# 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?

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- 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

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