Two capacitors of equal capacitance are connected to a battery as… - Vidyadip

Question

Two capacitors of equal capacitance are connected to a battery as shown. Switch $S$ is initially in closed state. Now the switch $S$ is opened and a material of dielectric constant $\epsilon_r=3$ is filled between the plates of the capacitor. Find the ratio of the values of electrical energy stored in the capacitors before and after placing the dielectric material.

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Answer

Initially, when the switch $S$ is closed, both the capacitors will be at the same potential $V$ because they are connected in parallel,
i.e., $ U_i =\frac{1}{2} C_1 V^2+\frac{1}{2} C_2 V^2$
$ =\frac{1}{2}\left(C_1+C_2\right) V^2=CV^2 $
$(i)$ When dielectric material is placed between the plates of the capacitor by keeping the switch $S$ open, then the capacitance of each capacitor will become $\epsilon_r$ times. And since capacitor $A$ is still connected to the battery. Therefore, the value of potential difference on it will be equal to the value $V$ of the initial potential difference, whereas the new value of potential difference on capacitor $B$ will become $V ^{\prime}$
$=\frac{ V }{\epsilon_r}$ and the charge will remain constant. Therefore in this situation the electric energy of the system will be :
$U _f=\frac{1}{2} C ^{\prime} V ^2+\frac{1}{2} C ^{\prime} V ^{\prime 2}$
$=\frac{1}{2} \epsilon_r CV ^2+\frac{1}{2} \epsilon_r C \left(\frac{ V }{\epsilon_r}\right)^2$
$\because V ^{\prime}=\frac{ V }{\epsilon_r}$ and $C ^{\prime}=\epsilon_r C$
$=\frac{1}{2} CV ^2\left(\epsilon_r+\frac{1}{\epsilon_r}\right)=\frac{1}{2} CV ^3\left(3+\frac{1}{3}\right)$
$=$$\frac{1}{2} CV ^2\left(\frac{10}{3}\right)$

On dividing eq.$(ii)$ by eq. $(i),$
Therefore, $ \frac{U_f}{U_i}=\frac{\frac{10}{6} CV^2}{CV^2} $ or $ \frac{U_f}{U_i}$
$=1.66$
$\because \epsilon_r=3$ Ans.

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