Question

A source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key ' $K$ ' is closed, the total energy stored across the combination is $E_1$. Now key ' $K$ ' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $E _2$. The ratio $E _1 / E _2$ will be :

• A. $\frac{1}{10}$
• B. $\frac{2}{5}$
• C. $\frac{5}{13}$
• D. $\frac{5}{26}$

Answer 1

Answer: (C)

(1) Switch is closed
$C _{ eq }=2 C$
Energy $ E _1=\frac{1}{2} C _{ eq } V ^2 $
$ =\frac{1}{2} 2 C \times V ^2 $
$ E _1= CV ^2$
(ii) When switch is opened charge on right capacitor remain $CV$ while potential on left capacitor remain same
Dielectric $K=5$
$ C^{\prime}=K C $
$ C^{\prime}=5 C $
$ E_2=\frac{1}{2}(5 C) V^2+\frac{(C V)^2}{2(5 C)} $
$ E_2=\frac{5 C^2}{2}+\frac{C^2}{10} $
$ E_2=\frac{13 V^2}{5} $
$ \frac{E_1}{E_2}=\frac{C^2}{13 C V^2}=\frac{5}{13} $
$ \frac{E_1}{E_2}=\frac{5}{13}$