Question

I have a capacitor who’s potential difference between the plates is maintained at 5V using a battery. If the capacitance is 5 F and a dielectric of relative permittivity 3 is placed between the plates such that it fully fills the space, What is the change in energy?

A
Increases by 6.24×104J
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Decreases by 6.24×104J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Increases by 9.36×104J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Decreases by 9.36×104J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A Increases by 6.24×10−4JHere the initial energy of the capacitor is E=12CV2 E=12×5×10−6×52 E=3.12×10−4J Now when calculating the final energy, we don’t have to change V since it is maintained at 5V by means of a battery. Only Capacitance becomes C’ = kC E′=12C′V2E′=123CV2E′=3EE′=9.36×10−4J Energy change is, E′−E=6.24×10−4J

Suggest Corrections
0
Join BYJU'S Learning Program
Select...
Join BYJU'S Learning Program
Select...