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Question

A parallel plate capacitor of plate area A and plate separation d is charged to p.d. V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively the magnitude of charge on each plate, the electric field between the plates after insertion of dielectric and work done on the system in the process of inserting the slab, then-
  1. Q=ϵ0AVd
  2. E=VKd
  3. Q=ϵ0KAVd
  4. W=ϵ0AV22d(11K)

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

Before insert dielectric, capacitance, C0=Aϵ0d
Magnitude of charge on each plate, Q=C0V=Aϵ0dV
Before disconnect the battery, filed between the plates, E0=Vd
Afe insert dielectric, electric filed, E=E0K=VKd
Before insert dielectric, work done, =Wi=12C0V2
After insert work done, Wf=12(KC0)(VK)2. (Cf=Aϵ0Kd,Vf=Ed)
So net work done,W=WiWf=12C0V212(KC0)(VK)2=12C0V2(11K)=Aϵ0V2d(11K)

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