Show that intensity of electric field E at a point in broadside on position is given by :
`" "E=(1/(4piepsilon_(0)))p/((r^(2)+l^(2))^(3//2))`
where the terms have their usual meaning.

Explanation

Electric field due to dipole (broadside on position) consider dipole of length 2l and moment `vec(p)`. From the Figure Resultant electric field intensity at C:
`vec(E )= vec(E )_(A) + vec(E )_(B)`
`|vec(E )_(A)|= (1)/(4pi epsi_(0)) (q)/((r^(2)+l^(2)))`
`|vec(E )_(B)|= (1)/(4pi epsi_(0)) (q)/((r^(2) + l^(2)))`

Sine component of `vec(E )_(A) and vec(E )_(B)` get cancelled each other as `|vec(E )_(A)|= |vec(E )_(B)|`. The cos components get added up to give the resultant field. i.e., `E= E_(A) cos theta + E_(B) cos theta`
`=(1)/(4pi epsi_(0)).(q)/((r^(2) + l^(2))) cos theta + (1)/(4pi epsi_(0)).(q)/((r^(2) + l^(2)))cos theta`
`=2.(1)/(4pi epsi_(0)) .(q)/((r^(2)+ l^(2))) cos theta`
`=2 xx (1)/(4pi epsi_(0)) (q)/((r^(2) + l^(2))) xx (l)/((r^(2) + l^(2))^(1//2))`
`= (1)/(4pi epsi_(0)) (q(2l))/((r^(2) + l^(2))^(3//2))`
`E= (1)/(4pi epsi_(0)) (p)/((r^(2) + l^(2))^(3//2))` as p= q (2l) The above expression gives teh magnitude of teh field.
The direction of Electric Field E at C is opposite to the direction of the dipole moment `vec(p)`