The capcitance of a parallel plate capacitor is `C_(0)` when the space between the plate is vacuum. Now this space is filled by a slab of dielectric constant `K`. The capacitor is connected to a battery emf `V` and the slab is taken out.
(A) Charge `(K_1)C_(0)V` flow through the battery
(B) Energy `(K-1)C_(0)V^(2)` is absorved by the cell
(c) The external agent has to do `(1)/(2)(K-1)C_(0)V^(2)` amount of work to take the slabe out
(D) the heat produced in connecting wires is `(1)/(2)(K-1)C_(0)V^(2)`

Options

A. `A,B,C`
B. `A,B,D`
C. `B,C,D`
D. A,B,C,D`

Answer

D

Explanation

`QgtQ'`, amount of charge absorbed by cell
`DeltaQ=(Q-Q')=(K-1)C_(0)V^(2),(D)` is O.K.
Change in stored energy in capacitor
`DeltaU=(1)/(2)(K-1)C_(0)V^(2)`
`W_(b)=DeltaU+HimpliesH=(1)/(2)(K-1)C_(0)V^(2),(D)` is O.K.
`W_(1rarr2)=|U_(2)-U_(1)|=(1)/(2)(K-1)C_(0)V^(2),(c)` is O.K.