What is the effect of presence of a dielectric medium on
(i) capacitance of a parallel plate capacitor
(ii) electrostatic force between two charges ?

Explanation

Consider a parallel-plate capacitor without a dielectric, of plate area A, plate separation d and capacitance `C_(0),` charged to a potential difference V and then isolated.
Suppose the free charges on its conducting plates are +Q and -Q, Fig. (a). The surface density of free charge is
`sigma =(Q)/(A) " "` ...(1)
If A is very large and d is very small, the electric field in the region between the plates is almost uniform, except at the edges. The magnitude of the electric field intensity is
`E_(0)=(V)/(d)=(sigma)/(epsilon_(0))=(Q)/(epsilon_(0)A) " " ` (2)
Now, suppose a dielectric completely fills the space the charged plates. A polarisation charge - `Q_(p)` appears on the exterior surface of the dielectric nearer to the positive plate while a polarisation charge `+Q_(p)` appears on its opposite face. Since the capacitor was isolated after charging, the free charge Q on the plates is the same as earlier.

Consider the Gaussian surface S as shown in Fig. (b). It encloses free charge +Q on the left plate and the bound polarisation charge - `Q_(p)` on the surface of the dielectric.
The net charge enclosed by `S=Q-Q_(p).`
`therefore` By Gauss's theorem, the TNEI over the Gaussian surface is
`epsilon_(0) underset(s) int vecE* d vecS=epsilon_(0)EA=Q-Q_(p)`
`therefore` The magnitude of the electric field intensity in the dielectric is
`E=(Q-Q_(p))/(epsilon_(0)A)=(Q)/(epsilon_(0)A)-(Q_(p))/(epsilon_(0)A)=(sigma)/(epsilon_(0))-(sigma_(p))/(epsilon_(0)) " " ` ...(3)
Writing `E=(E_(0))/(k),` where k is the relative permittivity (dielectric constant) of the medium, `(E_(0))/(k)=(sigma)/(epsilon_(0))-(sigma_(p))/(epsilon_(0))`
` therefore (sigma)/(k epsilon_(0))=(sigma)/(epsilon_(0))-(sigma_(p))/(epsilon_(0))`
`therefore (sigma)/(k)=sigma-sigma_(p)`
`therefore` The surface density of induced charge is
`sigma_(p)=sigma-(sigma)/(k)=sigma(1-(1)/(k)) " " ` ...(4)
Without the dielectric, the capacitance of the capacitor is, `C_(0)=Q//V.`
`therefore Q=C_(0)V=C_(0)E_(0)d " " ` ...(5)
Let the capacitance with the dielectric be C. Since the free charge Q remains the same, `Q=CEd " " `...(6)
Equating the right hand sides of Eqs. (5) and (6),
`CEd = C_(0)E_(0)d`
`therefore C=C_(0)((E_(0))/(E))=kC_(0) " " ` ...(7)
Thus, the capacitance increases by the factor of `k=E_(0)//E.`