A slab of material of dielectric constant k has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, when d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.

Answer

`(2K)/(K + 1) C_(0)`

Explanation

Without dielectric, `E_(0) = (V_(0))/(d)`
`V = E_(0) (d)/(2) + E. (d)/(2) = E_(0) (d)/(2) + (E_(0))/(K) (d)/(2)`
`= (E_(0) d)/(2) ((K + 1))/(K) = (V_(0) (K +1))/(2K)`
`C = (q_(0))/(V_(0)) = (2 kq_(0))/(V_(0) (K + 1)) = (2K)/(K + 1) C_(0)`