A parallel plate capacitor is charged and then isolated. Now a dielectric slab is Introduced in it. Which of the following quantities will remain constant ?

Options

A. Electric charge Q
B. Potential difference V
C. Capacitance C
D. Energy U

Answer

A, C

Explanation

When dielectric slab is introduced in a parallel plate capacitor, its capacitance C = CK, where K = dielectric constant of a slab. But when C. is
increased accroding to `V = (Q)/(C) ` the potential V is decreased .
Hence energy `U = (1)/(2)C^(1)V^(2)` also changes so charge Q remains constant.