A 6.93 µF capacitor charged to 94 V and a 1.76 µF capacitor charged to 81 V are connected to each other, with the two positive plates connected and the two negative plates connected. What is the final potential difference across the 1.76 µF capacitor? Answer in units of V.

225.56V

Explanation:

Q=cv

6.93 µF capacitor was charged to 94 V, therefore Q₁ = 6.93 * 10^-6 x 94 = 651.42µC

1.76 µF capacitor charged to 81 V, therefore Q₂ = 1.76 * 10^-6 x 81 = 142.56µC

When the capacitor are brought together, the charges on them move to maintain equilibrum, so therefore, each capacitor has 0.5 (142.56µC + 651.42µC) = 396.99µC

The final potential difference across the 1.76 µF capacitor = Q/c = 396.99µC/1.76 µF = 225.56V