A parallel capacitor of capacitance `C` is charged and disconnected from the battery. The energy stored in it is `E`. If a dielectric slab of dielectric constant `6` is inserted between the plates of the capacitor then energy and capacitance will become.

Options

A. 6E,6C
B. E,C
C. `E//6,6C`
D. E,6C

Answer

C

Explanation

`U=(q^(2))/(2C),U_(f)=(q^(2))/(2KC)`.