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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
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B
E, C
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C
E6,6C
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D
E, 6C
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Solution

## The correct option is C E6,6CLet C be charged to v⇒E=12×C×V2Now when dielectric is added.C increases by kC and as battery is disconnected υ is constant.∴ V decreases by k.∴C1=kC;V1=Vk⇒E1=12c1v21E1=12kc×vk×vkE1=12cv26c1=16⋅E∴E1=E6and C1=6C

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