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Question

In the figure shown, a parallel plate capacitor has a dielectric of width d/2 and dielectric constant K=2. The other dimensions of the dielectric are same as that of the plates. The plates P1 and P2 of the capacitor have area 'A' each. The energy of the capacitor is :
74450.jpg
  1. 2ϵ0AV2d
  2. ϵ0AV23d
  3. 32ϵ0AV2d
  4. 2ϵ0AV23d

A
ϵ0AV23d
B
32ϵ0AV2d
C
2ϵ0AV2d
D
2ϵ0AV23d
Solution
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Using

Energy=12×Cequivalent×V2

We need to find the equivalent capacitance of the capacitor.

Using law for series combination of capacitors,

1Cnet=1C1+1C2+1C3

where, C1=Aϵ0x ( x is the width between capacitor walls and dielectric on one side )

C2=Aϵ0κ(d2)=4Aϵ0d

C3=Aϵ0(d2x)

therefore,

1Cnet=x(Aϵ0)+(d2x)(Aϵ0)+d(4Aϵ0)=3d(4Aϵ0)Cnet=4Aϵ03d

Therefore energy =2Aϵ0V23d

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