CAIE A-Level Physics 19.1.2 Using C = Q/V

Practise using and rearranging C = Q/V to calculate capacitance, charge or potential difference, checking units and interpreting the capacitor relationship.

Exam points

calculate charge, potential difference or capacitance using C = Q/V
rearrange C = Q/V and use consistent charge, voltage and capacitance units
interpret the charge–potential relationship for a capacitor

[Maximum number: 2]

A capacitor of capacitance $470 \mu \mathrm{~F}$ is connected to a battery of electromotive force (e.m.f.) 24 V in the circuit of Fig. 5.1.

Fig. 5.1

The two-way switch S is initially at position X.
P and Q are identical long straight wires, each with a resistance of $5.6 \mathrm{k} \Omega$ . These wires are placed near to, and parallel to, each other. Wire Q is connected to a voltmeter.

At time t=0, switch S is moved to position Y so that the capacitor discharges through wire P .

Calculate the charge $Q_{0}$ on the capacitor at time t=0.

Q_{0}=

Q=C V

C1

$\begin{aligned} Q_{0} =24 \times 470 \times 10^{-6} =0.011 \mathrm{C} \end{aligned}$

A1

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19.1.2—C = Q / V question 1