The electric field at a point on the axis of an electric dipole, at a distance \(r\) from its centre, is
\( E=\displaystyle\frac{1}{4\pi\varepsilon_0}\,\frac{4qar}{(r^{2}-a^{2})^{2}} \)
Here \(q\) is the magnitude of each charge and \(2a\) is the distance between the charges. For \(r\gg a\), this field in terms of the dipole moment \(p\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\frac{2p}{4\pi\varepsilon_0 r^{3}}\)
Far from the dipole, \( a^{2} \) is very small compared with \( r^{2} \). So \( (r^{2}-a^{2})^{2} \) can be taken as \( r^{4} \). The dipole moment of the pair has magnitude \( p=q\times2a \).
\( \begin{aligned} E &= \frac{1}{4\pi\varepsilon_0}\,\frac{4qar}{r^{4}} \\ &= \frac{1}{4\pi\varepsilon_0}\,\frac{2(2qa)}{r^{3}} \\ &= \frac{2p}{4\pi\varepsilon_0 r^{3}} \end{aligned} \)
Therefore, the field on the axis decreases as \( 1/r^{3} \). The field of a single charge decreases more slowly, as \( 1/r^{2} \).
NCERT (2026-27):
The field of an electric dipole – “The dipole moment vector p of an electric dipole is defined by … that is, it is a vector whose magnitude is charge q times the separation 2a (between the pair of charges q, –q) and the direction is along the line from –q to q.”
The field of an electric dipole – “… In terms of p, the electric field of a dipole at large distances takes simple forms: At a point on the dipole axis”
\( \displaystyle E=\frac{2p}{4\pi\varepsilon_0 r^{3}} \)\( \quad (r\gg a) \qquad (1.20) \)