The given fig. shows two identical parallel plate capacitors connected to a battery with switch S is closed. The switch is now opened and free space between the plates of the capacitors is now filled with dielectric of K = 3. Then, ratio of energy stored in both capacitors before and after introduction of dielectric is ;
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a
53
b
35
c
75
d
57
answer is B.
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Detailed Solution
As switch S is closed; VA=VB=V ∴ Ei=12CV2+12CV2=CV2After introduction of slab as switch S is now opened, VA=V and CA=3C&CB=3C and Let VB′=new pot. difference across capacitor B. Acc. to law of conservation of charge in case of capacitor B;CV=3CVB′⇒ VB′=V3∴ Ef=12CAV2+12CBVB2=53CV2