Column I gives ertain situtaions situations in which capacitance of a capacitor is charge by different means. Column II gives resulting effect under different conditions. Match the proper entries from colum - 2to column -1 using the codes given below the columns,
`{:(," Column I",," Column II"),((P),"The plates of a plane parallel capacitor are slowly pulled apart. Then the magnitude of electric field intensity inside the capacitor.",(1),"Increases if the capacitor is maintained at constant5 charge."),((Q),"The paltes of a plane parallel plate capacitor are slowly pulled apart. Then the potential energy stored in the capacitor.",(2),"Decreases if the capacitor is maintained at constant charge"),((R),"The capacitance of an air filled plane parallel plate capacitor on insertion of dielectric",(3),"Increases if the capacitor is maintained at constant potential difference."),((S),"A dielectric on inserted an air filled plane parallel plate capacitor. The potential energy stored in the capacitor.",(4),"Decreases if the capacitor is maintained at constant potential difference."):}`

Options

A. `{:((P),(Q),(R),(S),),(1,2,4,3,):}`
B. `{:((P),(Q),(R),(S),),(4,1,3,2,):}`
C. `{:((P),(Q),(R),(S),),(1,2,3,4,):}`
D. `{:((P),(Q),(R),(S),),(4,1,2,3,):}`

Explanation

(A)At constant charge, the electric field within the capacitor remains same when plate separation is changed. The electric field in capacitor is `E=V/d`.Hence at constant potential difference the electric field decreases with increase in d.
(B)`U=1/2Q^2/C`.Hence at constant charge U increases with decrease in C.
`U=1/2CV^2`. Hence at constant potential difference U decreases with decrease in C.
( C)Capacitance increases on insertion of dielectric.
(d)As a result of insertion of dielectric the capacitance increases.
`U=1/2 Q^2/C`.Hence at constant charge U decreases with increase in C.
`U=1/2 Q^2/C`.Hence at constant potential difference U increases with increase in C.