Question

Each capacitor shown in the figure has a capacitance of $5.0\mu F$. The emf of the battery is $50V$. How much charge will flow through $AB$ if the switch $S$ is closed?

Answer 1

Initially when $S$ is not connected

$\begin{array}{c}{c}_{eff}\frac{2c}{3}q\\ =\frac{2c}{3}\times 50\\ =\frac{5}{2}\times {10}^{-4}\\ =1.66\times {10}^{-4}C\end{array}$

After the switch is made on

Then,

$\begin{array}{c}{c}_{eff}=2c={10}^{-6}\\ Q={10}^{-6}\times 50\\ =5\times {10}^{-4}C\end{array}$

Now the initial charge will remain stored and in the stored in the short capacitor

Therefore,

Net charge flowing is

$\begin{array}{c}5\times {10}^{-4}-1.66\times {10}^{-4}\\ =3.3\times {10}^{-4}C\end{array}$

Hence, the net charge flows is $3.3\times {10}^{-4}C$