Q.
Figure given below shows two identical parallel plate capacitors connected to a battery with switch $S$ closed. The switch is now opened and the free space between the plate of capacitors is filled with a dielectric of dielectric constant 3. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric
A) 3 : 1
B) 5 : 1
C) 3 : 5
D) 5 : 3
Explanation
(c) Initially potential difference across both the capacitor is same hence energy of the system is
$U_{1} = \frac{1}{2}CV^{2} + \frac{1}{2}CV^{2} = CV^{2}$ …(i)
In the second case when key $K$ is opened and dielectric medium is filled between the plates, capacitance of both the capacitors becomes $3C,$ while potential difference across $A$ is $V$ and potential difference across $B$ is $\frac{V}{3}$ hence energy of the system now is
$U_{2} = \frac{1}{2}(3C)V^{2} + \frac{1}{2}(3C)\left( \frac{V}{3} \right)^{2} = \frac{10}{6}CV^{2}$ …(ii)
So, $\frac{U_{1}}{U_{2}} = \frac{3}{5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.