Uniform fields between parallel plates
Key idea: Between large parallel plates, away from the edges, the field is uniform with E = V/d. A charge feels force qE; negative charges are forced against the field.
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Guided lessons · coming soon
Guided lesson
Guided steps
- Read how the central field relates to the potential gradient.
- Predict the field before changing the gap and probe position.
- Test your predictions using the gap, probe and edge controls.
- Explain the field readings, potential change and curved edge lines.
- Follow the gap conversion, field calculation and electron force.
- Find the gap, field, electron force and its direction.
- Compare the particles’ forces and accelerations.
- Find the voltage and proton acceleration, then compare your working.
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ContinueThe field of a point charge changes from place to place, which makes motion in it hard to predict. Two parallel metal plates connected to a supply give something much simpler: a field with nearly constant strength and direction in the central region between them. That is why parallel plates are used to accelerate and steer beams of charged particles.
The uniform field between parallel plates
Between two large parallel plates with a potential difference V across them, the field in the central region is uniform: it has the same strength and direction at every point. The equipotentials are evenly spaced planes parallel to the plates, and the field lines are parallel and evenly spaced, from the positive plate to the negative plate.
Because the field is uniform, the potential falls at a steady rate across the gap. The field strength is the size of the potential gradient, so for plates a distance d apart
This holds when the plates are much wider than the gap between them. Near the edges the field lines curve and the field is no longer uniform; this is called fringing.
Check your understanding 1
Before using the model, take a p.d. of 500 V across a 4.0 cm gap. Predict the central field strength. If only the gap doubles, by what factor does the field change? Does changing the probe position within the central region change the field?
Show the steps Optional working
Show answer
E = 500/0.040 = 1.25 × 10⁴ V m⁻¹. Doubling the gap gives half the field. In the ideal central region, moving the probe changes potential but not the uniform field strength.
Test the field and gap predictions
Select Parallel plates and leave Include edge effects off under More settings. Set p.d. between the plates to 500 V, compare Plate separation values 4.0 and 8.0 cm, and read E at the probe. Move the probe within the gap while keeping it far from the plate ends. Then turn on edge effects and compare the centre with the ends.
Parallel plates 4.0 centimetres apart with a p.d. of 500 volts, so E = 12500 volts per metre between them. An electron beam enters at 40.0 × 10⁶ metres per second and leaves deflected by 0.988 centimetres. The probe reads E = 12500 newtons per coulomb and V = 375 volts.
- E at the probe
- 12500 N C⁻¹
- Direction of E (from +x)
- −90 °
- V at the probe
- 375 V
- Distance to nearest charge
- 4.12 cm
- V / d
- 12500 V m⁻¹
- Deflection leaving the plates
- 0.99 cm
Try this
0 of 4 donePut two like charges on the stage and find the point where E = 0. (not done yet)
Between like charges the two fields point opposite ways and cancel at a neutral point, nearer the smaller charge.
With a single charge, record E at three or more distances. (not done yet)
E against 1/r² is a straight line through the origin: E = kQ/r², so doubling r makes E a quarter.
Make the field between the plates 2.0 × 10⁴ V m⁻¹. (not done yet)
E = V/d: 400 V across 2.0 cm gives the same field as 800 V across 4.0 cm.
Make the beam leave the plates just clear of a plate. (not done yet)
Inside the plates the particle has a constant acceleration qE/m across the field, so its path is a parabola: y = qEL²/(2mv²).
You’ve tried every task.
Your readings
| # | r / cm | E / N C−1 | V / V | Remove |
|---|---|---|---|---|
| No readings yet. Set up a measurement, then record it. | ||||
Check your understanding 2
Compare the two field readings and explain why moving the probe can change potential while leaving the central field unchanged. What changes near the edges when fringing is enabled?
Show answer
The ideal readings are about 1.25 × 10⁴ and 6.25 × 10³ V m⁻¹. Potential varies steadily across the gap, but its gradient is constant, so the central field is unchanged by probe position. With fringing, lines curve near the ends and the field becomes non-uniform; the central-region estimate remains the comparison model.
Force on a charge between the plates
A charge q in the field feels a force
Since E is the same throughout the central region, so is the force, and the charge has a constant acceleration a = qE/m (when its weight is negligible). A positive charge is pushed towards the negative plate, along the field. A negative charge is pushed towards the positive plate, against the field.
Worked example 1
Field and force between plates
Problem
Parallel plates 0.020 m apart have a p.d. of 500 V across them. Find the field strength between them, and the force on an electron in the central region.
Show full solution
Field
Method
Divide the p.d. by the separation.Reason
The field between the plates is uniform.Working
Answer
E = 500/0.020 = 2.5 × 10⁴ V m⁻¹, from the positive plate towards the negative plate
Force
Working
Answer
F = eE = (1.60 × 10⁻¹⁹)(2.5 × 10⁴) = 4.0 × 10⁻¹⁵ N, towards the positive plate, because the electron is negative
Quick check
The force is the same anywhere in the central region; it does not depend on how close the electron is to either plate.
Guided practice 2
A narrower gap
Problem
Parallel plates 8.0 mm apart have a p.d. of 240 V across them. Find the field strength and the force on an electron between them.
Find E and F
Show the steps Optional working
Show the steps Optional working
Hints
Hint 1
units
Convert 8.0 mm to metres before dividing.
Show full solution
Field and force
Working
Answer
E = 240/(8.0 × 10⁻³) = 3.0 × 10⁴ V m⁻¹ and F = (1.60 × 10⁻¹⁹)(3.0 × 10⁴) = 4.8 × 10⁻¹⁵ N, towards the positive plate
Check your understanding 3 A proton and an electron
A proton and an electron are placed in turn between the same pair of plates. Compare the forces on them and their accelerations.
Show answer
The forces have the same size, eE, in opposite directions: the proton is pushed along the field and the electron against it. Their accelerations are very different, because a = eE/m and the electron’s mass is about 1/1840 of the proton’s.
Exam-style question 1
A proton between plates
Examination question
Two large parallel plates are 12 mm apart. The field between them is 3.0 × 10⁴ V m⁻¹.
(a) Calculate the p.d. across the plates. [2 marks]
(b) Calculate the acceleration of a proton between the plates. The mass of a proton is 1.67 × 10⁻²⁷ kg. [2 marks]
Work on paper, then enter your results
Show solution step by step
(a) P.d.
2 marksMethod
Rearrange E = V/d, with d in metres.Reason
The field is in V m⁻¹.Working
Answer
V = Ed = (3.0 × 10⁴)(0.012) = 360 V
(b) Acceleration
2 marksWorking
Answer
a = eE/m = ((1.60 × 10⁻¹⁹)(3.0 × 10⁴))/(1.67 × 10⁻²⁷) = 2.9 × 10¹² m s⁻², along the field
Self-mark with the mark scheme
Tick each mark point that your answer includes.
Tick each point only if your answer states it clearly.
Common mistakes
- Using d in millimetres in E = V/d.
- Thinking the field is stronger near one plate; it is uniform in the central region.
- Applying E = V/d near the edges of the plates, or to plates that are not much wider than their separation.
- Giving the force on an electron the direction of the field.
- Expecting a proton and an electron to have the same acceleration because they feel the same size of force.
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Syllabus and review details
- GCE A-Level H2 Physics 2027 · 2027
Content Overview, PDF pages 9–10; Subject Content, PDF pages 11–30
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