Question

Derive an expression for electrostatic potential due to an electric dipole.

Answer 1

Electrostatic potential at a point due to an electric dipole: 

Consider two equal and opposite charges separated by a small distance 2a. The point P is located at a distance r from the midpoint of the dipole. Let 0 be the angle between the line OP and dipole axis AB. 

Let r₁ be the distance of point P from +q and r₁ be the distance of point P from -q.

Potential at P due to charge +q = \(\frac {1}{4πε_0}\)\(\frac{q}{r_1}\)

Potential at P due to charge -q =  \(\frac {1}{4πε_0}\)\(\frac{q}{r_2}\)

Total Potential at the point P,

V =   \(\frac {1}{4πε_0}\) q \((\frac{1}{r_1} - \frac{1}{r_2})\) .......... (1)

Suppose if the point P is far away from the dipole, such that r >> a, then equation can be expressed in terms of r. By the cosine law for triangle BOP,

Since the point P is very far from dipole, then r >> a. As a result the term \(\frac{a^2}{r^2}\) is very small and can be neglected. Therefore

since \(\frac {a}{r}\) << 1, we can use binominal theorem and retain the terms up to first order

\(\frac{1}{r_1}\) = (1 + \(\frac{a}{r}\cos\theta\)) .............. (2)

Similarly applying the cosine law for triangle AOP,

Using Binomial theorem, we get

Substituting equations (3) and (2) in equation 

(1)

But the electric dipole moment p = 2qa and we get,

V = \(\frac{1}{4πε_0} (\frac{p cos\theta}{r^2})\)

Now we can write p cos θ = \(\vec P\),\(\hat r\) where \(\hat r\) is the unit vector from the point O to point P. Hence the electric potential at a point P due to an electric dipole is given by

Equation (4) is valid for distances very large compared to the size of the dipole. But for a point dipole, the equation (4) is valid for any distance. 

Special cases: 

Case (I): If the point P lies on the axial line of the dipole on the side of +q, then θ = 0. Then the electric potential becomes

V =\(\frac {1}{4πε_0}\) \(\frac{p}{r^2}\)

Case (II): 

If the point P lies on the axial line of the dipole on the side of -q, then θ = 180° , then

 V= -\(\frac {1}{4πε_0}\) \(\frac{p}{r^2}\)

Case (III): 

If the point P lies on the equatorial line of the dipole, then θ = 90°. Hence, V = 0.