posted by BOB on .
A 170 -pF capacitor is connected in series with an unknown capacitance, and as a series combination they are connected to a battery with an emf of 25.0 V. If the 170 -pF capacitor stores 135 pC of charge on its plates, what is the unknown capacitance?
Q1 = C1V1.
V1 = Q1/C1 = 135/170 = 0.794 Volts.
V2 = 25 - 0.794 = 24.206 Volts.
C2 = (0.794 / 24.206) * 170pF=5.576pF.
NOTE: The largest voltage appears across the smallest capacitor.