Figure shows two identical parallel plate capacitors connected to a battery through a switch S. Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the initial total energy stored in the capacitors to the final total energy stored
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
a
9:16
b
5:9
c
2:3
d
3:5
answer is D.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Initial total energy =12CV2+12CV2=CV2When the switch is open, dielectric is introduced. Then capacitance, C = KC = 3C. Energy stored in C is =123CV2=32CV2Since the switch is open, charge will be the same in B so energy is B=12C3V2So total final energy =32CV2+16CV2=9CV2+1CV26=106CV2So required ratio =CV2106CV2=35=3:5
Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!
Figure shows two identical parallel plate capacitors connected to a battery through a switch S. Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the initial total energy stored in the capacitors to the final total energy stored