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Question

Space between the plates of a parallel plate capacitor is filled with a dielectric slab. The capacitor is charged and then the supply is disconnected to it. If the slab is now taken out , then:
  1. work is not done to take out the slab
  2. potential difference across the capacitor is decreased
  3. potential difference across the capacitor is increased
  4. energy stored in the capacitor reduces

A
potential difference across the capacitor is decreased
B
energy stored in the capacitor reduces
C
work is not done to take out the slab
D
potential difference across the capacitor is increased
Solution
Verified by Toppr

When a capacitor is charged and then the supply is disconnected charge present on it becomes constant (law of conservation of charge)
Initial capacitance C=kε0Ad (with slab)
Final capacitance C=ε0Ad (slab removed)
We know Q=CV
As C decreases and Q remains constant V has to increase.

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