Distance of point P from `+q` and `-q` are same.
`r_(+) = r_(-) = sqrt(r^(2) + a^(2))`
where, r = distance of P from 0.
The magnitudes of the electric fields due to the two charges `+q` and `-q` are equal.
`vec(E_(+q)) = q/(4pi epsilon_(0)).1/(r^(2) + a^(2)) = (kq)/(r^(2) + a^(2))`..........(1)
`vecE_(-q) =q/(4pi epsilon_(0)).1/(r^(2) + a^(2)) = (kq)/(r^(2) + a^(2))`............(2)
The distance of `vecE_(+q)` and `vecE_(-q)` are as shown in figure. The sine components will cancel each other.
The total electric field at P.
`therefore vecE =-(E_(+q) + E_(-q)) cos thetahatp = -(2kq)/(r^(2) + a^(2)). cos theta hatp`.............(3)
(`therefore` from equation (1) and (2))
Perpendicular components to dipole axis of `vecE_(+q)` and `vecE_(-q)` at P are `E_(+q) sin theta` and `E_(-q) sin theta`. They are equal in magnitude and opposite ir direction, so they will cancel the effects of each other.
But parallel components to dipole axis of `vecE_(+q)` and `vecE_(-q)` at P are `E_(+q) cos theta` and `E_(-q cos theta)`. And they are in same direction, so they will be added and resultant `vecE` will be in opposite to `hatp`
`therefore vecE = -(E_(+q) cos theta + E_(-q).cos theta).hatp`
`=-2E_(+q) cos theta hatp` `(therefore E_(+q) = E_(-q))`
`vecE = -(2kq)/(r^(2) + a^(2)) xx cos theta.hatp`...........(4)
(From equation (3))
Now, from figure,
`cos theta = a/(r^(2) + a^(2))^(2)`.............(5)
`cos theta = a/(r^(2) + a^(2))^(1/2)`
`therefore` From equation (4) and (5),
`vecE = (-2kq)/(r^(2) + a^(2)) xx a/(r^(2) + a^(2))^(2) hatp`.
`therefore vecE =-(2kqa)/(r^(2) + a^(2))^(2).hatp`
If `r gt gt a`, then `a^(2)` can be neglected.
`therefore vecE = -(2kqa)/r^(3).hatp`
But `2qa =p`
`vecE = -(kq)/r^(3)hatp` or `vecE =-p/(4piepsilon_(0)r^(3)).hatp`
