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Question

A parallel-plate air condenser of plate area A and separation d is charged to potential V and then the battery is removed. Now a slab of dielectric constant k is introduced between the plates. If Q, E and W denote respectively the magnitude of charge on each plate, the electric field between the plates (after introduction of dielectric slab) and work done on the system in the process of introducing the slab, then

A
W=ε0AVh22d(11/k)
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B
Q=ε0KAVd
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C
Q=ε0AVd
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D
E=Vkd
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Solution

The correct options are
B W=ε0AVh22d(11/k)
C Q=ε0AVd
D E=Vkd
As battery is removed so charge remains constant. Q0=C0V
here C0=Aϵ0d, V=Q0dAϵ0
after insert dielectric, the filed, E=Q0Akϵ0=Vkd
and charge Q=Q0=C0V=Aϵ0Vd and C=kC0
The work done , W= energy loss in capacitor =Q202C0Q202kC0
W=Q202C0(11k)=1/2C0V2(11k)=Aϵ0V22d(11k)
Ans :(C),(D)

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