Eduqas A-Level Physics A420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Conduction of Electricity
Resistance
D.C. Circuits
Capacitance
Solids Under Stress
Electrostatic and Gravitational Fields of Force
Using Radiation to Investigate Stars
Orbits and the Wider Universe
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Nuclear Decay
Particles and Nuclear Structure
Nuclear Energy
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
Energy stored in a capacitor
- 3 things to know
- Component 2
Overview
Recall questions on E = ½QV and its equivalent forms, why the factor of a half appears, the energy as the area under a charge–voltage graph, the fourfold rise when the voltage doubles, and worked energy calculations.
What you need to know
Answer each one before you reveal it.
Why is the energy stored in a capacitor ½QV rather than QV?
How is the energy stored obtained from a graph of charge against potential difference?
Questions from this subtopic
Each one with the model answer underneath.
- 1
State the equation for the energy stored in a capacitor in terms of charge and potential difference.
Model answer
The energy stored is U = ½QV, where U is in joules, Q is the charge on one plate in coulombs and V is the potential difference in volts.
- 2
Why is the energy stored in a capacitor ½QV rather than QV?
Model answer
The potential difference rises steadily from zero to V as the charge is transferred, so the average potential difference during charging is only ½V and the total work done is ½QV.
- 3
Give two further forms of the energy equation for a capacitor.
Model answer
Substituting Q = CV gives U = ½CV², and substituting V = Q/C gives U = Q²/2C.
- 4
How is the energy stored obtained from a graph of charge against potential difference?
Model answer
The energy is the area between the line and the potential difference axis. For the straight line through the origin that area is a triangle of value ½QV.
- 5
Calculate the energy stored in a 470 μF capacitor charged to 12 V.
Model answer
The energy is U = ½CV² = 0.5 × 470 × 10⁻⁶ × 144 = 3.4 × 10⁻² J. The charge is 5.64 mC, and ½QV = 0.5 × 5.64 × 10⁻³ × 12 gives the same 0.0338 J.
- 6
Calculate the energy stored in a 100 μF capacitor holding a charge of 2.0 mC.
Model answer
The energy is U = Q²/2C = (2.0 × 10⁻³)² ÷ (2 × 100 × 10⁻⁶) = 2.0 × 10⁻² J. The potential difference is 20 V, and ½QV = 0.5 × 2.0 × 10⁻³ × 20 gives the same 0.020 J.
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