The switch K is closed. Now,the switch K is o | vedclass

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(A) Initially,when the switch $K$ is closed,both capacitors $A$ and $B$ are in parallel with the battery of voltage $V$. The energy stored in the syst

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(A) Initially,when the switch $K$ is closed,both capacitors $A$ and $B$ are in parallel with the battery of voltage $V$. The energy stored in the system is:
$U_1 = \frac{1}{2}CV^2 + \frac{1}{2}CV^2 = CV^2$ ... $(i)$
When the switch $K$ is opened,capacitor $A$ remains connected to the battery,so its voltage remains $V_A = V$. Capacitor $B$ is disconnected,so its charge $Q_B$ remains constant. Initially,$Q_B = CV$. After filling with dielectric $K=3$,the new capacitance is $C' = KC = 3C$. The new voltage across $B$ is $V_B = \frac{Q_B}{C'} = \frac{CV}{3C} = \frac{V}{3}$.
The new energy of the system is:
$U_2 = \frac{1}{2}CV^2 + \frac{1}{2}(3C)\left(\frac{V}{3}\right)^2 = \frac{1}{2}CV^2 + \frac{1}{2}(3C)\frac{V^2}{9} = \frac{1}{2}CV^2 + \frac{1}{6}CV^2 = \frac{3+1}{6}CV^2 = \frac{4}{6}CV^2 = \frac{2}{3}CV^2$ ... $(ii)$
Taking the ratio $\frac{U_1}{U_2}$:
$\frac{U_1}{U_2} = \frac{CV^2}{\frac{2}{3}CV^2} = \frac{3}{2}$

The switch $K$ is closed. Now,the switch $K$ is opened and both capacitors are filled with a dielectric of constant $K = 3$. What is the ratio of the energy of the system when the switch is closed to when it is open,$\frac{U_1}{U_2}$?

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