Question

A series combination of two uncharged capacitors are connected to a 15-V battery, 300uJ of energy is drawn from the battery. If one of the capacitors has a capacitance of 5.0uF, what is the capacitance of the other?

Answer 1

The net capacitance

$$C=\frac{2W}{V^2}=\frac{2*300*10^{-6}}{15^2}=2.67*10^{-6}\rm\: F=2.7\: \mu F$$

On the other hand

$$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$$

$$C_2=\frac{1}{\frac{1}{C}-\frac{1}{C_1}}$$

$$C_2=\frac{1}{\frac{1}{2.7}-\frac{1}{5.0}}=5.9\:\rm \mu F$$