What is the electric potential due to a short electric dipole of moment p at a point on its axis at a distance r from its centre?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{p}{4\pi\varepsilon_0 r^{2}} \)
Take a short dipole, whose size is much smaller than the distance r. At a point whose position makes an angle \( \theta \) with the dipole moment \( \vec{p} \), the potential is
\( \displaystyle V=\frac{p\cos\theta}{4\pi\varepsilon_0r^{2}} \)
On the axis, on the side of the positive charge, \( \theta=0 \):
\( \begin{aligned} V &= \frac{p\cos0^\circ}{4\pi\varepsilon_0r^{2}} \\ &= \frac{p}{4\pi\varepsilon_0r^{2}} \end{aligned} \)
The dipole potential falls off as \( 1/r^{2} \). This is faster than the \( 1/r \) potential of a single charge. Far away, the two opposite charges nearly balance each other.
NCERT (2026-27):
Potential Due to an Electric Dipole – “… potential on the dipole axis (θ = 0, π) is given by
\( \displaystyle V=\pm\frac{1}{4\pi\varepsilon_0}\,\frac{p}{r^{2}} \)\( \qquad (2.16) \)
(Positive sign for θ = 0, negative sign for θ = π.)”