The separation between the plates of a parallel plate capacitor is made three times while it remains connected to a battery
(A) the battery absorbs some energy
(B) the electric field between plates remains same
(c) the charge on the capacitor decrease by `66.7%`
(D) some work has to be done by an external agent on the plates

Options

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

Answer

B

Explanation

`((Q_(2))/(Q_(1))-1)xx100=-66.7%%` i.e., charge decreases by `66.7%` (c)is O.K.
`Q_(2) lt Q_(1)` i.e., battery absorbs change `DeltaQ=Q_(1)-Q_(2)=(2CV)/(3)`
work done on the battery `W_(b)=DeltaQV=(2CV^(2))/(3)=` energy absorbed by battery `(A)` is O.K.
Electric field `E_(1)=(V)/(d),E_(2)=(V)/(3d), E_(2) lt E_(1)(B)` is wrong
`W_(1rarr2)=|U_(2)-U_(1)=|(1)/(2)(c)/(3)V^(2)-(1)/(2)CV^(2)|=(1)/(3)CV^(2),(D)` is O.K.