A parallel plate capacitor with air between the plates has a capacitance of 8 pF `(1pF = 10^-12 F)`. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Explanation

`C = 8 PF = 8 xx 10^(-12) F`
Let A = Area of each plate
d = distance between the plates
`epsilon_(0) = 8.854 xx 10^(-12) C^(2) N^(-1) m^(-2)`
`C = (epsilon_(0)A)/d = 8 PF`……(1)
When distance between the plates is halved then
`d. = d//2`
and K=6
Now new capacitance of the capacitor
`C.. = K(epsilon_(0)A)/d^(.)`
`C..=12 ((epsilon_(0)A)/d) = 12 xx 8`
`C.. = 96 PF` [Using eqn. (1)]