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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:

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

The correct option is C potential difference across the capacitor is increased
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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