Electric field 2

Dipole :- When two equal and opposite charge placed at very small distance then this formation is called dipole and distance between them is 2a.
here 2a = 1Å (One Angstrom ) = 10^-10 meter

Let two charges +q and -q are formed a dipole so

Dipole momentum :- It is the product of one charge of dipole and distance between them. It is denoted by 'p'.
p = 2*q = 2aq
It's unit is coulomb*meter (c*m).

Electric field due to a dipole :- There are two conditions to find electric field intensity due to a dipole -
1. Electric field intensity at axis of dipole :-
Let a dipole AB. A positive charge +q0 is present at point p at r distance from center of dipole O.

distance of p from +q is (r-a) and from -q (r+a) so
electric field intensity at p due to +q
E1 = kq / (r-a)^2          ....(1)
electric field intensity at p due to -q
E2 = kq / (r+a)^2         ....(2)
so total electric field intensity at p due to dipole
E = E1 - E2
E = [kq / (r-a)^2] - [kq / (r+a)^2]
E = kq [1 / (r-a)^2 - 1 / (r+a)^2]
E = kq [{(r+a)^2-(r-a)^2} / {(r-a)^2(r+a)^2}]
E = kq [{(r^2+a^2+2ar) - (r^2+a^2-2ar)} /  {(r-a)(r+a)}^2]
E = kq [(r^2+a^2+2ar - r^2-a^2+2ar) / (r^2 - a^2)^2}
E = kq [4ar / (r^2-a^2)^2}
since r >> a then r^2 >>>> a^2 so neglecting a^2
E = k [2*2aq*r / (r^2)^2]
since 2aq = p (dipole momentum )
E = 2kpr / r^4
E = 2kp / r^3                ....(3)
It is the electric field intensity at axis of dipole.
Here k = 1 / 4πε0  = 9*10^9 Nm^2/c^2

2. Electric field intensity at equator of dipole :-
 Let a dipole AB. A positive charge +q0 is present at r distance from mid point 'o' of dipole at point p.

now connect point A&B to P so AP & BP are hypotenuse so AP = BP = {(r^2+a^2)}^1/2so electric field intensity at p due to +q
E1 = kq / {(r^2+a^2)^1/2}^2
E1 = kq / (r^2+a^2)       ....(1)
electric field intensity at p due to -q
E2 = kq / (r^2+a^2)       ....(2)
here from equations (1) &  (2)
E1 = E2
so they cancel each other means E = 0 which is not possible.
so E1 & E2 are divide into components
E1 Sinθ & E2 Sinθ are equal and opposite so they cancel each other and E1 Cosθ & E2 Cosθ are in same direction so they add.
so resultant electric field intensity is given by
E = E1 Cosθ + E2 Cosθ
E = 2E1 Cosθ               ....(3)  
From triangle PBO
Cosθ = BO / BP
Cosθ = a / (r^2+a^2)^1/2  ....(4)
so again from (3)
E = 2[ {kq / (r^2+a^2)}{a / (r^2+a^2)^1/2}]
E = 2[kqa / (r^2+a^2)^3/2]
since 2aq = p (dipole momentum)
so E = kp / (r^2+a^2)^3/2
since r >>a so r^2 >>>> a^2 neglecting a^2
E = kp / (r^2)^3/2
E = kp / r^3        ....(5)
It is the electric field intensity at equator of dipole.
Here k = 1 / 4πε0  = 9*10^9 Nm^2/c^2