Calculate uniform electric fields
Find field strength between charged parallel plates and describe the force and motion of charged particles in that field.
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 18. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use force, charge, work, energy, potential difference, inverse-square relationships and SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Between ideal parallel plates, which way does the electric field point?
What you will learn
- Use E = ΔV/Δd between parallel plates
- Relate a charged particle's force to the uniform field
- Describe how a uniform field changes charged-particle motion
Potential difference sets the field between parallel plates
Away from the plate edges, oppositely charged parallel plates produce an approximately uniform electric field. The field is perpendicular to the plates and directed from the positive plate towards the negative plate.
For a uniform field of magnitude E across plate separation Δd and potential difference magnitude ΔV, E = ΔV/Δd. With volts and metres, E is in V m⁻¹, equivalent to N C⁻¹. Use the potential change in magnitude for the field strength.
Charge sign determines the force and deflection
A charge experiences F = qE and acceleration a = qE/m when the electric force is the resultant force. A positive particle accelerates along the field; a negative particle accelerates against it. The acceleration is constant in an ideal uniform field if the field and charge are constant.
A particle entering with velocity perpendicular to the field keeps its velocity component parallel to the plates while the field changes its perpendicular component. Its path bends towards the force direction; reversing the sign of charge reverses the deflection.
Pause and explain
A uniform field points right. How does an electron accelerate?
Worked example
A potential difference of 480 V is applied across parallel plates separated by 0.020 m. Find the field strength between them.
Show the worked solution
- Use E = ΔV/Δd.
- Divide 480 V by 0.020 m.
- Give either V m⁻¹ or N C⁻¹.
Answer E = 480/0.020 = 2.4 × 10⁴ V m⁻¹.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 600 V potential difference is maintained across plates 0.030 m apart. Calculate E.
Give me a hint
Divide the potential difference by the plate separation in metres.
Compare my reasoning
- E = ΔV/Δd.
- E = 600/0.030 = 20 000 V m⁻¹.
- The field points from the positive plate to the negative plate.
E = 2.0 × 10⁴ V m⁻¹, from the positive plate towards the negative plate.
Look for these in your work
- Uniform-field relation used
- Separation in metres used
- Magnitude and direction stated
An electron enters a region between parallel plates with velocity parallel to the plates. The electric field points downward. Describe its trajectory while inside the field.
Give me a hint
Resolve the motion parallel and perpendicular to the field.
Compare my reasoning
- The horizontal velocity component is unchanged if other forces are negligible.
- The electron's force and acceleration point upward, opposite the field.
- It therefore follows a curved path deflected upward while the field acts.
It curves upward: its horizontal motion continues while the downward field gives the electron an upward acceleration.
Look for these in your work
- Parallel component addressed
- Force direction correct
- Curved deflection explained
A proton and an electron enter the same uniform field with equal initial velocity perpendicular to it. Compare the initial force directions and acceleration magnitudes.
Give me a hint
Use F = qE, then a = F/m; compare charge signs and masses.
Compare my reasoning
- Their charges have equal magnitude and opposite sign, so the force directions are opposite and force magnitudes equal.
- The electron's mass is much smaller than the proton's mass.
- Since |a| = |q|E/m, the electron has the larger acceleration magnitude.
The forces are equal in magnitude and opposite in direction; the electron's acceleration magnitude is much greater because its mass is smaller.
Look for these in your work
- Force signs compared
- Force magnitudes compared
- Acceleration mass dependence used
Common mistakes
- Using plate area instead of separation in E = ΔV/Δd
- Forgetting to convert centimetres to metres
- Deflecting positive and negative charges in the same direction
- Assuming the charged-particle path is straight when the field is perpendicular to its velocity
What can you explain now?
Between ideal parallel plates, which way does the electric field point?
Compare with the explanation
From the positive plate to the negative plate
That is the direction for a positive test charge.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Why do equal-magnitude positive and negative particles bend in opposite directions in one uniform field?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.