1.60 × 10−19 = 3.004…, so 3 electrons
Watch out: the charge on a drop must be a whole-number multiple of e. If your answer is not close to a whole number, check your conversions.
Q8 (Exam) A charged dust particle of charge 6.4 × 10−19 C experiences a force of 4.8 × 10−14 N between two parallel plates 4.0 mm apart. (a) Calculate the electric field strength. (b) Calculate the potential difference across the plates. (c) The plate separation is doubled with the potential difference unchanged. Calculate the new force on the particle.
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Step 1 – (a) field strength
F: F = EQ
I: 4.8 × 10−14 = E × 6.4 × 10−19
F: E = 4.8 × 10−14 ÷ 6.4 × 10−19
A: 75 000 V m⁻¹ = 7.5 × 104 V m⁻¹
Step 2 – (b) potential difference
Convert first: 4.0 mm = 0.0040 m
V = Ed = 75 000 × 0.0040 = 300 V
Step 3 – (c) new force
E = V ÷ d, so doubling d halves E, and F = EQ halves too
A: F = 4.8 × 10−14 ÷ 2 = 2.4 × 10−14 N
Q9 (Stretch) An electron travelling horizontally at 2.0 × 107 m s⁻¹ enters the uniform field between two horizontal plates, midway between them, at right angles to the field. The plates are 5.0 cm long and 2.0 cm apart, with a potential difference of 200 V. Calculate the vertical deflection of the electron when it leaves the plates, to 2 significant figures. Ignore gravity. (e = 1.60 × 10−19 C, mass of electron = 9.11 × 10−31 kg)
Show solution
Convert first: 5.0 cm = 0.050 m; 2.0 cm = 0.020 m
Step 1 – field strength and vertical acceleration
E = V ÷ d = 200 ÷ 0.020 = 10 000 V m⁻¹; a = eE ÷ m = 1.60 × 10−19 × 10 000 ÷ 9.11 × 10−31 = 1.756… × 1015 m s⁻²
Step 2 – time between the plates
there is no horizontal force, so the horizontal speed is constant: t = 0.050 ÷ 2.0 × 107 = 2.5 × 10−9 s
Step 3 – vertical deflection
F: s = ut + ½at², with vertical u = 0
I: s = ½ × 1.756… × 1015 × (2.5 × 10−9)²
A: 5.488… × 10−3 m = 5.5 mm (2 s.f.), towards the positive plate. This is less than the 10 mm gap to the plate, so the electron leaves the field.
Watch out: the path is a parabola, like a projectile. Treat the horizontal and vertical motion separately; the field only changes the vertical velocity.
Q10 (Stretch) Electrons are accelerated from rest through a potential difference of 2500 V. They then enter, midway and at right angles to the field, the region between two horizontal plates 6.0 cm long and 12 mm apart, with a potential difference of 180 V across them. Calculate (a) the speed of the electrons as they enter the plates, (b) their vertical deflection on leaving the plates and (c) the angle their path makes with the horizontal as they leave. Give answers to 2 significant figures. Ignore gravity. (e = 1.60 × 10−19 C, mass of electron = 9.11 × 10−31 kg)
Show solution
Step 1 – (a) entry speed
eV = ½mv², so v = √(2eV ÷ m) = √(2 × 1.60 × 10−19 × 2500 ÷ 9.11 × 10−31) = 2.963… × 107 m s⁻¹
A: 3.0 × 107 m s⁻¹ (2 s.f.)
Convert first: 6.0 cm = 0.060 m; 12 mm = 0.012 m
Step 2 – vertical acceleration
E = 180 ÷ 0.012 = 15 000 V m⁻¹; a = eE ÷ m = 1.60 × 10−19 × 15 000 ÷ 9.11 × 10−31 = 2.634… × 1015 m s⁻²
Step 3 – time in the field
t = 0.060 ÷ 2.963… × 107 = 2.024… × 10−9 s
Step 4 – (b) deflection
s = ½at² = ½ × 2.634… × 1015 × (2.024… × 10−9)² = 5.4 × 10−3 m
A: 5.4 mm (2 s.f.), which is less than the 6.0 mm to the plate
Step 5 – (c) angle
vertical velocity = at = 2.634… × 1015 × 2.024… × 10−9 = 5.334… × 106 m s⁻¹; tan θ = 5.334… × 106 ÷ 2.963… × 107 = 0.18
A: θ = 10.2…° = 10° (2 s.f.)
Watch out: use the accelerating voltage (2500 V) for the entry speed and the deflecting plates (180 V across 12 mm) for the field. Mixing them up is a common error.
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