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Question

A parallel plate condenser is charged by connecting it to a battery. The battery is disconnected and a glass slab is introduced between the plates. Then,


  1. electric intensity incrases
  2. potential increases
  3. energy decreases
  4. capacity decreases

A
electric intensity incrases
B
energy decreases
C
potential increases
D
capacity decreases
Solution
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When the capacitor is charged and battery is removed the charge present on capacitor remains constant. (Law of conservation of charge)
We know C=ε0Ad
When slab is introduced, C=kε0Ad
C increases and Q is constant.
We know: Q=CV
V has to decrease.
We know energy store in capacitor =12QV
as V decreases, energy decreases.

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