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 $\frac{E_1}{E_2}$ will be

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

Answer 1

Answer: (C)

$\frac{5}{13}$

Answer 2

The ratio $\frac{E_1}{E_2}$ will be

$E_1 = \frac{1}{2}(2C)V^2$

$⇒ E_1 = CV^2...(i)$

$E_2 = \frac{1}{2}(5C)V^2+\frac{1}{2}\frac{(CV)^2}{5C}$

= $\frac{13}{5}CV^2$

$⇒\frac{E_1}{E_2}$

= $\frac{5}{13}$