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 to capacitor A will :
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a
decrease
b
increase
c
remain the same
d
be zero since circuit will not work
answer is B.
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Detailed Solution
Capacity of system increases so charge on A increases. Therefore, the energy of capacitor A will increase.