Derive the expression for the electric field at a point on the equtorial line of the dipole
Answers
Consider an electric dipole consists of charges −q and +q seperated by distance 2a and placed in vacuum .Let P be a point on the equatorial line of the dipole at distance r from it . Electric field at point P due to +q charge is E→+q=14π∈0.qr2+a2, directed along BP Electric field at point P due to −q charges is E→−q=14π∈0.qr2+a2 directed along PA.
Thus the magnitude of E→−q and E→+q normal to the dipole axis will cancel out .The components parallel to the dipole axis add up.The total electric field e→equa is opposite to p→.
$\therefore \overrightarrow{E}_{equa} = - (E_{-q} \cos\theta + E_{+q} \cos\theta) \hat p$
$\qquad= -2 E_{-q} \cos \theta \hat p$
E→equa=−14π∈0.p(r2+a2)3/2p^
Where p=2qa , is the electric dipole moment.
If the point p is located far away from the dipole r>>a, then
E→equa=−14π∈0.pr3p^
clearly the direction of electric field at any point on the equatorial line of dipole will be antiparallel to the dipole moment p→