Energy — OCR A-Level Physics
Test yourself on Energy with OCR A-Level practice questions.
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Energy explained
For a capacitor, the p.d. V across it is proportional to the charge Q stored, so a graph of V against Q is a straight line through the origin with gradient 1/C.
Read the full explanation
The energy stored is the work done in transferring charge against the increasing p.d., which equals the area under the V–Q graph. Because the graph is a triangle of base Q and height V, the area is ½QV. This geometric interpretation links directly to the energy formulae and explains why the factor ½ appears. A concrete check: if Q = 4.0 × 10⁻³ C and V = 20 V, the area is ½ × 4.0 × 10⁻³ × 20 = 4.0 × 10⁻² J.
(b) energy stored by capacitor; W = 1/2 QV, W = 1/2 CV² and W = 1/2 Q²/C
The energy stored by a capacitor can be written in three equivalent forms using Q = CV. Starting from W = ½QV, substituting Q = CV gives W = ½CV², and substituting V = Q/C gives W = ½Q²/C. Each form is useful depending on which quantities are known. For example, a 2.0 × 10⁻⁶ F capacitor charged to 12 V stores W = ½ × 2.0 × 10⁻⁶ × 12² = 1.44 × 10⁻⁴ J. The factor ½ arises because the p.d. rises from zero to V as charge is transferred, so the average p.d. during charging is ½V. Choosing the right form avoids unnecessary intermediate calculations.
(c) uses of capacitors as storage of energy.
A capacitor stores energy in the electric field between its plates, so it can act as a temporary energy store that is charged slowly and discharged quickly. The energy stored is W = ½QV, and substituting Q = CV gives W = ½CV² or W = ½Q²/C. For example, a 470 µF capacitor charged to 12 V stores ½ × 470 × 10⁻⁶ F × (12 V)² = 0.034 J. Uses exploit this: camera flashes charge a capacitor then release energy rapidly through a tube; defibrillators deliver a controlled pulse; smoothing circuits in power supplies hold charge between peaks; and backup or memory-hold circuits keep data alive briefly when power is interrupted.
Your focus
- Interpret a p.d.–charge graph for a capacitor.
- Explain why energy stored equals the area under the graph.
- Calculate energy stored from the triangular area ½QV.
Show all 9 objectives
- Recall and use the three equivalent expressions for energy stored by a capacitor.
- Derive W = ½CV² and W = ½Q²/C from W = ½QV.
- Select and apply the appropriate energy expression to solve numerical problems.
- Recall that capacitors store energy in the electric field between their plates.
- Select and apply the correct energy equation for a capacitor.
- Explain a named practical use in terms of charging and rapid discharge.
Energy exam tips
Quick Revision Summary (Key Takeaway)
In OCR A-Level Physics, energy is the capacity to do work, governed universally by the principle of conservation of energy across closed systems. Mastering this topic requires quantitative analysis of work done, kinetic energy, gravitational and elastic potential energy, alongside power and mechanical efficiency.
Topic Overview
Module 3 (Forces and Motion) of OCR A-Level Physics establishes energy as a unifying scalar quantity that links forces, motion, materials, and power. Students study quantitative formulations of work done, kinetic energy, gravitational potential energy, and elastic potential energy, developing the mathematical tools needed to model complex dynamic systems.
This foundational topic links directly to circular motion, simple harmonic motion, gravitational fields, and thermodynamics in Year 2. Mastering work-energy balances provides an alternative, often simpler approach to kinematics and dynamics than applying Newton's laws directly.
Key Concepts
- →Work done by a force is the product of magnitude of force and distance moved in the direction of the force: W = Fx cos(theta).
- →The principle of conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another within a closed system.
- →Kinetic energy (E_k = 0.5 m v^2) and uniform gravitational potential energy (E_p = m g h) allow interchange calculations in mechanical trajectories.
- →Elastic strain energy stored in a deformed linear material is equal to the area under a force-extension graph: E = 0.5 F x = 0.5 k x^2.
- →Power is the rate of energy transfer or rate of doing work (P = W / t), which for a moving body driven by force F at constant speed v translates to P = F v.
Marking Points
- State that p.d. is directly proportional to charge for a capacitor, giving a straight-line V–Q graph through the origin.
- Explain that the energy stored equals the area under the p.d.–charge graph.
- Calculate energy from the triangular area as ½QV.
- Relate the gradient of the V–Q graph to 1/C.
- State the three equivalent expressions W = ½QV, W = ½CV² and W = ½Q²/C for energy stored by a capacitor.
- Derive the alternative forms from W = ½QV using Q = CV.
- Select the most convenient form based on the quantities given in a problem.
- Calculate energy stored correctly, keeping charge in coulombs, p.d. in volts and capacitance in farads.
- States that a capacitor stores energy in the electric field between its plates.
- Recalls and applies W = ½QV, W = ½CV² or W = ½Q²/C correctly.
- Converts capacitance to farads and squares the voltage when using W = ½CV².
- Links a named use, such as a camera flash or defibrillator, to rapid discharge of stored energy.
- Recognises that a capacitor is a temporary store, not a source of continuous power.
Examiner Tips
- 💡Sketch the V–Q line and shade the triangular area to show where ½QV comes from.
- 💡Use the gradient to find 1/C and hence C if the graph is given.
- 💡Keep charge in coulombs and p.d. in volts so the area is in joules.
- 💡Write down the known quantities with units first, then choose the form of the energy equation that uses them directly.
- 💡Convert microfarads to farads (× 10⁻⁶) and nanocoulombs to coulombs (× 10⁻⁹) before substituting.
- 💡Check the answer is in joules and is reasonable for the size of capacitor and p.d. given.
- 💡Write the energy equation before substituting numbers so the factor of ½ is not lost.
- 💡Convert µF to F and check the power of ten in the answer.
- 💡When a question names a device, state the energy transfer and why rapid discharge matters.
- 💡Always state the fundamental conservation equation explicitly before rearranging terms or substituting numbers into physics problems.
- 💡Check your significant figures: OCR strictly penalises answers given to inappropriate significant figures (typically match the lowest number of significant figures supplied in the raw data, usually 2 or 3 s.f.).
- 💡When interpreting graphs, remember that the area under a force-displacement graph represents work done, while the gradient of an energy-time graph gives power.
Common Mistakes
- Treating the V–Q graph as a rectangle and using QV without the factor ½; the area is a triangle, so energy is ½QV.
- Thinking the graph curves; for an ideal capacitor V is proportional to Q, so the graph is straight.
- Confusing the area under a V–Q graph with the area under a V–I graph; they represent different quantities.
- Omitting the factor ½ and using W = QV or W = CV²; the correct expressions all include ½.
- Mixing units, such as using capacitance in microfarads without converting to farads, which gives an energy a million times too large.
- Substituting the wrong rearrangement, for example using V = Q/C when Q is unknown; choose the form matching the given data.
- Using W = CV² and omitting the factor of ½; the correct expression is W = ½CV².
- Treating capacitance in microfarads as farads without multiplying by 10⁻⁶, which makes the energy a million times too large.
- Assuming a capacitor can supply steady power indefinitely; it discharges as charge is removed, so its voltage falls.
- Assuming normal contact forces or centripetal forces do work: because these forces act perpendicular to the instantaneous displacement (cos 90 degrees = 0), they do zero work.
- Confusing energy and power units: students frequently treat Watts (J s^-1) as energy and Joules as power, or forget that 1 kWh = 3.6 MJ.
- Assuming spring potential energy scales linearly with extension: E is proportional to extension squared (x^2), meaning doubling extension quadruples the stored strain energy.
Revision Plan
- 1Day 1-3: Review definitions, units, and fundamental formulas for work done, kinetic energy, gravitational potential energy, and P = Fv.
- 2Day 4-6: Practice resolving non-parallel forces in work equations and calculating areas under non-linear and linear force-extension graphs.
- 3Day 7-9: Solve integrated conservation of energy problems involving both elastic deformations and resistive losses (friction/drag).
- 4Day 10-14: Complete OCR past exam questions under timed conditions, cross-referencing mark schemes specifically for command words and correct significant figures.
Exam Question Types
- 📋Calculation questions: Multi-step quantitative problems requiring calculation of energy changes, resistive work, or motor power with unit handling.
- 📋Graphical interpretation: Determining work done from the area under a force-displacement graph or spring extension plot.
- 📋Structured 6-mark explanations: Describing an energy conversion experiment (such as validating E_p to E_k conversion with light gates) including uncertainties and evaluation.
Command Word Expectations (OCR)
Provide a complete algebraic or numerical derivation setting out every intermediate step; never jump straight to the given value and never round numbers prematurely.
Obtain an answer through numerical calculation or graphical extraction; working must be shown clearly to gain method marks even if the final calculation is incorrect.
Provide reasoning based on fundamental physical principles (e.g., quoting conservation of energy or work done against drag) rather than merely restating the observation.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A toy car of mass 0.250 kg is released from rest at the top of a curved track of vertical height 0.800 m. It reaches the bottom of the track with a speed of 3.60 m s^-1. Calculate the average resistive force acting on the car along the track given that the total distance along the track is 1.40 m. Take g = 9.81 m s^-2.
- 1.Step 1: Calculate the initial gravitational potential energy (E_p) at the top: E_p = m * g * h = 0.250 kg * 9.81 m s^-2 * 0.800 m = 1.962 J.
- 2.Step 2: Calculate the final kinetic energy (E_k) at the bottom: E_k = 0.5 * m * v^2 = 0.5 * 0.250 kg * (3.60 m s^-1)^2 = 1.620 J.
- 3.Step 3: Determine the work done against friction (W): W = E_p - E_k = 1.962 J - 1.620 J = 0.342 J.
- 4.Step 4: Use work done = force * distance along track to solve for average resistive force (F): F = W / d = 0.342 J / 1.40 m = 0.24428... N.
Question: An electric winch powered by a 230 V motor operating at 4.50 A lifts a load of mass 65.0 kg vertically through a height of 8.00 m in 12.0 s at constant velocity. Calculate the efficiency of the winch system. Take g = 9.81 m s^-2.
- 1.Step 1: Calculate electrical power input using P_in = I * V: P_in = 4.50 A * 230 V = 1035 W. Total electrical energy input E_in = P_in * t = 1035 W * 12.0 s = 12420 J.
- 2.Step 2: Calculate useful gravitational potential energy output: E_out = m * g * h = 65.0 kg * 9.81 m s^-2 * 8.00 m = 5101.2 J.
- 3.Step 3: Alternatively calculate useful power output: P_out = E_out / t = 5101.2 J / 12.0 s = 425.1 W.
- 4.Step 4: Calculate efficiency as (useful energy output / total energy input) * 100%: Efficiency = (5101.2 J / 12420 J) * 100% = 41.07%.