Question

A battery with an emf of $60V$is connected to the two capacitors shown in FIGURE $P26.62$. Afterward, the charge on capacitor $2$is $450\mu C$. What is the capacitance of capacitor $2$?

Short answer

The capacitance of capacitor$2$is$20\mu C$.

Step-by-step solution

Given information

We have given that the charge on capacitor $2$is $450\mu F$.

We need to find the capacitance of capacitor$2$.

Simplify 

Considering this charge will be $q$, we have

$C_1{U}_1=q=C_2(U-U_1)$,

Where we consoder$U$be the emf of the battery, which since the capacitors are in series, leads to $U_2=U-U_1$. Now we can express the capacity of the second capacitor as

$C_2=\frac{q}{U-U_1}$

Putting the value of$U_1$expressed with the charge and emf

localid="1648642860563" $$\begin{gathered} C_2=\frac{q}{U-{\displaystyle \frac{q}{C_1}}}=\frac{q}{{\displaystyle \frac{U{C}_1-q}{C_1}}} \\ =\frac{q{C}_1}{U{C}_1-q} \end{gathered}$$

Using the numerical values

$C_2=\frac{4.5·{10}^{-4}·1.2·{10}^{-5}}{60·1.2·{10}^{-5}-4.5·{10}^{-4}}=2·{10}^{-5}C$