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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance2 Marks
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 $\in_r=3$ 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. Image
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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.,
$ \begin{aligned} 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 \end{aligned} $
(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 $\in_r$ 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 $V ^{\prime}$ $=\frac{ V }{\in_r}$ and the charge will remain constant. Therefore in this situation the electric energy of the system will be :
$U _f=\frac{1}{2} C ^{\prime} V ^2+\frac{1}{2} C ^{\prime} V ^{\prime 2}$
$=\frac{1}{2} \in_r CV ^2+\frac{1}{2} \in_r C \left(\frac{ V }{\in_r}\right)^2$
$\because V ^{\prime}=\frac{ V }{\in_r}$ and $C ^{\prime}=\in_r C$
$=\frac{1}{2} CV ^2\left(\in_r+\frac{1}{\in_r}\right)=\frac{1}{2} CV ^3\left(3+\frac{1}{3}\right)$
$=$$\frac{1}{2} CV ^2\left(\frac{10}{3}\right)$
Image
On dividing eq.(ii) by eq. (i),
Therefore, $ \frac{U_f}{U_i}=\frac{\frac{10}{6} CV^2}{CV^2} $ or $ \begin{array}{l} \frac{U_f}{U_i} \\ =1.66 \\
\because \in_r=3 \quad \text { Ans. } \end{array} $

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