Question 89 of 180

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 $\dfrac{{E}_{1}}{{E}_{2}}$ will be

A$\dfrac{1}{10}$B$\dfrac{2}{5}$C$\dfrac{5}{13}$D$\dfrac{5}{26}$

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Answer: (C) $\dfrac{5}{13}$

(i) When the switch is closed ${C}_{eq}=2C$
Charge on each capacitor will be $q=CV$.

Energy ${E}_{1}=\dfrac{1}{2}{C}_{eq}{V}^{2}$ $=\dfrac{1}{2}2C\times {V}^{2}$
${E}_{1}=C{V}^{2}$(ii) When the switch is opened charge on the right capacitor remain $CV$ while the potential on the left capacitor remains the same.
Dielectric $k=5$
${C}^{'}=kC$ ${C}^{'}=5C$
Now to calculate ${E}_{2}$,
${E}_{2}=\dfrac{1}{2}{C}^{'}{V}^{2}+\dfrac{{q}^{2}}{2{C}^{'}}$
$\Rightarrow {E}_{2}=\dfrac{1}{2}\left(5C\right){V}^{2}+\dfrac{{\left(CV\right)}^{2}}{2\left(5C\right)}$
$\Rightarrow {E}_{2}=\dfrac{5C{V}^{2}}{2}+\dfrac{C{V}^{2}}{10}$
$\Rightarrow {E}_{2}=\dfrac{13C{V}^{2}}{5}$
$\Rightarrow \dfrac{{E}_{1}}{{E}_{2}}=\dfrac{5}{13}$