Question

The label rubbed off one of the capacitors you are using to build a circuit. To find out its capacitance, you place it in series with a $10\mu \mathrm{F}$ capacitor and connect them to a $9.0\mathrm{V}$ battery. Using your voltmeter, you measure $6.0\mathrm{V}$ across the unknown capacitor. What is the unknown capacitor's capacitance?

Short answer

The unknown capacitor's capacitance is$5\mu F.$

Step-by-step solution

Given Information

Capacitor $=10\mu F$

$V_1=9.0V$

$V_2=6.0V$

Explanation 

In order to solve this problem, you need to understand that both capacitors will accumulate the same amount of charge. Electrons from both plates reach the other plate at the same rate as they leave the plates of one. As soon as you understand this, solving the problem should be straightforward. This accumulated charge can be written as follows based on the definition of capacitance:

$C=\frac{q}{\Delta V}\Rightarrow q=C\Delta V$

Since, as we said, the charge will be the same for both capacitors, we can establish the following relation

$C_1\Delta V_1=q=C_2\Delta V_2,$

which allows us to express the unknown capacitance $C_2$as

localid="1648639155543" $C_2=\frac{\Delta V_1}{\Delta V_2}{C}_1$

It is important to remember that$\Delta V_2$is the voltage across the unknown capacitor and$V$ is the voltage supplied by the battery.
We can therefore express the potential difference across the known capacitor because both capacitors are in series,

$\Delta V_1=V-\Delta V_2$

Substituting this we reach to a final expression, in which we will only need to substitute numerically:

$C_2=\frac{V-\Delta V_2}{\Delta V_2}{C}_1$

Applying numerically for our given situation, we will have

localid="1648636548826" $C_2=\left(\frac{9V-6V}{6V}\times 10\mu F\right)=5\mu \mathrm{F}$

Hence, the unknown capacitor's capacitance is$5\mu F$