Uniform fields, force and charged-particle motion
Key idea: H2 Physics lessons on point-charge interactions, uniform-field motion and capacitor energy.
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The core idea
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Big question: How does a uniform electric field steer a charged particle?
Between parallel plates away from edges, E = V/d is approximately uniform and a charge experiences constant force F = qE. Its acceleration is qE/m, with direction reversed for a negative charge. With an initial velocity perpendicular to the field, one component is uniform while the field component accelerates, producing a parabola.
Find the constant acceleration
Between large parallel plates away from edges, field is approximately uniform with E = ΔV/d. A charge has F = qE and acceleration a = qE/m. The field points from the positive plate to the negative plate, while an electron accelerates oppositely.
The model assumes plate separation is small compared with plate dimensions and ignores fringing. Use the perpendicular separation d, not the particle's sloping path length.
Check your understanding: What happens to an electron's acceleration if plate p.d. doubles at fixed d?
Its acceleration magnitude doubles and its direction remains opposite to E.
Separate perpendicular motion components
If a particle enters with velocity perpendicular to E, its velocity parallel to the plates remains constant while the field-direction component changes uniformly. The path is parabolic, just as horizontal projectile motion combines constant and accelerated components.
Electric work qΔV changes kinetic energy: ΔK = −qΔV when only the electric force acts. Component kinematics and energy are alternative approaches; choose the one matching the requested quantity.
Check your understanding: Does the field change the velocity component perpendicular to E?
No, not in the uniform-field model with no other forces, because there is no force in that direction.
Key ideas to keep
- Use plate separation measured perpendicular to the plates.
- Electron force is opposite to the field direction.
- Electric force can change speed and therefore kinetic energy.
See the reasoning
Worked example
Predict an electron's path from perpendicular components
Question: An electron enters horizontally at 3.0 × 10⁷ m s⁻¹ between horizontal plates where E = 2.0 × 10⁴ N C⁻¹ downward. Find its vertical acceleration and describe its path.
Step 1: Find the force direction
Why: An electron's negative charge reverses the field direction.
Working: With E downward, F = qE is upward.
Step 2: Calculate constant acceleration
Why: The uniform field gives constant force.
Working: a = eE/m = 3.51×10¹⁵ m s⁻² upward.
Step 3: Combine components
Why: Horizontal velocity remains constant while vertical velocity changes uniformly.
Working: The path is parabolic while the electron remains between the plates.
Answer: The electron force is opposite the field, upward. Its magnitude is eE = 3.20 × 10⁻¹⁵ N, so a = 3.51 × 10¹⁵ m s⁻² upward. Horizontal velocity stays constant while vertical velocity changes uniformly, producing a parabolic path within the uniform field.
Check: The path curves opposite to E for a negative particle.
Use a hint if needed
Practise with support
Try this
A −3.0 nC charge is between plates with E = 2.0 × 10⁵ N C⁻¹ to the right. Find its force.
Hint: The field is defined for a positive test charge; retain the charge sign.
Check your answer
F = qE = (−3.0 × 10⁻⁹)(2.0 × 10⁵) = −6.0 × 10⁻⁴ N, meaning 6.0 × 10⁻⁴ N to the left.
Now work without the hint
Practise independently
Your turn
A positive particle enters a uniform field perpendicular to its initial velocity. Explain its motion and how the answer changes for a negative particle.
Check your answer
The constant force qE gives constant acceleration parallel to E for positive q, while the perpendicular velocity component stays constant, producing a parabola. For negative q the acceleration and curvature reverse; the path need not follow a field line.
Avoid these traps
Common mistakes
Common mistake
E = ΔV/d applies to every field.
What is wrong with this reasoning?
Show better thinking
It applies directly only across a uniform field; a non-uniform field requires a local gradient.
Common mistake
A charged particle must follow an electric field line.
What is wrong with this reasoning?
Show better thinking
The field fixes acceleration; an initial transverse velocity produces a curved path.
Write for the examiner
Exam guidance
Treat the two perpendicular motion components separately, exactly as for projectile motion.
Exam-style practice [6 marks]
Plates 8.0 mm apart have 240 V across them. Find E and the force on an electron, then describe its acceleration direction.
Plan before you answer
- Use E = V/d.
- Find force magnitude.
- Reverse direction for an electron.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
E = 240/0.0080 = 3.0 × 10⁴ V m⁻¹. Force magnitude is eE = 4.8 × 10⁻¹⁵ N, opposite E because the electron is negative; its acceleration has the same direction as its force.
Come back in three days
Check what stayed with you
Recall question
A charged particle starts from rest in a uniform field. State the form of its motion and one condition under which that description ends.
Check the answer
It has constant acceleration a = qE/m in a straight line parallel or antiparallel to E. The description ends on leaving the uniform region or when relativistic effects, collisions or another force becomes significant.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 14 states no explicit exclusions. Coulomb-force, point-field and point-potential equations are used for point charges in free space or air. Electric field and force are vectors; potential and potential energy are scalars. E = ΔV/d is restricted to a uniform field, while E = −dV/dr is the local negative potential gradient.
- GCE A-Level H2 PhysicsTopic 14(g) / Topic 14(h) / Topic 14(i) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027