An electric dipole is made of charges \(+5\ \mathrm{nC}\) and \(-5\ \mathrm{nC}\) placed 2 mm apart. Points A and B are each 3 cm from the centre of the dipole. A lies on the axis of the dipole, on the side of the positive charge, and B lies on its equatorial plane. The potential difference \(V_A-V_B\) is (Take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(100\ \mathrm{V}\)
The dipole moment is the charge times the distance between the two charges, \( p=q\times2a \). Both points are far from the dipole compared with its size. On the axis \( V=p/4\pi\varepsilon_0r^{2} \), while on the equatorial plane the potential is zero.
\( \begin{aligned} p &= 5\times10^{-9}\times2\times10^{-3} \\ &= 1\times10^{-11}\ \mathrm{C\,m} \end{aligned} \)
\( \begin{aligned} V_A &= \frac{1}{4\pi\varepsilon_0}\,\frac{p}{r^{2}} \\ &= \frac{9\times10^{9}\times1\times10^{-11}}{(0.03)^{2}} \\ &= 100\ \mathrm{V} \end{aligned} \)
\( V_B=0 \)
\( \begin{aligned} V_A-V_B &= 100-0 \\ &= 100\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Potential Due to an Electric Dipole – “… potential on the dipole axis (θ = 0, π ) is given by”
\( V=\pm\frac{1}{4\pi\varepsilon_0}\,\frac{p}{r^{2}} \)\( \qquad (2.16) \)
“(Positive sign for θ = 0, negative sign for θ = π.) The potential in the equatorial plane (θ = π/2) is zero.”