Derive an expression for the capacitance of a parallel plate capacitor having two identical plates, each of area A and separated by a distance d, when the space between the plates is filled with a dielectric medium.

Consider, a capacitor consisting of two thin conducting plates 1 and 2, each of area 'A' held parallel to each other, at a distance 'd' apart. 
One of the plates is insulated and the other plate is earth connected. 
When a charge +Q is given to the insulated plate , then a charge of -Q is induced on the nearer face of the plate 2 and +Q is induced on the farther face of plate 2. As the plate 2 is earthed charge +Q being free, flows to the earth. 

Surface charge density of plate 1 , $\mathrm{\sigma } = \frac{\mathrm{Q}}{\mathrm{A}}$ 
Surface charge density of plate 2 = - $\mathrm{\sigma }$ 
Electric field intensity in between the plates, E = $\frac{\mathrm{\sigma }}{{\mathrm{\epsilon }}_{\mathrm{o}}} = \frac{Q}{{\epsilon }_o A}$

Taking this field localised between the plates as uniform throughout, 
Potential difference between the plates, V = $\mathrm{E}\times \mathrm{d} = \frac{\mathrm{Q}}{{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{A}}\mathrm{d}$ 

Capacity of a parallel plate capacitor is given by, C = $\frac{\mathrm{Q}}{\mathrm{V}}$= $\frac{\mathrm{Q}}{\mathrm{Qd}/{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{A}}$ = $\frac{{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{A}}{\mathrm{d}}$

Now, when the plates of the capacitor are seperated by a dielctric medium of relative permittivity ${\mathrm{\epsilon }}_{\mathrm{r}} = \mathrm{K}$ then, 

Capacity, C_m = $\frac{\mathrm{\epsilon A}}{\mathrm{d}}=\frac{{\epsilon }_r{\epsilon }_{o}A}{d}= {\epsilon }_r{C}_o = K C_o$ 

i.e.,                    C_m = K C_o 

Therefore, the capacity becomes K times the initial capacity when dielectric medium is inserted in between the plates.