CCEA A-Level Physics 1210
Physical Quantities, Scalars and Vectors
Moments and Equilibrium
Linear Motion and Dynamics
Newton's Laws of Motion
Linear Momentum and Impulse
Work, Energy and Power
Current, Charge and Potential Difference
Resistance and Resistivity
Internal Resistance and Potential Dividers
Waves
Refraction
Lenses
Superposition, Interference and Diffraction
Quantum Physics
Wave-Particle Duality
Astronomy
Practical Skills and Data Analysis
Deformation of Solids
Thermal Physics
Uniform Circular Motion
Simple Harmonic Motion
The Nucleus and Radioactive Decay
Nuclear Energy, Fission and Fusion
Gravitational Fields
Electric Fields
Capacitors
Magnetic Fields
Charged Particles and Accelerators
Capacitors
Energy stored in a charged capacitor
- 3 things to know
- Unit A2 2
Overview
Recall questions on E = ½QV and its two equivalent forms, why the factor of ½ appears, the energy as the area under a charge against potential difference graph, the fourfold rise when the voltage doubles, worked energy calculations, why only half the battery's energy is stored, and the defibrillator as a worked case.
What you need to know
Answer each one before you reveal it.
Why is the energy stored ½QV rather than QV?
A 100 μF capacitor charged to 50 V stores 0.125 J by ½CV². Check that energy using the charge stored
Questions from this subtopic
Each one with the model answer underneath.
- 1
Give the equation for the energy stored in a charged capacitor in terms of charge and potential difference.
Model answer
E = ½QV, where E is the energy in joules, Q the charge in coulombs and V the potential difference in volts.
- 2
Why is the energy stored ½QV rather than QV?
Model answer
The potential difference rises from zero to V as the capacitor charges, so the average potential difference during charging is only half the final value.
- 3
How is the energy stored found from a graph of charge against potential difference?
Model answer
It is the area under the line, which for a straight line through the origin is the area of a triangle, ½QV.
- 4
Calculate the energy stored by a 100 μF capacitor charged to 50 V.
Model answer
E = ½CV² = 0.5 × 100 × 10⁻⁶ × 50², which is 0.5 × 100 × 10⁻⁶ × 2500, giving 0.125 J.
- 5
Check that energy using the charge stored.
Model answer
Q = CV = 100 × 10⁻⁶ × 50 = 5.0 × 10⁻³ C, and ½QV = 0.5 × 5.0 × 10⁻³ × 50 = 0.125 J, which agrees.
- 6
The potential difference across a capacitor is doubled. What happens to the energy stored?
Model answer
It becomes four times as great, because energy is proportional to the square of the potential difference.
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