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secret5437
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Homework Statement
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)
Homework Equations
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.