A slab of material of dielectric constant K has the same area as the plates of a parallel 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)/(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 unchanged. The capacitance thus increases
`C = (Q_(0))/(V) = (4K)/(K+3)(Q_(0))/(V_(0)) = (4K)/(K+3)C_(0)`