Question

A parallel plate capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E. If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become :

A

6E, 6C

B

E, C

C

D

E, 6C

Medium

Solution

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Correct option is C)

Let C be charged to v

Now when dielectric is added.
C increases by kC and as battery is disconnected is constant.
V decreases by k.






and

Solve any question of Electrostatic Potential and Capacitance with:-

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