Question

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

Answer 1

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 θ be the angle between the line OP and dipole axis AB.

Potential due to electric dipole

Let r₁ be the distance of point P from +q and r₂ be the distance of point P from -q.

Potential at P due to charge +q = `1/(4 pi ε_0) "q"/"r"_1`

Potential at P due to charge –q = `−1/(4 pi ε_0) "q"/"r"_1`

Total potential at the point P

V = `1/(4 pi ε_0) "q"(1/"r"_1 - 1/"r"_2)`

By the cosine law for triangle BOP

r₁² =r² + a² – 2ra cos θ

`"r"_1^2 = "r"^2 (1 + "a"^2/"r"^2 - "2a"/"r" cos theta)`

`"a"^2/"r"^2` is very small, and can be neglected.

`"r"_1^2 = "r"^2(1 - "2a" (cos theta)/"r")` (or)

`"r"_1 = "r"(1 - "2a"/"r" cos theta)^2`

`1/"r"_1 = 1/"r" (1 - "2a"/"r" cos theta)^(-1/2)`

Since `"a"/"r"` << 1, we can use binomial theorem and retain the terms up to first order

`1/"r"_1 = 1/"r" (1 + "a"/"r" cos theta)`   ....(2)

Similarly applying the cosine law for triangle A0P,

r₂² = r² + a²  – 2racos(180-θ)

since cos(180-θ) = – cos θ

r₂² = r² + a² +2ra cos θ

Neglecting `"a"^2/"r"^2 "r"^_2 = "r"^2 (1 + (2"a" cos theta)/"r")`

`"r"_2 = "r" (1 + (2"a" cos theta)/"r")^(1/2)`

Using Binomial theorem, we get

`1/"r"_2 = 1/"r" (1 - "a" (cos theta)/"r")`    ....(3)

Sun (2) and (3) in eqn (1), we get

V = `"q"/(4piε_0) (1/"r"(1 + "a"(cos theta)/"r") - 1/"r" (1 - "a" (cos theta)/"r"))`

V = `"q"/(4piε_0) (1/"r"(1 + "a" (cos theta)/"r" - 1 + "a" (cos theta)/"r"))`

V = `1/(4piε_0) (2"aq")/"r"^2 cos theta`

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

V = `1/(4piε_0) (("p" cos theta)/"r"^2)`

p cos θ = `vec"p" * hat"r"`, where `hat "r"` is the unit vector from the point O to point P.

V = `1/(4piε_0) ((vec"p" * hat"r")/"r"^2)`

Special Cases:

If the point lies near	θ	V
+q	O°	`"p"/(4 piε_0"r"^2)`
-q	180°	-`"p"/(4 piε_0"r"^2)`
equatorial point	90°	0