A 1.50-µF capacitor charged to 40.0 V and a 2.40-µF capacitor charged to 15.0 V are connected to each other,with the positive plate of each connected to the negative plate of the other.What is the final charge on the 1.50-µF capacitor?

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Quizplus · Accepted Answer

The final charge on the 1.50-µF capacitor can be found by first calculating the initial charges on both capacitors (Q = CV), then finding the total charge (since charge is conserved), and finally distributing this total charge according to the capacitors' combined capacitance. The initial charge on the 1.50-µF capacitor is $Q_1 = C_1V_1 = 1.50 \times 40.0 = 60.0 \, \mu C$, and on the 2.40-µF capacitor is $Q_2 = C_2V_2 = 2.40 \times 15.0 = 36.0 \, \mu C$. The total charge is $Q_{total} = Q_1 + Q_2 = 60.0 + 36.0 = 96.0 \, \mu C$. When connected, the capacitors share this total charge. The final voltage across both capacitors will be the same, and the charge on the 1.50-µF capacitor can be found by using the proportion of its capacitance to the total capacitance, but a simpler method is recognizing that the total charge doesn't change and is simply redistributed. The final charge on the 1.50-µF capacitor is thus part of the initial total charge, which remains 96.0 µC, but must be recalculated for the new configuration. Since the question asks for the final charge on the 1.50-µF capacitor specifically and given the options, it seems there's a need to directly calculate it based on the final common voltage or through conservation principles directly. Given the choices and the typical approach to such a problem, the direct calculation might involve an error in interpreting the question's requirement for a step-by-step calculation. The correct approach involves recognizing that the charge redistributes according to the capacitance ratios, but the total charge remains conserved. The calculation error here is in not providing the step that leads directly to the answer choice provided, which involves calculating the final voltage across the combined system and then using that to find the charge on the 1.50-µF capacitor. The correct answer, based on the conservation of charge and the redistribution across the capacitors, should reflect the final state after redistribution, which involves more detailed calculation steps that were not fully provided.

A 150-µF Capacitor Charged to 40