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Question

A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of capacitor so as to fill the space between the plates. If Q,E and W denote respectively, the magnitude of charge on each plate electric field between the plates (after the slab is inserted), and work done on the system, in question, in the process of inserting the slab, then which is wrong?

A
Q=ϵ0AVd
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B
Q=ϵ0KAVd
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C
E=VKd
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D
W=ϵ0AV22d(11K)
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Solution

The correct option is B Q=ϵ0KAVd
Initial capacitance, C0=Aϵ0d
So initial charge on the capacitor is: Q0=C0V=AVϵ0d
As the battery is disconnected so charge remains unchanged.
So, Q=Q0=AVϵ0d
After insert the dielectric the capacitance becomes, C=KC0 where K dielectric constant.
Now potential becomes, V=Q/C=Q0/KC0=V/K
Field, E=V/d=VKd
Initial energy, U0=12C0V2=AV2ϵ02d and
final energy Uf=12CV2=(1/2)(KC0)(V/K)2=AV2ϵ02Kd
Since, work done =Decrease in Potential Energy
Work done, W=ΔU=(UfU0)=U0Uf=12AV2ϵ0d(11/K)

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