Shows two charges of magnitude q but of opposite sign, separated by a distance 2d.

The electric field due to the dipole at a point P along its axis at a distance r from the centre O of the dipole. Let the medium be air or vacuum.



The electric field at P due to the positive charge is given by

    E1 = ____1_____ _____q_____
        ______4π ε_o    ___ (r + d)²

E1 and E 2 act in opposite directions and E1 > E 2.

Therefore, the magnitude of the electric field at P is
E = E1 + E2
   = ____1_____ _____q_____ _ ____1_____ _____q_____
    ____4π ε_o ____    (r - d)²     ____    4π εo  ___       (r + d)²

= ____q_____ [_____1_____ _ ____1____]
    ___ 4π ε_o  ____   (r - d)²___         (r + d)²
= ____q_____  _____4rd_____
     __ 4π ε_o  ____   (r² – d²)²

If P is far from the dipole, then r >> d.

.:     E = ____q_____ _____4rd_____ = ____q_____ _____4rd_____
              ___4π ε_o_____        r⁴      _______4π ε_o         ______r³

But q x 2d = p = Electric dipole moment of the dipole.

. :    E = ____1 _____ __2p__
________4π ε_o  ___       r³

The direction of E is in the direction of the dipole moment vector p.

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