A parallel plate capacitor of capacitance `10muF` is charged to potental `50V` and then isolated.
(A) After filling the space by a dielectric slab, the potential of capacitor becomes `25V`. The dielectric constant is `2`
(B) If plate separation is reduced to half, the potential of capacitor become `25V`
(c) The force between plates after indriduction of dilectric or after change in separation between plates remains same

Options

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

Answer

D

Explanation

Change remain same,
`Q=CV=KCV'impliesV'=(V)/(K)`
`25=(50)/(K)implies K=2`
(B) charge remain same, `Q=CV=C'V'impliesCv=2CV'`
`V'=(V)/(2)=(50)/(2)=25V`
(c) Electric field due to `(1)` at`(2)`, `E=(Q)/(2Ain_(0))`
Force between plates `F=QE`
Electric field at `(2)` due to induced charges `+Q_(in)` and `-Q_(in)` cancels each other hence `E.F`. at `(2)` remains same and hence force.
If separation is changed, `E` remains same.