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

B

energy stored in the capacitor reduces

C

potential difference across the capacitor is decreased

D

potential difference across the capacitor is increased

Easy

Solution

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Correct option is D)

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

Solve any question of Electrostatic Potential and Capacitance with:-

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