Question

Figure (31-E26) shows two identical parallel plate capacitors connected to a battery through a switch S. Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the initial total energy stored in the capacitors to the final total energy stored.

Answer 1

When the switch is closed, both capacitors are in parallel.
$\Rightarrow$The total energy of the capacitor when the switch is closed is given by

E_i$=\frac{1}{2}C{V}^2+\frac{1}{2}C{V}^2=C{V}^2$

When the switch is opened and the dielectric is induced, the capacitance of the capacitor A becomes
$C\text{'}=KC=3C$

The energy stored in the capacitor A is given by
$$\begin{gathered} E_{\mathrm{A}}=\frac{1}{2}C\text{'}{V}^2 \\ \Rightarrow E_{\mathrm{A}}=\frac{1}{2}\left(3C\right)V^2=\frac{3}{2}C{V}^2 \end{gathered}$$

The energy in the capacitor B is given by
$E_{\mathrm{B}}=\frac{1}{2}\times \frac{C}{3}\times V^2$

∴ Total final energy

$$\begin{gathered} E_{\mathrm{f}}=E_{\mathrm{A}}+E_{\mathrm{B}} \\ \Rightarrow E_{\mathrm{f}}=\frac{3}{2}C{V}^2+\frac{1}{6}C{V}^2 \\ \Rightarrow E_{\mathrm{f}}=\frac{9C{V}^2+1C{V}^2}{6}=\frac{10}{6}C{V}^2 \end{gathered}$$
Now,
Ratio of the energies, $\frac{E_1}{E_2}=\frac{C{V}^2}{{\displaystyle \frac{10}{6}C{V}^2}}=\frac{3}{5}$