A parallel plate capacitor has 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?
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a
l/4
b
l/2
c
2l/3
d
l/3
answer is D.
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Detailed Solution
Energy stored in capacitor (i) V1=12C1V2=12ε0AdV2=12ε0lwdV2 Similarly inii,parts with air and dielectric can be considered in parallelU2=12C2V2=12ε0A1d+ε0A2dV2=12ε0l−xwd+4ε0xwdV2 & also given U2=2U1⇒l−x+4x=2l⇒x=l/3