Question

Figure 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.

• A. 9 : 16
• B. 5 : 9
• C. 2 : 3
• D. 3 : 5

Answer 1

The correct option is D 3 : 5

Initial condition when switch S is closed
$E_i=E_{Ai}+E_{Bi};E_{Ai}=\frac{1}{2}C{V}^2;E_{Bi}=\frac{1}{2}C{V}^2$
$E_i=\frac{1}{2}C{V}^2+\frac{1}{2}C{V}^2$
$E_i=C{V}^2$

Now, in second case the switch is disconnected and dielectric of constant 3 is inserted in both capacitors

$E_f=E_{A_f}+E_{B_f}$
$E_{A_f}=\frac{1}{2}\times 3\times C\times V^2=\frac{3}{2}C{V}^2$

In B the potential changes as dilectric is inserted to a free charged capacitor, which is given as $\frac{V}{K}$.


$\begin{array}{cc}{E}_{B_f}& =\frac{1}{2}\times 3C\times {\left(\frac{V}{3}\right)}^2\\ & =\frac{1}{2}\times 3C\frac{V^2}{9}\\ E_{Bf}& =\frac{1}{2}\times \frac{1}{3}{CV}^2\end{array}$

$\begin{array}{cc}{E}_f& =\frac{3}{2}C{V}^2+\frac{1}{6}C{V}^2\\ & =\frac{1}{2}C{V}^2[3+\frac{1}{3}]\\ & E_f=\frac{1}{2}\times \frac{10}{3}\times C{V}^2\end{array}$

Therefore, the ratio of energies is given as:
$\frac{E_i}{E_f}=\frac{C{V}^2}{\frac{1}{2}\times \frac{10}{3}\times C{V}^2}=3:5$