In given circuit
Diffcult
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$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{a}}=\frac{120}{6} \times 1=20 \mathrm{\,V}$

$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{b}}=\frac{120}{6} \times 2=40 \mathrm{\,V}$

$\therefore \mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}=40-20=20 \mathrm{\,V}$

charge on $\mathrm{C}_{1}$ is $\mathrm{q}_{1}=\mathrm{C}_{1}\left(\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}\right)$

$=40 \times 10^{-6} \times 20=80 \,\mu C$

$\mathrm{V}_{\mathrm{c}}=\mathrm{V}_{\mathrm{d}}$ and $\mathrm{V}_{\mathrm{e}}=\mathrm{V}_{\mathrm{f}}$

so charge on $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are zero.

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