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