Motion in a uniform electric field
Key idea: A uniform electric field gives a charged particle constant acceleration. The charge sign sets the acceleration direction, while the initial velocity sets the shape of the path.
Continue where you stopped
The core idea
H1 Physics 8867 · Lesson 1 of 3
Check your understandingBy the end of this lesson, you should be able to
- Calculate electric force and acceleration in a uniform field.
- Predict motion for positive and negative charges.
- Analyse straight and parabolic paths from the relative directions of velocity and acceleration.
Learn the idea
Big question: How does the sign of a charge change its path in a uniform electric field?
Move from field to force to acceleration
Electric field direction is defined by the force on a positive test charge. F = QE therefore points along E for positive charge and opposite E for negative charge.
In a uniform field, E is constant, so a = QE/m is constant. A stationary charge accelerates in a straight line; a charge entering sideways combines uniform motion in one direction with constant acceleration in the other.
Check your understanding: Does a stationary electron experience force in an electric field?
Yes. Electric force depends on charge and field, not on the charge already moving.
Recognise projectile-like deflection
When initial velocity is perpendicular to E, the component parallel to the plates remains constant while the field produces a t² displacement across them. The trajectory is parabolic within an ideal uniform field.
Electric force can do work. If force has a component along velocity, the particle's speed and kinetic energy change, unlike deflection by a magnetic force alone.
Check your understanding: An electron beam enters horizontally between plates and curves upward. Which plate is positive?
The lower plate. The electric field points downward from positive to negative, so the electron force is upward.
Key ideas
- Use the signed charge for direction and charge magnitude for force magnitude.
- Uniform field means constant acceleration, not constant velocity.
- A field acts on moving and stationary charges.
Relationships to know
F = QEa = QE/m
Follow the reasoning
Worked example
Calculate beam deflection between plates
Question: An electron enters horizontally at 3.0 × 10⁷ m s⁻¹ between plates 0.080 m long. A uniform field of 2.0 × 10⁴ N C⁻¹ acts downward. Find its vertical deflection on leaving. Use e = 1.60 × 10⁻¹⁹ C and me = 9.11 × 10⁻³¹ kg.
Step 1: Find time in the field
Why: Horizontal velocity is unchanged in the ideal model.
Working: t = L/vx = 0.080/(3.0 × 10⁷) = 2.67 × 10⁻⁹ s.
Step 2: Find vertical acceleration
Why: The electron force is opposite the downward field.
Working: a = eE/m = (1.60 × 10⁻¹⁹)(2.0 × 10⁴)/(9.11 × 10⁻³¹) = 3.51 × 10¹⁵ m s⁻² upward.
Step 3: Use vertical motion
Why: Initial vertical velocity is zero and acceleration is constant.
Working: y = ½at² = ½(3.51 × 10¹⁵)(2.67 × 10⁻⁹)² = 0.0125 m.
Answer: The electron is deflected about 1.25 cm upward.
Check: The direction is opposite the downward field, and the deflection is small compared with plate length, consistent with the sketch.
Now try it with support
Practise with support
A proton is released in a uniform field of 5.0 × 10³ N C⁻¹. Find its acceleration using Q = 1.60 × 10⁻¹⁹ C and m = 1.67 × 10⁻²⁷ kg.
Hints
- Use a = QE/m.
- The proton accelerates with the field.
View the guided answer
a = (1.60 × 10⁻¹⁹)(5.0 × 10³)/(1.67 × 10⁻²⁷) = 4.79 × 10¹¹ m s⁻² along the field.
Your turn
Practise independently
An electron enters a uniform downward electric field with horizontal velocity. State its acceleration direction and describe its path, with justification.
Check your answer
An electron is negative, so its force and acceleration are upward, opposite the downward electric field. Its horizontal velocity remains constant while upward velocity grows, so the path curves parabolically upward.
Common mistakes and exam guidance
Watch out for
- Making every charge accelerate along the field direction.
- Saying a horizontally moving electron is unaffected because its velocity is perpendicular to E.
In an exam
- Write the field direction, charge sign and force direction as three separate steps.
- Describe both velocity components when explaining a curved trajectory.
Put the ideas together
Exam-style practice [7 marks]
A proton enters a 5.0 × 10³ N C⁻¹ uniform field at 2.0 × 10⁵ m s⁻¹ parallel to the field and travels 0.040 m. Find its acceleration and final speed. State how an electron entering with the same initial velocity would differ qualitatively. Use proton mass 1.67 × 10⁻²⁷ kg and e = 1.60 × 10⁻¹⁹ C.
Plan before you answer
- Use charge sign for direction.
- Find constant acceleration, then a kinematics equation.
- For the electron, consider both sign and much smaller mass.
View the marking points and model answer
Marking points
- Finds proton force 8.0 × 10⁻¹⁶ N.
- Finds acceleration 4.79 × 10¹¹ m s⁻² along E.
- Uses v² = u² + 2as.
- Obtains v ≈ 2.80 × 10⁵ m s⁻¹.
- States electron force is opposite E.
- States it slows initially/reverses if field region is sufficient.
- Notes much larger acceleration magnitude due to smaller mass.
Model answer
F = eE = 8.0 × 10⁻¹⁶ N, so a = F/m = 4.79 × 10¹¹ m s⁻² along the field. v² = (2.0 × 10⁵)² + 2(4.79 × 10¹¹)(0.040), giving v = 2.80 × 10⁵ m s⁻¹. An electron's force is opposite the field, so it would slow and may reverse; its acceleration magnitude is much larger because its mass is far smaller.
Finish from memory
Three-question recap
How is electric field direction defined?
Check
The force direction on a positive test charge.
What path results from perpendicular entry into a uniform field?
Check
A parabola in the ideal model.
Can an electric field change kinetic energy?
Check
Yes, when electric force does work.
Continue with the next resource in this course.
Course and syllabus information
- Course
- GCE A-Level H1 Physics
- Edition
- GCE A-Level H1 Physics 2027