A parallel plate capacitor with only air between the plates has a capacitance of `8pF`. 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

Given `C=8xx10^(-12)F,epsi_(r)=k=6,d'=d//2`
The new capacitance, `C'=(epsi_(0)epsi_(r)A)/(d')`
i.e., `C'=(epsi_(0)epsi_(r)A)/(d//2)=2epsi_(r)((epsi_(0)A)/(d))=2epsi_(r)C_(air)=2xx6xx8xx10^(-12)F`
i.e., `C'=96xx10^(-12)F=96pF`.