8 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 d is the separation of the plates How is the capacitance
changed when the slab is inserted between the plates Solution Let  E0 = V0/d be the electric field between the plates when
there is no dielectric and the potential difference is V0 If the dielectric
is now inserted, the electric field in the dielectric will be E = E0/K