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 ls inserted between the plates ?

Explanation

Let `E_(0) = (V_(0))/(d) be the electric field between the plates when there is vacuum (air) and here the potential difference is V 0.
If the dielectric is now inserted, the electric field in the dielectric will be `E = (E_(0))/(K)` where K is dielectric constant
The potential difference will then be
`V= (V_(0))/(4) +(3)/(4) .(V_(0))/(K)`
`= (E_(0)d)/(4) +(3)/(d) .(E_(0)d)/(K)`
`=E_(0)d[(1)/(4)+(3)/(4K)]`
`= V_(0)[(K+3)/(4K)]`
Hence, the potential difference decreases by the factor `(K+3)/(4K)` while the free charge `Q_(0)` on the plates remains unchanged. The capacitance thus increase,
`C = (Q_(0))/(V) = (4K)/(K+3) .(Q_(0))/(V_(0))`
`:. C = (4K)/(K+3).C_(0)[ because (Q_(0))/(V_(0))=C_(0)]`