How Does Charge Distribution Affect Potential in Electromagnetic Fields?

[solipsis]

Homework Statement

A hoop of radius R and total charge -q is oriented with its normal vector along the z-axis.
A positive charge +q is placed at the centre of the hoop. The potential at a distance r along the z-axis is:

$$V(r) = (\frac{q}{4\pi\epsilon_0}(\frac{1}{r} - \frac{1}{\sqrt{r^2 + R^2}})$$

In the case of azimuthal symmetry, the general solution to Laplace's equation $$\nabla^2 V = 0$$ is:

$$V(r,\theta) = \sum^{\infty}_{l=0}(A_l r^l + \frac{B_l}{r^{l+1}})P_l cos(\theta)$$

Assuming r >> R, use equations (1) and (2) to obtain an expression for the leading term of the potential at points off the axis.

Homework Equations

The Attempt at a Solution

I know that on the axis, theta = 0, so

$$V(r,\theta) = \sum^{\infty}_{l=0}(A_l r^l + \frac{B_l}{r^{l+1}})P_l$$

I tried expanding equation (1), I think I have to get it into powers of R/r, but not entirely sure how to do that.

Any pointers/hints to get me started would me much appreciated :)

 

[diazona]

You've got the right idea. I would start by pulling out a factor of R from the denominator in the second term of the on-axis potential.

$$\sqrt{r^2 + R^2} = R(\cdots ?\cdots)$$

 

[gabbagabbahey]

You've got the right idea. I would start by pulling out a factor of R from the denominator in the second term of the on-axis potential.

$$\sqrt{r^2 + R^2} = R(\cdots ?\cdots)$$

You mean pull out a factor of $r$, right? (Since $\frac{R}{r}\ll 1$ )

$$\sqrt{r^2 + R^2} = r\sqrt{\cdots ?\cdots}$$

 

[diazona]

You mean pull out a factor of $r$, right? (Since $\frac{R}{r}\ll 1$ )

$$\sqrt{r^2 + R^2} = r\sqrt{\cdots ?\cdots}$$

Ah, yes, thanks for catching that.