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 difference will then be
`V = E_(0)(1/4d) + 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)/4K while the
free charge `Q_(0)` 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)`.