Obtain an expression for the electric field intenstiy at a point on the equatorial line of an electric dipole.

Explanation

Consider an electric dipole of dipole moment `P= qxx2a` placed in vaccum.
Where q= magnitude of either charge
2a - distance between the charges
Let p be the point on the equatorial line of the dipole at a distance r from the centre of the dipole O (fig )
Electric field at P due to the charge +q is
`vecE_(1)= ((1)/(4piepsilon_(0)))(q)/((r^(2)+a^(2)))hatP`
along Ap .... (1)
Electric field at p due to the charge - q is
`vecE_(2)((1)/(4piepsilon_(0)))(q)/((r^(2)+a^(2)))hatP`
along PB .... (2)
Electric fileds `vecE_(1)` and `VecE_(2)` can be resolve into two rectangular components (Fig0 THe components `vecE_(1)` and `sintheta` and `vecE_(2)sintheta` are equal in magnitude but acting in opposite direction , so they cancel each other . The component `vecE_(1)costheta` and `vecE_(2)costheta` are acting parallel to the dipole axis gets at P is
`vec = (E_(1)costheta + E_(2)costhat) hatp`
`vecE= - 2E_(1)costhetahatP`
`thereforeE_(1)=E_(2)`
From equation (1) we get
`vecE=-2[((1)/(4piepsilon_(0)))(q)/((r^(2)+a^(2)))]costhetahatP`
`=-2 [((1)/(4piepsilon_(0)))(q)/((r^(2)+a^(2)))](a)/((r^(2)+a^(2))^(1//2))hatp`
` = - ((1)/(4piepsi_(0))) (2aq)/((r^(2) + a^(2))^(3//2))hatp`
`vecE = - ((1)/(4piepsi_(0))) (p)/((r^(2) + a^(2))^(3//2))`
If the dipole is short (i.e., a `lt lt` r) n`a^(2)` can be neglected.
`vecE = - ((1)/(4piepsi_(0))) (p)/((r^(2))^(3//2)))`
`vecE = - ((1)/(4piepsi_(0)))(p)/(r^(3))`