Question

Each capacitor shown in figure (31-E10) has a capacitance of 5⋅0 µF. The emf of the battery is 50 V. How much charge will flow through AB if the switch S is closed?

Answer 1

Initially, when the switch S is open, the equivalent capacitance is given by
$$\begin{gathered} C_{\mathrm{eq}}=\frac{2C\times C}{3\mathrm{C}} \\ \Rightarrow C_{\mathrm{eq}}=\frac{2}{3}C=\frac{2}{3}\times 5.0\mathrm{\mu F} \end{gathered}$$

The charge supplied by the battery is given by
$$\begin{gathered} Q=C_{\mathrm{eq}}\times V \\ \Rightarrow Q=\frac{2}{3}\times (5.0\mathrm{\mu F})\times (50\mathrm{V}) \\ \Rightarrow Q=\frac{500}{3}\mathrm{\mu C} \end{gathered}$$
When the switch S is closed, no charge goes to the capacitor connected in parallel with the switch.
Thus, the equivalent capacitance is given by

$C_{\mathrm{eq}}=2C=2\times 5.0=10\mathrm{\mu F}$
The charge supplied by the battery is given by
$Q=10\mathrm{\mu F}\times 50=500\mathrm{\mu C}$
The initial charge stored in the shorted capacitor starts discharging."?
Hence, the charge that flows from A to B is given by
$$\begin{gathered} Q_{\mathrm{net}}=500\mathrm{\mu C}-\frac{500}{3}\mathrm{\mu C} \\ \Rightarrow Q_{\mathrm{net}}=3.3\times {10}^{-4}\mathrm{C} \end{gathered}$$