A short electric dipole has a dipole moment of \(2\times10^{-9}\ \mathrm{C\,m}\). A point P is 0.3 m from the centre of the dipole. The line from the centre to P makes an angle of \(60^\circ\) with the dipole moment. The electric potential at P is (take \(1/4\pi\varepsilon_0=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(100\ \mathrm{V}\)
Far from a short dipole, the potential depends on both the distance \( r \) and the angle \( \theta \) between \( \mathbf{p} \) and the line to the point. Since \( \mathbf{p}\cdot\hat{\mathbf{r}}=p\cos\theta \), the potential is \( V=p\cos\theta/4\pi\varepsilon_0r^{2} \).
\( \begin{aligned} V &= \frac{1}{4\pi\varepsilon_0}\,\frac{p\cos\theta}{r^{2}} \\ &= \frac{9\times10^{9}\times2\times10^{-9}\times\cos60^\circ}{(0.3)^{2}} \\ &= \frac{9}{0.09} \\ &= 100\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Potential Due to an Electric Dipole – “The electric potential of a dipole is then given by”
\( \displaystyle V=\frac{1}{4\pi\varepsilon_0}\,\frac{\mathbf{p}\cdot\hat{\mathbf{r}}}{r^{2}} \)\( \quad (r\gg a) \qquad (2.15) \)