Question

You have 2 capacitors, one with a capacitance of 10.7 c 10^-6 F and the other with an unknown capacitance. You connect the two capacitors in series with a voltage of 301 V applied across the capacitor pair. You discover that, as a result, the unknown capacitor has a charge of 0.00161 C. Find it’s capacitance

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Answer 1

Series equivalent

$C_{eq}=\frac{C_1C_2}{C_1+C_2}$

In series, the charge on each capacitor is same as total charge that flows.

$Q=VC_{eq}=V\frac{C_1C_2}{C_1+C_2}$

$\frac{1}{C_1}+\frac{1}{C_2}=\frac{V}{Q}$

Putting the values you get

$C_2=10.7\times10^{-6}\textup{ F}$