The figure shows two identical parallel plate capacitors connected to a battery with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant (or relative permittivity) 3. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
( 
)
Sol. Before opening the switch potential difference across both the capacitors is V, as they are in parallel. Hence, energy stored in them is,
U A = U B = 
CV 2
∴ U Total = CV 2 = U i ........... (1)
After opening the switch, potential difference across it is V and its capacity is 3C
∴ U A = 
(3C)V 2 = 
CV 2
In case of capacitor B, charge strored in it is q = CV and its capacity is also 3C. Therefore,
U B = 
= 
∴ U Total = 
+ 
= 
CV 2 = 
= U f ......... (2)
From Eqs.(1) and (2)
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems