The capacitance of a parallel plate capacitor is `C` when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant `k.` The capacitor is connected to a cell of `emfE`, and the slab is taken out

Options

A. charge `EC_(0)(K-1)` flows through the cell
B. energy `E^(2)C_(0)(K-1)` is absorbed by the cell
C. the energy stored in the capacitor is reduced by `E^(2)C_(0)(K-1)`
D. the external agent has to do `E^(2)(C_(0)(K-1)` amount of work to take out the slab.

Answer

A, B

Explanation

Initial charge on capacitor `= KC_(o) E`
Charge after removing slab `= C_(o)E`
Amount of charge flows through the cell
`=KC_(o)E-C_(o)E=C_(o)E(K-1)`
Energy absorbed by cell `= C_(0)E(K-1)E_(0)=C_(0)E^(2)(K-1)`
Initial energy stored in capacitor `= 1//2 kC_(0) E^(2)`
Final energy stored in capacitor `= 1//2 C_(0) E^(2)`
Energy reduces in capacitor by
`1//2C_(0)E^(2)(k-1)`
Work done by external agent `(1)/(2)E^(2)C_(0)(K-1)`.