A parallel-plate air capacitor of capacitance C0 is connected to a cell of emf V and then disconnected from it. A dielectric constant K, which can just fill the air gap of capacitor, is now inserted in it. Which of the following is incorrect?
A
The potential difference between the plates decreases K times.
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B
The energy stored in the capacitor decreases K times.
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C
The change in energy is 12C0ε2(K−1).
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D
The change in energy is 12C0ε2(1−1k).
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Solution
The correct option is A The potential difference between the plates decreases K times.
As the capacitor is disconnected from cell so charge will remain constant .
here Q=C0ϵ
Before inserting dielectric potential difference between the plates is V0=QdAϵ0
after inserting, potential V=QdAkϵ0=V0k
before inserting dielectric the energy is U0=Q22C0
after inserting, the energy is U=Q22kC0=U0k
Change in energy =U0−U=Q22C0−Q22kC0=Q22C0(1−1k)=(C0ϵ)22C0(1−1k)=12C0ϵ2(1−1k)