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AS 91526 · 2023 paper · capacitors

Question 1(a) Merit

Practises Capacitors

Kate investigates a capacitor found in a camera, labelled 185 nF (1.85 × 10⁻⁷ F). The camera also contains a 1.50 V “AA” battery.

C185 nF
V1.50 V

Show that the energy stored by the capacitor, fully charged by the battery, is 2.08 × 10⁻⁷ J.

The 2023 paper, page 2 ↗ Marking schedule, page 1 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

  1. 1 Once it is fully charged, what voltage is across the capacitor?

    The battery’s full 1.50 V. Charging has stopped, so no current flows and nothing is lost across the battery’s internal resistance.

  2. 2 How much charge does that put on the plates?

    Q = CV = 1.85 × 10−7 × 1.50 = 2.775 × 10−7 C, or 2.78 × 10−7 C.

  3. 3 Which energy formula fits the values you have? It is a “show that”, so keep more figures than the question gives.

    E = ½CV² = ½ × 1.85 × 10−7 × 1.50² = 2.081 × 10−7 J. ½QV gives the same.
Answer 2.081 × 10−7 J, which rounds to the 2.08 × 10−7 J asked for.

What the marker wanted

  • Achieved The charge on the plates, Q = 2.78 × 10−7 C.
  • Merit The energy, worked out to more figures than the question gives: 2.081 × 10−7 J.

In our words, from NZQA’s 2023 marking schedule.

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Common mistake

Writing 2.08 × 10−7 J as your answer. That is the number you were given; the schedule’s Merit column is the 2.081 × 10−7 J that shows you got there yourself.

How did it go?

Got it moves it further out. Review brings it back tomorrow.