WJEC A-Level Physics 1420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Solids Under Stress
Using Radiation to Investigate Stars
Particles and Nuclear Structure
Conduction of Electricity
Resistance
D.C. Circuits
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Nuclear Decay
Nuclear Energy
Capacitance
Electrostatic and Gravitational Fields of Force
Orbits and the Wider Universe
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
Capacitance
The uniform field between capacitor plates
- 3 things to know
- A2 Unit 4 Section A
Overview
Recall questions on The uniform field between capacitor plates.
What you need to know
Answer each one before you reveal it.
State the equation for the force on a charge in a uniform electric field, defining every symbol with its unit.
Show that calculating the work done on a charge crossing the gap as QV agrees with calculating it as force times distance.
Questions from this subtopic
Each one with the model answer underneath.
- 1
State two equivalent units for electric field strength and show that they are the same.
Model answer
Field strength may be given in V m⁻¹ or in N C⁻¹. A volt is a joule per coulomb and a joule is a newton metre, so V m⁻¹ becomes N m C⁻¹ m⁻¹ and the metres cancel to leave N C⁻¹.
- 2
Why is the field near the edges of a pair of parallel plates not uniform?
Model answer
The plates are of finite size, so at the edges the field lines bulge outwards and are no longer straight, parallel and equally spaced.
- 3
Calculate the electric field strength between two plates 5.0 mm apart with 250 V across them.
Model answer
Using E = V/d gives 250 ÷ 5.0 × 10⁻³ = 5.0 × 10⁴ V m⁻¹.
- 4
State the equation for the force on a charge in a uniform electric field, defining every symbol with its unit.
Model answer
The equation is F = EQ, where F is the force in newtons, E is the field strength in V m⁻¹ and Q is the charge in coulombs.
- 5
An electron is placed in a uniform field of strength 5.0 × 10⁴ V m⁻¹. Calculate the force on it.
Model answer
The force is 5.0 × 10⁴ × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻¹⁵ N.
- 6
In which direction does the force act on a negative charge placed between charged parallel plates?
Model answer
It acts in the direction opposite to the field, so the negative charge is pushed towards the positive plate.
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