Question:medium

Derive the expression for Electric Field Intensity on the axial and equatorial line of a dipole.

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Key dipole results to remember: - Axial field: \( E = \dfrac{1}{4\pi \varepsilon_0} \dfrac{2p}{r^3} \) - Equatorial field: \( E = \dfrac{1}{4\pi \varepsilon_0} \dfrac{p}{r^3} \) - Both decrease as \( 1/r^3 \), faster than a point charge field.
Updated On: Mar 5, 2026
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Solution and Explanation

Electric Dipole
An electric dipole consists of two equal and opposite charges \(+q\) and \(-q\) separated by a small distance \(2a\). The product of charge and separation is called the electric dipole moment and is given by
\[ p = q \times (2a) \] The direction of the dipole moment is from the negative charge to the positive charge.

Electric Field on the Axial Line of a Dipole
Consider a point \(P\) on the axial line at a distance \(r\) from the centre of the dipole. The electric fields due to charges \(+q\) and \(-q\) act along the same straight line.

Electric field at \(P\) due to charge \(+q\):
\[ E_1 = \frac{1}{4\pi\varepsilon_0}\frac{q}{(r-a)^2} \] Electric field at \(P\) due to charge \(-q\):
\[ E_2 = \frac{1}{4\pi\varepsilon_0}\frac{q}{(r+a)^2} \] The resultant electric field along the axial line is:
\[ E = E_1 - E_2 \] \[ E = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{(r-a)^2}-\frac{q}{(r+a)^2}\right] \] For points far away from the dipole where \(r \gg a\), the expression simplifies to:
\[ E_{axial} = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3} \] Thus, the electric field on the axial line of a dipole is directly proportional to the dipole moment and inversely proportional to the cube of the distance from the dipole.

Electric Field on the Equatorial Line of a Dipole
Consider a point \(P\) on the equatorial line at a distance \(r\) from the centre of the dipole. The electric fields produced by the two charges have equal magnitude but different directions. Their vertical components cancel each other while the horizontal components add together.

The electric field at point \(P\) due to each charge is:
\[ E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2 + a^2} \] After resolving the components and combining them, the resultant electric field along the equatorial line becomes:
\[ E_{equatorial} = \frac{1}{4\pi\varepsilon_0}\frac{p}{(r^2 + a^2)^{3/2}} \] For points far away from the dipole where \(r \gg a\), the expression simplifies to:
\[ E_{equatorial} = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3} \] The direction of the electric field on the equatorial line is opposite to the direction of the dipole moment.

Conclusion
For an electric dipole, the electric field on the axial line is
\[ E_{axial} = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3} \] and on the equatorial line it is
\[ E_{equatorial} = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3} \] Both expressions show that the electric field decreases rapidly with the cube of the distance from the dipole.
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