Explain in detail the effect of a dielectric placed in a parallel plate capacitor.

Explanation

A dielectric is an insulating material in which all the electrons are bound to the nucleus of the atom. In this material no free electrons to carry current.
(ii) When dielectric is introduced between the two plates of parallel plate capacitor, the capacitance of the capacitor increases.
Capacitance of a parallel plate capacitor with a dielectric medium:
(i) Consider a parallel plate capacitor having two conducting plates X and Y each of area A, separated by a distance d part. X is given a positive charge so that the surface charge density on it is `sigma` and Y is earthed.
(iii) Let a dielectric slab of thikness t and relative permittivity `epsi_(r)` be introduced between the plate figure.
(iii) Thickness of air gap =(d-t). Electric field at any point in the air between the plates, `E=(sigma)/(epsi_(0))`
(iv) Electric field at any point, in the dielectric slab `E.=(sigma)/(epsi_(0))`

(v) The total potential difference between the plates, is the work done in crossing unit positive charge from one plate to another in the field E over a distance (d-t) and in the field E. over a distance t, then
`V=E(d-t)+E.t`
`V=(sigma)/(epsi_(0))(d-t)+(sigmat)/(epsi_(0)epsi_(r))`
`=(sigma)/(epsi_(0))[(d-t)+(t)/(epsi_(r))]`
(v) The charge on the plate `X,q=sigmaA` Hence the capacitance of the capacitor is,
`C=(q)/(V)=(sigmaA)/((sigma)/(epsi_(0))[(d-t)+(t)/(epsi_(r))])=(epsi_(0)A)/((d-t)+(t)/(epsi_(r)))`
Effect of dielectric: In capacitors, the region between the two plates is filled with dielectric like mica or oil.
(i) The capacitance of the air filled capacitro, `C=(epsi_(0)A)/(d)`
The capacitance of the dielectric filled capacitor, `C.=(epsi_(r)epsi_(0)A)/(d):.(C.)/(C)=epsi_(r)`
or `C.=epsi_(r)C`
since, `epsi_(r)gt1` for any dielectric medium other than air, the capacitance increases, when dielectric is placed.