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?

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 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_(a)` on the plates remains unchanged. The capacitance thus increases.
`C=Q_(0)/V=(4K)/(K+3) (Q_(0))/(V_(0))=(4K)/(K+3) C_(0)`