A 190 -pF capacitor is connected in series with an unknown capacitance, and as a series combination they are c?
If the 190 -pF capacitor stores 125 pC of charge on its plates, what is the unknown capacitance?
- JacksonLv 79 years agoBest Answer
You are missing a critical piece of information.
You need the Voltage of the Source.
The Formula is Q = CV
So Q1 = (C1) x (V1)
For the first capacitor Q1 = 125 pC = (190pF) (V)
So the voltage across the 190 pF capacitor is 0.658 Volts
Since this is a series circuit, the Charge Q is the same across each capactitor.
So the charge Q across the 2nd unknown capacitor is the same --- 125 pC
So Q2 = (C2) x (V2)
Therefore, Q2 = 125 pC = (C2) ((V2 which is the Source Voltage minus 0.658 Volts)
Voltage (source) = V1 + V2
Therefore --- you need the source voltage to calculate C2