Dipole Electric Field: Definition, Electric field at equatorial Field

The electric field of a dipole is an important concept in electrostatics, forming the basis for understanding electric interactions in both natural and engineered systems. A dipole consists of two equal but opposite charges separated by a small distance. This configuration creates a characteristic electric field pattern that varies in strength and direction depending on the position relative to the dipole.Electric dipoles are found in many real-world applications — from polar molecules in chemistry and biology to components in antennas and sensors in electrical engineering. Studying the dipole electric field helps explain how forces act on charged particles, how molecules align in electric fields, and how electric energy is distributed in space.

1.0Definition of Electric Dipole

• A system of two equal and opposite charges placed at very small separation is known as electric dipole.
• Every dipole has a characteristic property called dipole moment.

2.0Electric Dipole Moment

• The dipole moment of a dipole is equal to the product of magnitude of either charge and separation between the charges.
  $\vec{p}=q\vec{l}$
• It is a vector quantity whose direction is from (-q) to (+q)
• SI Unit $\to C-m$
• Practical Unit $\to Debye\left(3.33\times 10^{-30}C-m\right)$

Illustration-1.Find electric dipole moment of the given arrangement.

1)

2)

Solution:

(1)

Angle between both the dipole moment vectors is 60°. 

$P_{res}=\sqrt{3}p=\sqrt{3}(ql)$

(2) 

Angle between both the dipole moment vectors is 90°.

$P_{net}=\sqrt{2}p=\sqrt{2}(ql)$

3.0Electric Field Due to a Dipole

• The dipole produces an electric field, which is the vector sum of the fields due to +q and –q. The net field at any point depends on the location relative to the dipole

(1) At axial / End on position:The axial line is the line that passes through both charges (along the dipole axis).

$$\begin{gathered} Here{E}_A=\frac{kq}{(r+a{)}^2}\text{and}{E}_B=\frac{kq}{(r-a{)}^2}\text{Net electric field at point P on axis is} \\ {E}_{\text{axis}}=E_B-E_A=\frac{kq}{(r-a{)}^2}-\frac{kq}{(r+a{)}^2}=kq\left(\frac{4ar}{(r^2-a^2{)}^2}\right) \\ Forr\gg a:r^2-a^2\approx r^2 \\ {E}_{\text{axis}}=kq\left(\frac{4ar}{r^4}\right)=\left(\frac{2k(q2a)}{r^3}\right)=\left(\frac{2kp}{r^3}\right) \\ \text{For axial point electric field vectors and dipole moment vectors both are parallel to each other, or angle between both vectors is zero degree.} \\ \text{In vector form }{\vec{E}}_{\text{axis}}=\left(\frac{2kp}{r^3}\right)\hat{r}=\frac{2k\vec{p}}{r^3} \end{gathered}$$

(2) At equator / Broad side on Position :The equatorial or broadside-on position is the location in the plane perpendicular to the dipole axis and equidistant from both charges, where the electric field is evaluated due to the dipole.

 $$\begin{gathered} E_A=E_B(\text{in magnitude}) \\ {E}_A=E_B=\frac{kq}{{\left(\sqrt{a^2+r^2}\right)}^2}=\frac{kq}{a^2+r^2} \\ \text{Net electric field at equator} \\ {E}_P=E_B\cos \theta +E_A\cos \theta \\ \cos \theta =\frac{a}{\sqrt{a^2+r^2}} \\ {E}_P=2E_B\cos \theta =2E_A\cos \theta \\ =2\left(\frac{kq}{a^2+r^2}\right)\frac{a}{\sqrt{a^2+r^2}} \\ {E}_P=\left(\frac{kq(2a)}{a^2+r^2}\right)\frac{1}{\sqrt{a^2+r^2}}=\frac{kp}{(a^2+r^2{)}^{3/2}} \\ Forr\gg a,r^2+a^2\approx r^2 \\ {E}_P=\frac{kp}{r^3} \\ \text{In vector form,} \\ {\vec{E}}_P=\frac{kp}{r^3}(-\hat{i}) \\ \text{Negative sign is used because electric field vector and dipole moment vector both are in opposite directions.} \end{gathered}$$

3. General Position :The general position is any point in the space around the dipole where the electric field is evaluated at a distance r from the center of the dipole, making an angle θ with the dipole axis.    

$$\begin{gathered} \text{Net electric field at that point will be} \\ E=\sqrt{E_1^2+E_2^2} \\ E=\frac{kp}{r^3}\sqrt{(2\cos \theta {)}^2+(\sin \theta {)}^2}=\frac{kp}{r^3}\sqrt{3{\cos }^2\theta +1} \\ \text{Net electric field is making angle}(\alpha +\theta )\text{from the direction of dipole moment vector.} \\ \tan \alpha =\frac{E_2}{E_1}=\frac{\frac{kp\sin \theta }{r^3}}{\frac{2kp\cos \theta }{r^3}}=\frac{1}{2}\tan \theta \Rightarrow \tan \alpha =\frac{1}{2}\tan \theta \end{gathered}$$