A-LevelPhysicsCapacitance9702/42 · May/June 2021 · Q68 Marks

Two flat metal plates are held a small distance apart by means of insulating pads, as shown in [Figure 6.1].

(a)Explain how the plates could act as a capacitor.[2]
(b)(i)Show that the average current I in the ammeter is given by I = CVf.[2]
(b)(ii)For a potential difference V of 180 V and a frequency f of switching of 50 Hz, the average current I in the ammeter is 2.5 µA. Calculate the capacitance, in pF, of the parallel plates.[2]
(c)A second capacitor is connected into the circuit of [Figure 6.2]. The two capacitors are connected in parallel. State and explain the change, if any, in the average current in the ammeter.[2]

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Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

[1] potential difference applied between the plates
[1] causes charge separation (between the plates) OR causes energy to be stored (between the plates)
[1] I = Q/t
[1] clear substitution of Q = CV and f = 1 / t, leading to I = fCV
[1] 2.5 x 10⁻⁶ = 50 × C × 180
[1] C = 280 pF
[1] (total) capacitance increases
[1] greater charge (for each cycle/discharge) so greater (average) current OR V and f are constant so (average) current increases OR I is (directly) proportional to C so (average) current increases

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