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From OAC,
 =
=
i.e., OC = a
. Similarly OD = a
.
If r>>a, then PA = PC = PO + OC = r + a, and PB = PD = PO – ODÂ = r – a
.
If V is the potential at a point P due to electric dipole then,
V = k[]
= k[ –
]
= kq[]
=
Or, V = Â =
 [p = q2a = dipole moment and k =
]
If r>>a, then is neglected. So V =
.
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Potential energy due to electric dipole:Â
Let us consider an electric dipole consists of -q and +q at points A and B respectively separated by a distance a is placed in the electric field of intensity  at an angle θ. The dipole moment is
= qa
.
The magnitude of torque acting on the dipole is  = pE
.
If dθ is the angular displacement of the dipole in electric field then the work done stored as the potential energy of the dipole as dw = dθ = pE
dθ. Then the total work done stored into the dipole to rotate it from θ1 to θ2 is W =
dW = pE
 = – pE[cos θ2 – cos θ1] = pE[cos θ1 – cos θ2]
If θ1 = 900 and θ2 = θ then w = – pEcosθ
1. If θ = 00 then, W = – pEcos00 = -pE. This is stable equilibrium.
2. If θ = 900 then, w = – pEcos900 = 0
3. If θ = 1800 then, w = – pEcos1800 = pE. This is unstable equilibrium.
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Equipotential surface:
An equipotential surface is defined as the surface, on which all points lying are at the same potential.
An equipotential surface is the locus of all points in a medium at which electric potential due to a charge distribution is same.
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Electric field is perpendicular to the equipotential surface:
Let us consider A and B are very closed point on the equipotential surface (AB = dr). Let electric field creates angle θ with this surface. Ecosθ is the component of the field along the surface. The potential difference between points A and B is dV = 0. The work done to move a charge q from A to B is dw = q.dV = 0.
Similarly, dw = F.dr = qEcosθdr = 0. As q, E and dr 0 then cosθ = 0
Or, cosθ = cos900 Â
θ = 900 so, electric field is perpendicular to the equipotential surface.
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