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.

 

[paranormal]

I would suggest first looking at the equation for the potential at a distance r along the z-axis (equation 1) and considering the condition r >> R. This means that R is much smaller than r, so we can assume that the term \frac{1}{\sqrt{r^2 + R^2}} is negligible compared to \frac{1}{r}. Therefore, we can simplify equation (1) to:

V(r) = \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r} - 0\right) = \frac{q}{4\pi\epsilon_0}\frac{1}{r}

We can then substitute this simplified expression into equation (2) for the general solution to Laplace's equation. This will give us:

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

Since we are interested in the leading term, we can focus on the first term in the summation, where l = 0. This will give us:

V(r, \theta) \approx A_0 + \frac{B_0}{r}

We can then use this expression to find the leading term of the potential at points off the axis, by substituting in different values for \theta. For example, when \theta = \frac{\pi}{2}, we get:

V(r, \frac{\pi}{2}) \approx A_0 + \frac{B_0}{r}\cos(\frac{\pi}{2}) = A_0

Therefore, the leading term of the potential at points off the axis is simply A_0. This term will vary depending on the specific boundary conditions and geometry of the problem, but this method can be used to find the leading term in general. I hope this helps!