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-

Options

A. decreases
B. increases
C. Remains the same
D. Be zero since circuit will not work

Answer

B

Explanation

`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)`