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 capacitor is represented as
`C=(epsilon_(0)A)/(d)`
Initially, `C=8pF`
New capacitance on inserting the dielectric slab with dielectric constant K and reducing the plate separation can be calculated as:
`C.=(epsilon_(0)KA)/(d.) rArr C.=(epsilon_(0)KA)/((d)/(2))`
`rArr C.=(6xx2) (epsilon_(0)A)/(d)=6xx2xx8pF=96pF`