A parallel plate capacitor with air between the plates has a capacitance of 8 pF. The separation between the plates is now reduced by half and the space between them is filled with a medium of dielectric constant K. Calculate the value of the capacitance of the capacitor in the second case.

Explanation

`C = 8PF = 8 xx 10^(-12) F`
Let A = Area of each plate of capacitor. d = distance between the plates of the capacitor
`C = (epsilon_(0)A)/d = 8PF`
Now, `d. =d/2`
`K=5`
Then new capacitance of the capacitor is
`C.. = K (epsilon_(0)A)/(d.) = K(epsilon_(0)A)/(d//2)`
`C.. = 2K (epsilon_(0)A)/d`
`C.. = 2KC =2 xx 5 xx 8`
`C.. = 80 PF`