Two identical capacitors A and B are connected as shown in the circuit. Initially the switch 'S' is closed.…
Two identical capacitors A and B are connected as shown in the circuit. Initially the switch 'S' is closed. Now the switch is opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. The ratio of total electrostatic energy stored in the capacitors before and after introduction of the dielectric is
$3: 1$
$5: 1$
$3: 5$
$5: 3$
Solution
Let the capacitance of each capacitor be $\mathrm{C}$.
Total electrostatic energy stared in capacitor
$
\begin{aligned}
\mathrm{U}_1 & =\frac{1}{2} \mathrm{CE}^2+\frac{1}{2} \mathrm{CE}^2 \\
\mathrm{U}_1 & =\mathrm{CE}^2
\end{aligned}
$
Now dielectric intorduces after pening the switch s.
Energy stored in $\mathrm{A}=\frac{1}{2} \mathrm{KCE}^2$
Let the charge stared in capacitor B initially be Q. In an open circuit, the charge will remain same.
$
\mathrm{Q}=\mathrm{EC}
$
Energy stored in $B=\frac{1}{2} \frac{Q^2}{K C}$
$
=\frac{1}{2} \frac{(\mathrm{EC})^2}{\mathrm{KC}}=\frac{1}{2} \frac{\mathrm{E}^2 \mathrm{C}^2}{\mathrm{KC}}=\frac{1}{2} \frac{\mathrm{CE}^2}{\mathrm{~K}}
$
Total energy, $\mathrm{U}_2=\frac{1}{2} \mathrm{KCE}^2+\frac{1}{2} \frac{\mathrm{CE}^2}{\mathrm{~K}}$
$
=\frac{1}{2} \mathrm{CE}^2\left(\mathrm{~K}+\frac{1}{\mathrm{~K}}\right)
$
Ratio, $\frac{\mathrm{U}_1}{\mathrm{U}_2}=\frac{\mathrm{CE}^2}{\frac{1}{2} \mathrm{CE}^2\left(\mathrm{~K}+\frac{1}{\mathrm{~K}}\right)}$
Here, $\mathrm{K}=3$
$
=\frac{2}{3+\frac{1}{3}}=\frac{6}{10}=\frac{3}{5}
$
