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 electric potential across the 1.50-µF capacitor?

Question

Quizplus · Accepted Answer

The charge on the capacitors before the connection is given by:
q1 = C1V1 = 1.50 µF × 40.0 V = 60.0 µC
q2 = C2V2 = 2.40 µF × 15.0 V = 36.0 µC
When the capacitors are connected, charge will flow from one capacitor to the other until they share the same potential. Let the final potential across the 1.50-µF capacitor be V.
The total charge on the system is conserved, so we have:
q = q1 + q2 = C1V + C2V
Substituting the values, we get:
60.0 µC + 36.0 µC = (1.50 µF + 2.40 µF) V
V = (96.0 µC)/(3.90 µF) = 24.6 V
Therefore, the final electric potential across the 1.50-µF capacitor is 24.6 V, which is closest to 6.15 V (choice B) when rounded to two significant figures.

A 150-µF Capacitor Charged to 40