The electric potential due to a short electric dipole is 40 V at a point on its axis, at distance \(r\) from its centre. The potential at a point at distance \(2r\) from the centre, on a line making an angle of \(60^\circ\) with the dipole moment, is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 5 V
Far from a dipole, the potential is \( V=p\cos\theta/4\pi\varepsilon_0r^{2} \). Here \( \theta \) is the angle between the dipole moment and the line to the point. On the axis, \( \theta=0 \) and \( \cos\theta=1 \). So the potential varies as \( \cos\theta/r^{2} \).
\( \begin{aligned} \frac{V^{\prime}}{V} &= \frac{\cos60^\circ}{\cos0^\circ}\times\frac{r^{2}}{(2r)^{2}} \\ &= \frac{1}{2}\times\frac{1}{4} \\ &= \frac{1}{8} \end{aligned} \)
\( \begin{aligned} V^{\prime} &= \frac{V}{8} \\ &= \frac{40}{8} \\ &= 5\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Potential Due to an Electric Dipole – “… and p = 2qa, we get”
\( \displaystyle V=\frac{p\cos\theta}{4\pi\varepsilon_0r^{2}} \)\( \qquad (2.14) \)
Potential Due to an Electric Dipole – “(ii) The electric dipole potential falls off, at large distance, as 1/r2, not as 1/r, characteristic of the potential due to a single charge.”