A parallel plate air capacitor of capacitance C is connected to a cell o f 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. (A) The energy stored in the capacitor decreases K times.
B. (B) The change in energy stored is
`(1)/(2) CV^(2) ((1)/(K) -1)`
C. (C) The charge on the capacitor is not conserved.
D. (D) The potential difference between the plates decreases K times.

Answer

A, C, D

Explanation

Once the capacitor is being charged, its charge remains constant. Charge Q = CV When dielectric slab is placed, its equivalent capacitance, C. = KC Initially, the energy stored in capacitor,
`U = (Q^(2))/(2C)`
Energy stored after placing dielectric slab
`U = (U)/(K) ` and ` V =(Q)/(C)`
and p.d. after placing dielectric slab
`V = (V)/(K)`