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(C) Let the potential at point $B$ be $V_B = 0 \, V$. Moving from $B$ to $A$ through the circuit components: $V_A - V_B = V_{capacitor1} + V_{battery1

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$8$

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(C) Let the potential at point $B$ be $V_B = 0 \, V$. Moving from $B$ to $A$ through the circuit components: $V_A - V_B = V_{capacitor1} + V_{battery1} + V_{capacitor2} + V_{battery2}$ However,since the capacitors are in series and there is no steady current,the charge $q$ on both capacitors must be the same. Let the charge on each capacitor be $q$. The potential difference across the $4 \, \mu F$ capacitor near $B$ is $V_1 = q / C = q / 4$. The potential difference across the $4 \, \mu F$ capacitor near $A$ is $V_2 = q / C = q / 4$. Starting from $B$ and moving to $A$: $V_A - V_B = (q / 4) + 2 + (q / 4) + 6 = 4$ $q / 2 + 8 = 4$ $q / 2 = -4$ $q = -8 \, \mu C$. The magnitude of the charge is $8 \, \mu C$.

$A$ part of a circuit is shown in the figure. If $V_A - V_B = 4 \, V$,then find the charge on the $4 \, \mu F$ capacitor connected to point $A$. (in $, \mu C$)

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