Potential and Field due to an Electric Dipole

A dipole consisting of two charges + q and – q separated by a distance 2d.
Dipole moment of the electric dipole = p = q x 2d.
Direction of p is form – q to + q.

P is a point at a distance r from the centre O.
∟POB = θ. Let us calculate the potential at P.
AC and BD are drawn perpendicular to OP.

Potential at P due} = ___1___ (q/PB)
to the charge + q            4πε₀

Potential at P due} = ___1___ (q/PA)
to the charge - q            4πε₀

Let us assume r >> d. Then
PA = PC = PO + OC = r + d cos θ
PB = PD = PO + OD = r – d cos θ

The resultant potential at P is
    V = __1__ [_____q______ - _____q_____]
            4πε₀      (r – d cos2θ)       ( r + d cos θ)

    = __1__ [_2d cos θ _____]
        4πε₀     (r² – d² cos²θ)

d² cos²θ << r² and so it can be neglected. Also, q X 2d = P.
. :    V = __1__ p cos θ)
_______4πε₀       r²
or     V = __1__ p.r
_______4πε₀    r²

Here, r is the unit vector along r. θ is the angle between p and r (i.e., OP).

Special cases (i) When the point P lies on the axial line of the dipole on the side of the positive charge, θ = 0 and cos θ = 1.
    V = ___1___ p
______4πε₀     r²

(ii) When the point p lies on the axial line of the dipole on the side of the negative charge, θ = 180 and cos θ = - 1.
    V = ___1___ (-p)
______4πε₀       r²

(iii) When the point P lies on the equatorial line of the dipole, θ = 90 and cos θ = 0. Therefore V = 0.

Electric Field. The field at P is calculated using the relation E = - Δ V.
The components of the field along the radius vector r is
    Er = - dV = - d (___1___ p cos θ) = ___1___ 2p cos θ
               dr     dr    4πε₀       r²    _______4πε₀         r³

Along a direction perpendicular to r, field component
E_θ = - 1 dV = - 1 [d (___1___ p cos θ) ] = ___1___ p sin θ
           r d θ        r  dθ    4πε₀       r²  _______ 4πε₀    r³

The magnitude of the resultant field at P is
E = √(E²_r + E²_θ) = ___1___ p (√(3cos² θ + 1)
          ___________ 4πε0    r3

The angle Ф which the resultant field E makes with r is
    Ф = tan^-1 (E _θ / E_r) = tan^-1 (1/2 tan θ)

Special cases. (i) If P is on the axial, line, θ = O.
E is in the direction of p and has magnitude 2p / (4πε₀r³)

(ii) If P is on the equatorial line, θ = 90.
E is in the direction of – p and has magnitude p / (4πε₀r³).



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