A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness (3/4)d, where is the separa - tion of the plates. How is the capacitance when the slab is inserted between the plates ?

Explanation

Let `E_(0)=V_(0)//d` be the electric field between the plates when there is no dielectric and the potential difference is `V_(0)`. If the dielectric is now inserted, the electric field in the dielectric will be `E=E_(0)//K`, The potential difference will then be
`V=E_(0)((1)/(4)d)+(E_0)/(K)((3)/(4)d)`
`=E_(0)d((1)/(4)+(3)/(4K))=V_(0)(K+3)/(4K)`
THe potential difference decreases by the factor `(K+3)//K` while the free charge `Q_0` on the plates remains uncharged . The capacitance thus increases ,
`C=(Q_0)/(V)=(4K)/(K+3)(Q_0)/(V_0)=(4K)/(K+3)C_0`