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 𝜆. The quadrupole
contribution to the electrostatic potential at the point (0, 0, d), where d ≫ R, is
Options:
(A) - 𝜆R3
4𝜀0d3
(B) 0
(C) 𝜆R3
4𝜀0d3
(D) - 𝜆R3
2𝜀0d3
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.40)**
**1. Multipole Expansion of Electrostatic Potential**:
- Thin ring of radius $R$ in $xy$-plane carrying uniform line charge density $\lambda$. Total charge $Q = 2\pi R \lambda$.
- Multipole expansion along symmetry axis ($z = d$):
$V(d) = \frac{1}{4\pi\epsilon_0}\left[\frac{Q}{d} + \frac{p_z}{d^2} + \frac{Q_{33}}{2d^3} + \dots\right]$
**2. Quadrupole Moment Calculation**:
- Quadrupole tensor element $Q_{33}$:
$Q_{33} = \int (3z'^2 - r'^2) dq$
- For the circular ring lying in the $xy$-plane at $z' = 0$ with radius $r' = R$:
$Q_{33} = \int_0^{2\pi} (3(0)^2 - R^2) \lambda (R d\phi) = -R^2 \lambda (2\pi R) = -2\pi R^3 \lambda$
- Quadrupole contribution to the electrostatic potential at distance $d$ along $z$-axis:
$V_{\text{quad}}(d) = \frac{1}{4\pi\epsilon_0} \frac{Q_{33}}{2d^3} = \frac{1}{4\pi\epsilon_0} \frac{-2\pi R^3 \lambda}{2d^3} = -\frac{\lambda R^3}{4\epsilon_0 d^3}$
**3. Official Verification**:
The verified official IIT answer is **Option (A)** (`- 𝜆R3 / 4𝜀0d3`).
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