GATE 2025 PH – Question 40
A thin circular ring of radius $R$ lies in the xy plane centered at the origin, carrying uniform line charge density $\lambda$. The quadrupole contribution to electrostatic potential at $(0,0,d)$, where $d\gg R$, is
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: (A) $-\lambda R^3/(4\epsilon_0d^3)$
Explanation
**Potential on the axis of a ring.** For a ring of radius $R$ with uniform line charge density $\lambda$, the total charge is $Q=2\pi R\lambda$ and every point of the ring is at the distance $\sqrt{d^2+R^2}$ from the point $(0,0,d)$:
$$V(d)=\frac{1}{4\pi\epsilon_0}\frac{Q}{\sqrt{d^2+R^2}}=\frac{\lambda R}{2\epsilon_0\sqrt{d^2+R^2}} .$$
**Expand for $d\gg R$:**
$$\frac1{\sqrt{d^2+R^2}}=\frac1d\left(1-\frac{R^2}{2d^2}+\cdots\right).$$
$$V=\frac{\lambda R}{2\epsilon_0d}-\frac{\lambda R^3}{4\epsilon_0d^3}+\cdots$$
- The first term ($\propto1/d$) is the monopole.
- There is no dipole term (the $1/d^2$ term vanishes by symmetry).
- The second term ($\propto1/d^3$) is the **quadrupole** contribution:
$$V_{quad}=-\frac{\lambda R^3}{4\epsilon_0d^3}\quad(\text{option A}).$$