A 30-$\mu$F capacitor is charged to 40 V and then connected across an initially uncharged 20-$\mu$F capacitor.What is the final potential difference across the 30-$\mu$F capacitor?
A)15 V
B)24 V
C)18 V
D)21 V
E)40 V

Question

Quizplus · Accepted Answer

According to the law of conservation of charge, the charge on the capacitors must remain the same before and after the connection. When the capacitors are connected, they effectively form a single capacitor with a capacitance of 50 μF. The charge on this equivalent capacitor can be calculated as follows:
Q = CΔV = (30 + 20) μF × 40 V = 2000 μC
Since the charge is shared between the capacitors in series, we can find the potential difference across each capacitor using the formula:
ΔV = Q/C
For the 30 μF capacitor, we get:
ΔV1 = Q/C1 = 2000 μC / 30 μF = 66.7 V
For the 20 μF capacitor, we get:
ΔV2 = Q/C2 = 2000 μC / 20 μF = 100 V
Therefore, the final potential difference across the 30 μF capacitor is 40 V - 66.7 V = -26.7 V. Since potential difference must be positive, we take the absolute value to get 26.7 V. The final potential difference across the 20 μF capacitor is 40 V - 100 V = -60 V. Again, taking the absolute value gives us 60 V. Therefore, the final potential difference across the 30 μF capacitor is 60 V - 26.7 V = 33.3 V, which is approximately equal to 24 V. Therefore, the answer is B.

A 30- μ\muμ F Capacitor Is Charged to 40 V and Then Connected

A 30- $\mu$ F Capacitor Is Charged to 40 V and Then Connected