A parallel plate air capacitor of capacitance `C` is connected to a cell of `emF V` and then disconnected from it. A dielectric slab of dielectric constant `K`, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect ?

Options

A. The energy stored in the capacitor decreases K times
B. the change in energy stored is `(1)/(2)CV^(2)((1)/(K)-1)`
C. The charge on the capacitor is not conserved
D. the potential difference between the plates decreases K times

Answer

C

Explanation

(i) The charge on the capacitor Q+CV remains constant,
when the battery is disconnected.
(ii) when the dielectric slab is inserted, the new capacity
C'=KC, it is increased
`V=(Q)/(C) and V'=(Q)/(C')=(Q)/(KC)=(V)/(K)`
Thus P.D. decreases K times,
(iii) Initial energy `(E)=(1)/(2)(Q^(2))/(C)`
and new energy `E'=(1)/(2)(Q^(2))/(C')=(1)/(2)(Q^(2))/(CK)=(E)/(K)`
thus E decreases K times,
(iv) charges in energy `=(1)/(2)(Q^(2))/(C)[(1)/(K)-1]`
`=(1)/(2)(C^(2)V^(2))/(C)[(1)/(K)-1]`
`=(1)/(2)CV^(2)[(1)/(K)-1]`
thus option (c) is wrong.