Electric Dipole - Electric Field at Any Point

Description:

We know electric field due to a dipole in axial and equatorial position .We also know the dipole moment is a vector quantity which is directed from –q to +q and represented by P→ as shown in figure.

To calculate electric field created by a dipole at any general position (neither on the axial position nor on the equatorial position) −

A point P is taken at any general position which is at position ‘r→’ making an angle ‘θ’ with the dipole moment P→.

On resolving the dipole moment P→ into two components, (PCosθ) along r→ and (P Sinθ) normal to r→.

From figure, PCosθ has a moment along r→ and point P is also on the same line. So, we can say that PCosθ lies on the axial line.

Similarly, Psinθ has a moment normal to ‘r’ and point P is at the equatorial position with respect to the Psinθ component of dipole.

Electric field due to component of dipole PCosθ (or p_parallel (P₁₁)) at the point P can be calculated by using the formula for electric field due to a dipole at a point on the axial line and we get −

E due to P₁₁ 1/4πε₀ 2Pcosθ/r³ (axial line) ........(1)

Note

Direction of  Electric field (E₁) due to P₁₁ is along the direction of dipole moment.

Similarly, electric field due to component of dipole P Sinθ (or p_{perpendicular} (P⊥)) at the point P can be calculated by using the formula for electric field due to a dipole at a point on the equatorial line and we get −

E due to P₁₁ 1/4πε₀ PSinθ/r³ (equatorial line) ........(2)

Note

Direction of  Electric field(E₂) due to P₁ is opposite to the direction of dipole moment.

So, resultant electric field E due to P₁₁ and P⊥ will be −

E = (E₁² + E₂²)^1/2

E = √(1/4πε₀ 2Pcosθ/r³)² + (1/4πε₀ PSinθ/r³)²

E = 1/4πε₀ P/r³ √4 Cos²θ + Sin²θ

E = 1/4πε₀ P/r³√3Cos²θ + 1

Direction of Electric field at point P

For direction we fix up a reference line say ‘r→’. This resultant will make an angle α with the position vector r→

Finding angle α −

Join resultant electric field E_R and E₁₁.

Now, in Δ E_RE₁₁P

tanα = E_{perpendicular} / E_parallel

from eq (1) and (2)

tan α = sinθ / 2cosθ

This implies tanα = 1/2 tanθ

Note − Here θ is the angle of position vector with P.

The direction of resultant will be −

α_r = tan^-1 (1/2 tanθ)