Question

Consider the situation shown in figure (31-E23). The switch S is open for a long time and then closed. (a) Find the charge flown through the battery when the switch S is closed. (b) Find the work done by the battery.

Figure

(c) Find the change in energy stored in the capacitors.
(d) Find the heat developed in the system.

Answer 1

Since the switch is opened for a long time, the capacitor is in series.
∴ $C_{eq}=\frac{C}{2}$

When the switch is closed, the charge flown from the battery is given by

$q=\frac{\mathrm{C}}{2}\times \epsilon =\frac{\mathrm{C}\epsilon }{2}$

(b) Work done, W$=q\times V$
$\Rightarrow q=\frac{\mathrm{C}\epsilon }{2}\times \epsilon =\frac{\mathrm{C}{\epsilon }^2}{2}$

(c) The change in the energy stored in the capacitors is given by
$$\begin{gathered} ∆E=\frac{1}{2}{\mathrm{C}}_{eq}\times v^2 \\ =\frac{1}{2}\times \frac{\mathrm{C}}{2}\times {\epsilon }^2=\frac{1}{4}\mathrm{C}{\epsilon }^2 \end{gathered}$$

(d) The heat developed in the system is given by

$$\begin{gathered} H=∆E \\ \Rightarrow H=\frac{1}{4}\mathrm{C}{\epsilon }^2 \end{gathered}$$