Energy stored by a capacitor (A-level only) — AQA A-Level Physics

This topic covers how capacitors store energy in the electric field between their plates, and the mathematical relationships that describe this energy. You will learn the three equivalent forms of the energy equation: W = 1/2 QV, W = 1/2 CV^2, and W = 1/2 Q^2/C, and understand when to use each. The topic also explores the graphical representation of energy as the area under a charge-voltage graph, and the concept of energy density in an electric field.

Understanding energy storage in capacitors is crucial for analysing circuits involving capacitors, especially in timing and filtering applications. It builds directly on earlier work on capacitance, electric fields, and potential difference, and it links to practical applications such as camera flashes, defibrillators, and energy recovery systems. This is an A-level only topic, so it appears in Paper 2 and Paper 3 of the AQA A-level Physics exams, often combined with other capacitor concepts.

• →The energy stored by a capacitor is given by W = 1/2 QV, where Q is the charge and V is the potential difference. This is derived from the area under a charge-voltage graph, which is a straight line through the origin.
• →Alternative forms: W = 1/2 CV^2 (when voltage is known) and W = 1/2 Q^2/C (when charge is known). These are all equivalent because Q = CV.
• →Energy is stored in the electric field between the plates, not on the plates themselves. The energy density (energy per unit volume) in a vacuum is 1/2 epsilon_0 E^2, where E is the electric field strength.
• →When a capacitor is disconnected from a battery, charge Q remains constant. If the capacitance changes (e.g., by separating plates), the voltage and energy change accordingly.
• →The charging and discharging of a capacitor involves energy transfer: the battery does work to move charge, and half of the energy supplied by the battery is stored in the capacitor (the other half is dissipated as heat in the circuit resistance).

• The graph of charge against potential difference is a straight line through the origin because Q = CV.
• The area under the graph has units of charge times potential difference, which are joules.
• The area under the graph represents the energy stored by the capacitor.
• For a straight-line graph through the origin, the area is ½ × base × height, giving E = ½QV.
• Substituting Q = CV or V = Q/C gives the equivalent forms E = ½CV² and E = ½Q²/C.
• The gradient of the graph is the capacitance C.

• 💡Sketch the Q against V graph and shade the triangular area to show clearly what is being calculated.
• 💡State the units of the area as coulombs times volts, then convert to joules to justify the interpretation.
• 💡Choose the energy form that matches the data given: use E = ½QV when Q and V are known, E = ½CV² when C and V are known, and E = ½Q²/C when C and Q are known.
• 💡Always write down the formula you are using before substituting numbers. This helps you choose the correct form (W = 1/2 CV^2, W = 1/2 QV, or W = 1/2 Q^2/C) and shows the examiner your method.
• 💡Pay attention to units: convert microfarads to farads (multiply by 10^-6) and ensure all quantities are in SI units before calculating.
• 💡For 'explain' questions, use precise language: 'energy is stored in the electric field between the plates' and refer to the area under the charge-voltage graph as the energy stored.

• Using the whole rectangle QV rather than the triangle ½QV: the error is forgetting the factor of one half; the correction is that the area under a straight line through the origin is a triangle, so E = ½QV.
• Treating the area as Q/V or V/Q: the error is confusing area with gradient; the correction is that the gradient is C = Q/V, while the area is the energy.
• Using E = ½QV when the graph is not a straight line through the origin: the error is applying the triangle formula outside its valid range; the correction is that the formula applies to a capacitor obeying Q = CV.
• Misconception: Energy is stored on the plates. Correction: Energy is stored in the electric field between the plates. The plates are just the conductors that create the field.
• Misconception: Using W = 1/2 CV instead of W = 1/2 CV^2. Correction: The energy depends on the square of the voltage, so forgetting to square V leads to a large error.
• Misconception: When a capacitor is disconnected and the plates are separated, the energy stored decreases. Correction: The energy actually increases because work must be done against the attractive force between the plates (the field does negative work).

1. 1Day 1: Review the definition of capacitance and the charge-voltage graph. Derive the energy formula W = 1/2 QV from the area under the graph.
2. 2Day 2: Learn the three forms of the energy equation and practise converting between them. Solve basic calculation problems with different given quantities.
3. 3Day 3: Study the energy density of an electric field and understand that energy is stored in the field. Complete a few conceptual questions on where the energy is stored.
4. 4Day 4: Work through problems involving capacitors that are disconnected and then modified (e.g., plate separation changed). Focus on constant charge scenarios.
5. 5Day 5: Attempt exam-style questions, including 6-mark structured questions. Review mark schemes to understand the level of detail required.
6. 6Day 6-7: Create a summary sheet with key formulas, common pitfalls, and a flowchart for choosing the right formula. Test yourself with active recall prompts.

• 📋Calculation questions: Given two of C, Q, V, calculate the energy stored. Advice: Identify which quantities are constant and choose the appropriate formula. Show all steps and units.
• 📋Graph questions: Sketch or interpret a charge-voltage graph and calculate energy from the area. Advice: Remember the area under the graph is the energy stored; for a straight line through the origin, area = 1/2 QV.
• 📋Explanation questions: Explain why energy increases when plates are separated after disconnecting the battery. Advice: Refer to work done against the electric field and the constant charge condition.
• 📋Practical-based questions: Describe an experiment to determine the energy stored by a capacitor, e.g., using a joule meter or by discharging through a resistor. Advice: Include measurements of C and V, and use W = 1/2 CV^2.