The capacitor in Fig. 25-25 has a capacitance of 27 µF and is initially uncharged. The battery provides a potential difference of 118 V. After switch S is closed, how much charge will pass through it? (picture is in the attachment)
C = Q/V
U = Q2/2e = QV/2 = CV2/2
The Attempt at a Solution
I think this is really easy, but for some reason, i just can't get it.