Question

When switch $S$ in figure is pushed to the left, the plates of capacitor $C_1$ acquire a potential difference $V$. Initially, capacitors $C_2$ and $C_3$ are uncharged. The switch is now pushed to the right. What are the final charges $q_1,q_2$ and $q_3$ on the corresponding capacitors.

Answer 1

Applying loop rule:

$\frac{q_0-q}{C_1}-\frac{q}{C_3}-\frac{q}{C_2}=0$

or $\frac{q_0-q}{C_1}=q(\frac{1}{C_2}+\frac{1}{C_3})=2\left(\frac{C_2{+C}_3}{C_2{C}_3}\right)$

$\therefore q=\left(\frac{C_2{C}_3{q}_0}{C_1{C}_2+C_1{C}_3+C_2{C}_3}\right)$

$\therefore q_1=q_0-q=C_{1}V-\left(\frac{C_2{C}_3{q}_0}{C_1{C}_2+C_1{C}_3+C_2{C}_3}\right)$

$=\frac{{C_1}^{2}V\left(C_2{+C}_3\right)}{C_1{C}_2+C_1{C}_3+C_1{C}_3}$

Also

$q_2=q_3=q=\frac{C_2{C}_3{q}_0}{C_1{C}_2+C_1{C}_3+C_1{C}_3}$

$=\frac{C_2{C}_3{C}_{1}V}{C_1{C}_2+C_1{C}_3+C_2{C}_3}$