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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 remain connected then the energy of capacitor A will
  • a)
    decrease
  • b)
    Increase
  • c)
    remain the same
  • d)
    be zero since circuit will not work
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Two identical capacitors A and B shown in the given circuit are joined...
The energy stored in capacitor of capacity C and potential V is given as

Now when slab of dielectric constant K is introduced without disconnecting the battery the equivalent capacities becomes


∵  energy increases.
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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 remain connected then the energy of capacitor A willa)decreaseb)Increasec)remain the samed)be zero since circuit will not workCorrect answer is option 'B'. Can you explain this answer?
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