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, 

A

potential increases

B

electric intensity incrases

C

energy decreases

D

capacity decreases

Easy

Solution

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Correct option is C)

When the capacitor is charged and battery is removed the charge present on capacitor remains constant. (Law of conservation of charge)
We know
When slab is introduced,
increases and is constant.
We know:  
has to decrease.
We know energy store in capacitor
as V decreases, energy decreases.

Solve any question of Electrostatic Potential and Capacitance with:-

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