Question

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:

• A. \(\mathbf{-\frac{\lambda R^3}{4 \epsilon_0 d^3}}\)
• B. \( 0 \)
• C. \( \frac{\lambda R^3}{4 \epsilon_0 d^3} \)
• D. \( -\frac{\lambda R^3}{2 \epsilon_0 d^3} \)

Hint:

For a ring of charge, the monopole and dipole contributions to the potential vanish, and the quadrupole term dominates for distant points.

Answer 1

The Correct Option is A

We are asked to find the quadrupole contribution to the electrostatic potential at a point along the \( z \)-axis at a distance \( d \) from the origin, where \( d \gg R \). 
1. The quadrupole moment of the ring:
The electrostatic potential due to a charge distribution can be written as a series expansion. For a point far away from the charge distribution (i.e., \( d \gg R \)), the potential is dominated by the monopole, dipole, and quadrupole moments. However, in this case, since the charge distribution is symmetric, the monopole (total charge) and dipole moments vanish, leaving the quadrupole moment as the leading term. The quadrupole moment \( Q_{ij} \) for a continuous charge distribution is given by: \[ Q_{ij} = \int (\delta_{i} \delta_{j} - \hat{r}_i \hat{r}_j) \lambda \, d\ell \] where \( \hat{r}_i \) and \( \hat{r}_j \) are the unit vectors in the \( x, y, z \) directions. 
2. The potential from the quadrupole moment:
The potential due to the quadrupole moment at a point far along the \( z \)-axis (at \( (0, 0, d) \)) is given by: \[ V_{{quad}} = \frac{Q_{zz}}{4 \pi \epsilon_0 d^3} \] where \( Q_{zz} \) is the quadrupole moment along the \( z \)-direction. 
3. Calculating \( Q_{zz} \) for the ring:
For a ring of radius \( R \) with uniform charge density \( \lambda \), the quadrupole moment in the \( z \)-direction (since the ring lies in the \( xy \)-plane) is: \[ Q_{zz} = \lambda R^2 \] 
4. Final expression for the potential:
Using the above expressions, we find the quadrupole contribution to the potential at the point \( (0, 0, d) \) to be: \[ V_{{quad}} = - \frac{\lambda R^3}{4 \epsilon_0 d^3} \] Thus, the quadrupole contribution to the electrostatic potential is \( -\frac{\lambda R^3}{4 \epsilon_0 d^3} \), which corresponds to option (A).