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Question

A parallel plate capacitor has a plate of length l, width w and separation of plates is d. It is connected to a battery of emf V. A dielectric slab of the same thickness d and of dielectric constant k=4 is being inserted between the plates of the capacitor. At what length of the slab inside plates, will the energy stored in the capacitor be two times the initial energy stored?


A

2l/3

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B

l/2

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C

l/4

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D

l/3

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Solution

The correct option is D

l/3


The explanation for the correct option:

Step 1: Given data:

where length l, width w and separation of plates is d

thickness of dielectric slab is also d and its dielectric constant is k=4.

UinitialUfinal=2

where, U is the energy stored in the capacitor

Step 2: Finding the length at which initial energy will be twice of the stored energy:

Using the equation for energy stored in a capacitor, U=12CV2…………1

Also using capacitance in parallel plate capacitor is given by C=ε(lw/d)

where, C is the capacitance

ε is the permittivity of free space

Ais area of the plate

Comparing above equations and our given condition,

UinitialUfinal=2C1+C2Cstored=2kx+l-xl=24x+l-x=2lx=l3

Hence, option D is correct.


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