A parallel plate capacitor with air between the plates has a capacitance of 8 pF.`(1 pF = 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

Capacitance of parallel plate air capacitor `C=(epsi_(0)A)/(d)=8pF` . .. (i)
When separation between the plates becomes `(d)/(2)` and the space between the plates is filled with dielectric (K=6), then new capacitance
`C'=(Kepsi_(0)A)/(d//2)=(2Kepsi_(0)A)/(d)` . .. (ii)
`implies (C')/(C)=2K or C'=2KC=2xx6xx8pF=96pF`