Write the expression for electric field at a point on the axis of a short electric dipole.

Explanation

Figure
Expression for `E_(1)`
Expression foar `E_(2)` Substitution in `E=E_(2)+E_(1)`
Arriving at final expression
`E=(1)/(4piepsi_(0))*(2pr)/((r^(2)-a^(2))^(2))`
Detailed Answer:

Consider an electric dipole consisting of +q and -q charges.
2l is the length of the dipole consider a point P at distance r from the centre .O. of the dipole on the axial line of the dipole.
Let a unit positive charge be placed at point P.
Now electric field intensity at P due to +q charge is given by
`vecE_(+)=(1)/(4piepsi_(0))(q)/((r-l)^(2))hati`
Electric field intensity at .P. due -q charge is given by
`vecE_(-)=(-q)/(4piepsi_(0)(r+l)^(2))hati`
According to the principle of superposition
`vecE=(vecE_(+)+vecE_(-))`
`=(q)/(4piepsi_(0))[(1)/((r-l)^(2))-(1)/((r+l)^(2))]hati`
`=(q)/(4piepsi_(0))[(4rl)/((r^(2)-l^(2))^(2))]hati`
`=(1)/(4piepsi_(0))xx(2r(qxx2l))/((r^(2)-l^(2))^(2))hati` `[becausep=qxx2l]`.
`vecE=(2rp)/(4piepsi_(0)(r^(2)-l^(2))^(2))hati`.