In Fig.25-37 $\mathbf{V}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{V}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{C}}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{\mu F}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{C}}_{\mathbf{2}}\mathbf{=}{\mathbf{C}}_{\mathbf{3}}\mathbf{=}\mathbf{20}\mathbf{\mu F}$.Switch S is first thrown to the left side until capacitor 1 reaches equilibrium. Then the switch is thrown to the right. When equilibrium is again reached, how much charge is on capacitor 1?

$$\begin{gathered} \mathrm{V}=10\mathrm{V} \\ {\mathrm{C}}_1=10\mathrm{\mu F} \\ {\mathrm{C}}_2={\mathrm{C}}_3=20\mathrm{\mu F} \end{gathered}$$

For solving the given problem, use the concept of conservation of charge. First, find the charge due to the battery when the switchis connected to the left. After the switch is connected to the right and equilibrium is reached, find the charge on capacitor1by the concept of conservation of charge.

Formulae are as follows:

$\mathrm{q}=\mathrm{CV}$

Where C is capacitance, V is the potential difference, and q is the charge.

Here, consider the conservation of charge. Initially S is connected to the left.

The total charge is given by,

$$\begin{gathered} \mathrm{Q}={\mathrm{C}}_1\mathrm{V} \\ =10\mathrm{\mu F}\times 10\mathrm{V} \\ =100\mathrm{\mu C} \end{gathered}$$

If ${\mathrm{q}}_1,{\mathrm{q}}_2$and ${\mathrm{q}}_3$are the charge on the capacitors ${\mathrm{C}}_1,{\mathrm{C}}_2$and ${\mathrm{C}}_3$respectively, after the switch is thrown to the right and equilibrium is reached, then according to charge conservation,

$\mathrm{Q}={\mathrm{q}}_1+{\mathrm{q}}_2+{\mathrm{q}}_3............................\left(1\right)$

Since ${\mathrm{C}}_2\mathrm{and}{\mathrm{C}}_3$are identical, so ${\mathrm{q}}_2={\mathrm{q}}_3$. They are in parallel with ${\mathrm{C}}_1$so that${\mathrm{V}}_1={\mathrm{V}}_3$

i.e.$\frac{{\mathrm{q}}_1}{{\mathrm{C}}_1}=\frac{{\mathrm{q}}_3}{{\mathrm{C}}_3}$

$$\begin{gathered} {\mathrm{q}}_1=\frac{{\mathrm{q}}_3}{{\mathrm{C}}_3}{\mathrm{C}}_1 \\ {\mathrm{q}}_1=\frac{10\mathrm{\mu F}\times {\mathrm{q}}_3}{20\mathrm{\mu F}} \\ {\mathrm{q}}_1=\frac{{\mathrm{q}}_3}{2} \end{gathered}$$

And,${\mathrm{q}}_2={\mathrm{q}}_3$

Therefore, the equationcan be written as,

$$\begin{gathered} \mathrm{Q}={\mathrm{q}}_1+{\mathrm{q}}_2+{\mathrm{q}}_3 \\ =\frac{{\mathrm{q}}_3}{2}+{\mathrm{q}}_3+{\mathrm{q}}_3 \\ =\frac{5{\mathrm{q}}_3}{2} \end{gathered}$$

Solving for ${\mathrm{q}}_3$,

${\mathrm{q}}_3=\frac{2\mathrm{Q}}{5}$

By substituting the value of Q , find${\mathrm{q}}_3$,

$$\begin{gathered} {\mathrm{q}}_3=\frac{2\times 100\mathrm{\mu C}}{5} \\ =40\mathrm{\mu C} \end{gathered}$$

Therefore,

$$\begin{gathered} {\mathrm{q}}_1=\frac{{\mathrm{q}}_3}{2} \\ {\mathrm{q}}_1=\frac{40\mathrm{\mu C}}{2} \\ =20\mathrm{\mu C} \end{gathered}$$

Hence, the charge on the capacitor 1 is ${\mathrm{q}}_1=20\mathrm{\mu C}$

Therefore, by using the concept of charge conservation, find the charge on the capacitor 1 .