A parallel plate capacitor of capacitance C is charged to a potential V and then disconnected with the battery. If separation between the plates is decreased by `50%` and the space between the plates is filled with a dielectric slab of dielectric constant `K=10`. then

Options

A. P.d. between the plates increases by `95%`
B. P.d. between the plates decreases by `50%`
C. P.d. between the plates increases by `50%`
D. P.d. between the plates decreases by `95%`

Answer

D

Explanation

New capacity becomes
`C^(')=K(epsilon_(0)A)/((d)/(2))=20(epsilon_(0)A)/(d)`
`C^(')=20C`
`becauseq=` constant
`CV=` constant
`Vprop(1)/(C)`
`becauseV^(')=(V)/(20)`
`%` decrease `=((V-(V)/(20))/(V))xx100=95%`