A-Level Physics May/June 2024 Q6(b)(ii): Calculate the total energy, in mJ, stored in the capacitors when C_Q has its maximum va…

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  Step 1: Calculate the total capacitance

  Capacitors in parallel add up. When $C_Q$ is at its maximum value, the total capacitance $C_{total}$ is: $C_{total} = C_P + C_{Q(max)}$ $C_{total} = 200 \text{ µF} + 400 \text{ µF} = 600 \text{ µF}$

  Step 2: Calculate the total energy stored

  The formula for energy stored in a capacitor is $E = \frac{1}{2}CV^2$. $E = \frac{1}{2} \times (600 \times 10^{-6} \text{ F}) \times (5.0 \text{ V})^2$ $E = \frac{1}{2} \times (600 \times 10^{-6}) \times 25$ $E = 7.5 \times 10^{-3} \text{ J}$

  Step 3: Convert the energy to millijoules (mJ)

  The question asks for the answer in mJ. $E = 7.5 \text{ mJ}$

  Total Energy = 7.5 mJ

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  How the marks are awarded

  • C1 — The first mark is for correctly calculating the total capacitance of the parallel combination when C_Q is at its maximum value, resulting in 600 µF.
  • C1 — The second mark is for correctly substituting the total capacitance (600 × 10⁻⁶ F) and the potential difference (5.0 V) into the energy formula E = ½CV².
  • A1 — The final mark is for the correct final answer of 7.5 mJ, including the correct unit prefix as required by the question.

  Common mistakes

  • Calculating the total capacitance as if the capacitors were in series (using 1/C_total = 1/C_1 + 1/C_2).
  • Forgetting to convert the capacitance from microfarads (µF) to farads (F) before substituting into the energy formula.
  • Calculating the correct energy in joules (7.5 x 10⁻³ J) but failing to convert it to millijoules (mJ) for the final answer.
  • Forgetting to square the voltage term in the energy formula, i.e., using V instead of V².

  Examiner tip: Always double-check the units required for the final answer and ensure all quantities are converted to base SI units within your calculation steps unless a formula specifically allows for prefixes.

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