An electric dipole consists of two charges 2 μC and -2 μC separated by 2.4 cm. The magnitude of the potential on its axis at a distance of 0.6 m from the center of the dipole is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1200 V
This is the potential on the axis of an electric dipole. The distance 0.6 m is much larger than the half-separation 1.2 cm, so the short-dipole formula applies.
\( \displaystyle p = q \times 2a = 2\times10^{-6} \times 2.4\times10^{-2} \)
\( \displaystyle p = 4.8\times10^{-8}\,\mathrm{C\,m} \)
\( \displaystyle V = \frac{1}{4\pi\varepsilon_0}\,\frac{p}{r^{2}} \)
\( \displaystyle V = \frac{9\times10^{9} \times 4.8\times10^{-8}}{(0.6)^{2}} \)
\( \displaystyle V = \frac{432}{0.36} \)
\( \displaystyle V = 1200\,\mathrm{V} \)
The exact expression \( \frac{p}{4\pi\varepsilon_0 (r^{2} - a^{2})} \) gives 1200.5 V, the same to the accuracy of the options.