Question 19N.2.HL.TZ0.9

$E=\frac{1}{2}\frac{Q^2}{C}$ OR $V=\frac{Q}{C}$✔

$E=«\frac{1}{2}\frac{\left(48\times {10}^{-6}\right)}{18\times {10}^{-6}}=» 6.4\times {10}^{-5} «\mathrm{J}»$✔

ALTERNATIVE 1
$Q_{\mathrm{X}}+Q_{\mathrm{Y}}=48$ ✔

$\frac{Q_{\mathrm{X}}}{18}=\frac{Q_{\mathrm{Y}}}{12}$ ✔

solving to get $Q_{\mathrm{X}}=29 «\mu \mathrm{C}» Q_{\mathrm{Y}}=19 «\mu \mathrm{C}»$ ✔

ALTERNATIVE 2

$48=18 \mathrm{V}+12 \mathrm{V}\Rightarrow V=1.6 «\mathrm{V}»$✔

$Q_{\mathrm{X}}=«1.6\times 18=» 29 «Q_{\mathrm{X}}=1.6\times 18=29 «\mu \mathrm{C}»$ ✔

$Q_{\mathrm{Y}}=«1.6\times 12=» 19 «\mu \mathrm{C}»$ ✔

NOTE: Award [3] for bald correct answer

ALTERNATIVE 1

$E_{\mathrm{T}}=\frac{1}{2}\frac{{\left(29\times {10}^{-6}\right)}^2}{18\times {10}^{-6}}+\frac{1}{2}\frac{{\left(19\times {10}^{-6}\right)}^2}{12\times {10}^{-6}}$✔

$=3.8\times {10}^{-5}«\mathrm{J}»$✔

ALTERNATIVE 2

$E_{\mathrm{T}}=\frac{1}{2}\times 18\times {10}^{-6}\times 1.6^2+\frac{1}{2}\times 12\times {10}^{-6}\times 1.6^2$ ✔

$=3.8\times {10}^{-5}«\mathrm{J}»$✔

NOTE: Allow ECF from (b)(i)
Award [2] for bald correct answer
Award [1] max as ECF to a calculation using only one charge

charge moves/current flows «in the circuit» ✔
thermal losses «in the resistor and connecting wires» ✔

NOTE: Accept heat losses for MP2