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 `CE(k-1)` flows through the cell
B. enegry `E^(2)C(k-1)` is obserobed by the cell
C. the enegry stored in the capacitor is reduced by `E^(2)C(k-1)`
D. the external agent has to do `(1)/(2)E^(2)C(k-1)` amount of work to take the slab out.

Answer

A, B, D

Explanation

(A) `Deltaq = KCE - KC = CE (K -1)`
(B) Enegry absorbed by cell `= E Deltaq = CE^(2) (K - 1)`
(C ) `U_(i) = (1)/(2)KCE^(2) &U_(f) = (1)/(2)CE^(2)`
`DeltaU = U_(i) - U_(f) = (1)/(2)CE^(2) (K-1)`
(D) `U_(i) +` work done by External agent `=` enegry aboserbedby cell `+U_(f)`
work done by External agent `=` enegry aboserbed cell `+U_(f) -U_(i)`
`= CE^(2) (K-1) +(1)/(2)CE^(2) -(1)/(2)KCE^(2) = (1)/(2)CE^(2)(K-1)`