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

Answer

D

Explanation

Capacitance of the capacitor, `C = (Q)/(V)`
After inserting the dielectric, new capacitance
`C_(1) = K.C`
New potential difference `V_(1) = (V)/(K)`
Hence option (a) is the correct statement.
`U_(i) = (1)/(2)CV^(2) = (Q^(2))/(2C)` (`:' Q = CV)`
`U_(f) = (Q^(2))/(2C_(1)) = (Q^(2))/(2KC) = (C^(2) V^(2))/(2KC) = ((U_(i))/(K))`
Hence option (b) is also correct statement.
`DeltaU = U_(f) - U_(i) = (1)/(2) CV^(2) {(1)/(K) -1}` Hence option (c) is also correct statement.
After diconnecting the battery, the capacitor is isolated on inserting the dielectric slab the charge on the plates remains conserved.
Hence option (d) is not the correct statement.