Uniform electric fields Cambridge International AS & A Level Physics revision
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In plain words
Connect a battery across two parallel metal plates and the space between them fills with an even electric field, the same strength everywhere. A charged particle fired through it is pushed steadily sideways, just as a thrown ball is pulled steadily downwards by gravity, so it follows the same kind of curve.
4 things to know
- Between parallel plates with a p.d. V across them, a distance d apart: E = V ÷ d, in V/m.
- The force on a charge is the same everywhere between the plates: F = QE = QV ÷ d.
- A charged particle that enters at right angles to the field keeps a constant velocity in its original direction and has a constant acceleration along the field. Its path is a parabola, like a projectile's.
- A charge Q that moves through a p.d. V gains a kinetic energy QV.
Worked example
Two parallel plates 0.020 m apart have a p.d. of 500 V across them. Find the field strength, and the force on an electron between them.
- E = V ÷ d = 500 ÷ 0.020 = 2.5 × 10⁴ V/m.
- F = QE = 1.60 × 10⁻¹⁹ × 2.5 × 10⁴.
- = 4.0 × 10⁻¹⁵ N, towards the positive plate.
Tips and tricks
- Put the plate separation into metres.
- Treat a particle crossing the field like a projectile: use the constant velocity to find the time, then use a = QE ÷ m across the field.
It lands in your notebook with its questions as flashcards.
Uniform electric fields: 5 questions and answers
These are the quiz’s questions. Do the quiz first, then come back here for the ones that got you.
Two plates 0.010 m apart have a p.d. of 200 V across them. What is the field strength between them?
E = V ÷ d.
An electron enters a uniform electric field at right angles to the field. What shape is its path?
A constant velocity one way and a constant acceleration the other, like a projectile.
The plates are moved further apart with the same p.d. across them. What happens to the field strength?
E = V ÷ d.
How does the force on a charge change as it moves about between the plates?
The field is uniform.
How much kinetic energy does a charge Q gain when it moves through a p.d. V?
The p.d. is the energy transferred per unit charge.
Quiz
5 questions
Tap an answer and you’ll see straight away whether it’s right, and why.
Worksheet
4 questions, 11 marks. Write your answers on paper, then check them.
Uniform electric fields
Cambridge International AS & A Level Physics 9702 · 11 marks · papermunch.org
Name ______________________________ Date ______________
Find the field strength between two plates 5.0 mm apart with a p.d. of 100 V across them.[2]
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2.0 × 10⁴ V/m.
An electron is accelerated from rest through a p.d. of 200 V. Find its final speed. (Mass of an electron = 9.11 × 10⁻³¹ kg.)[3]
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8.4 × 10⁶ m/s. It gains 3.2 × 10⁻¹⁷ J, and v = √(2 × 3.2 × 10⁻¹⁷ ÷ 9.11 × 10⁻³¹).
An electron enters a uniform electric field at right angles to the field lines. Describe its path, and explain it.[3]
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A parabola, curving towards the positive plate. There is no force along its original direction, so its velocity that way is constant. There is a constant force, and so a constant acceleration, along the field.
A charged oil drop of weight 3.2 × 10⁻¹⁴ N is held still between horizontal plates where the field strength is 1.0 × 10⁵ V/m. Find the charge on the drop.[3]
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3.2 × 10⁻¹⁹ C. The electric force balances the weight: Q = 3.2 × 10⁻¹⁴ ÷ 1.0 × 10⁵.
Answers: Uniform electric fields
- 1. 2.0 × 10⁴ V/m.
- 2. 8.4 × 10⁶ m/s. It gains 3.2 × 10⁻¹⁷ J, and v = √(2 × 3.2 × 10⁻¹⁷ ÷ 9.11 × 10⁻³¹).
- 3. A parabola, curving towards the positive plate. There is no force along its original direction, so its velocity that way is constant. There is a constant force, and so a constant acceleration, along the field.
- 4. 3.2 × 10⁻¹⁹ C. The electric force balances the weight: Q = 3.2 × 10⁻¹⁴ ÷ 1.0 × 10⁵.