`{:(,"Column-I",,"Column-II"),("(A)",underset("Plate air core capacitor are slowly pulled apart")"Plates of an isolated charged parallel",,underset("increases in the process")"(P)Electric energy stored inside capacitor increases in the process"),("(B)",underset("air cored capacitor to completely fill the space between plates")"A dielectric is slowly inserted inside an isolated and charged parallel plate",,underset("capacitor remain unchanged")"(Q)Force between the two plates of the"),("(C)",underset("connected across a battery are slowly pulled apart")"Plates of a parallel plate capacitor",,underset("plates remain unchanged")"(R) Electric field in the region between"),("(D)",underset("a battery to completely fill the space between plates")"A dielectric slab is slowly inserted inside a parallel plate capacitor conneted across",,underset("capacitor decreases in the process")"(S)total electric energy stored inside"),(,,,"(T)Electriec field in the region between plates decreases"):}`

Answer

[(A)PQR(B)QST(C)ST(D)PR]

Explanation

(A) Field remains same as charge is same Hence energy stored increases (P) & (Q) as field is same and (R)
(B) Since distance between plate and charge remains same and field decreases (s&T)
`F=(Q^(2))/(2 in_(0) A)` charge is same hence force remain same (Q)
(C) Since connected to battery P.D remains same and as distance increases the electric field decreases (T) Also as capacitance decreases the charge also decreases decreasing the force of interaction
(D) Capacitance increases and P.D remaining same, thus charge and energy increases (P) and force of interaction increases Field remains same as distance and P.D remains same (R) and energy increases as