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In the circuit shown in the figure, the total charge is 750 $$\mu C$$ and the voltage across capacitor $$C_2$$ is 20 V. Then the charge on capacitor $$C_2$$ is:
Given total charge: $$Q = 750\ \mu\text{C}$$
Since $$C_2$$ and $$C_3$$ are in parallel, their voltages are equal: $$V_3 = V_2 = 20\ \text{V}$$
Charge on capacitor $$C_3$$: $$Q_3 = C_3 V_3 = 8\ \mu\text{F} \times 20\ \text{V} = 160\ \mu\text{C}$$
Charge conservation for the parallel combination: $$Q = Q_2 + Q_3$$
$$750\ \mu\text{C} = Q_2 + 160\ \mu\text{C} \implies Q_2 = 590\ \mu\text{C}$$
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