Discussion
A thin circular ring of radius R lies in the \(xy\) plane with its centre coinciding with the origin. The ring carries a uniform line charge density \(\lambda\). The quadrupole contribution to the electrostatic potential at the point \((0, 0, d)\), where \(d \gg R\), is
Answer: A
The quadrupole term is \(-\frac{\lambda R^{3}}{4\varepsilon_{0}d^{3}}\).
On the axis the exact potential of the ring (charge \(Q = 2\pi R\lambda\)) is expanded in powers of \(\frac{R}{d}\):
\[ \begin{aligned} V &= \frac{Q}{4\pi\varepsilon_{0}\sqrt{d^{2} + R^{2}}} \\ &= \frac{\lambda R}{2\varepsilon_{0}d}\left(1 - \frac{R^{2}}{2d^{2}} + \cdots\right) \end{aligned} \]
The first term is the monopole, there is no dipole term (the ring's dipole moment is zero by symmetry), and the next term, falling as \(\frac{1}{d^{3}}\), is the quadrupole contribution:
\[ \begin{aligned} V_{\text{quad}} &= -\frac{\lambda R}{2\varepsilon_{0}d}\cdot\frac{R^{2}}{2d^{2}} \\ &= -\frac{\lambda R^{3}}{4\varepsilon_{0}d^{3}} \end{aligned} \]
The positive version has the wrong sign, the \(\frac{1}{2\varepsilon_{0}}\) version drops the factor \(\frac{1}{2}\) of the binomial expansion, and 0 would be the dipole term, not the quadrupole.
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