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

Answer

C

Explanation

Once the capacitor is charged, its charge will be constant, `Q = CV`. This is because the cell is disconnected.
when dielectric slab is inserted `C' = KC`
As `E = (Q^(2))/(2C) :. E' = (1)/(K) E`
As `V = (Q)/(C ) :. V' (V)/(K)`
S charge is not conserved is incorrect statment.