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Two identical capacitors A and B shown i...

Two identical capacitors A and B shown in the given circuit are joined in series with a battery. If a dielectric slab of dielectric constant K is slipped between the plates of capacitor B and battery remains connected, then the energy of capacitor A will-

A

decreases

B

increases

C

Remains the same

D

Be zero since circuit will not work

Text Solution

Verified by Experts

The correct Answer is:
B
`q_(A)=q_(B)=q`
`q=C/2 V`
`q'=(KC)/(K+1) VrArr q' gt q`
therefore, `U_(A)=(q^(2))/(2C)`
`U_(2)^(1) gt U_(A)`
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