B Method of images and spherical coordinates

[josephsanders]

TL;DR Summary

I am having trouble with the magnitude of distance vector in spherical coordinates.

I am finding the potential everywhere in space due to a point charge a distance 'a' on the z-axis above an infinite xy-plane held at zero potential. This problem is fairly straight forward; place an image charge q' = -q at position -a on the z-axis. I have the solution in cartesian coordinates but I am having trouble writing it in spherical coordinates and this bothers me.

Suppose the position vector of the real charge is R = a\hat{z} and the image charge is R'= -a\hat{z}. So in spherical coordinates the magnitude of |r-R'| = (r^2 + R'^2 - 2rR'cosθ)^(1/2). My question lies in the last term. If "2rR'" refers to the magnitude of r times the magnitude of R' that would imply that the potential is zero everywhere since |R'| = |R|. So then I thought maybe it would make more sense to write |r-R'| = (r^2 + R'^2 - 2r⋅R'cosθ)^(1/2). Because then we would get that the potential is only zero on the xy-plane as desired. However when we are deriving the distance formula in spherical coordinates from the cartesian coordinates we have that (x,y,z) = (rsinθcosφ,rsinθsinφ,rcosθ) where the 'r' here does refer to the magnitude so I don't see how we can get the dot product that we would need to have the potential not vanish.

I hope I explained what I was confused on well enough! I know that this is the right answer for the method of images problem with a conducting sphere so I must be thinking about something wrong.EDIT: Actually I understand my error now. I wasn't thinking of θ as the angle between R' and r. So the angle in the two magnitude calculations is not actually the same angle. Thank you to those who helped!

 

[anuttarasammyak]

The difference in the two formula you show is 2r⋅R'cosθ or 2rR'cosθ. They have a same value.

 

[josephsanders]

The difference in the two formula you show is 2r⋅R'cosθ or 2rR'cosθ. They have a same value.

I don't think they are the same since if we use the dot product definition r⋅R' = rR'cosθ, then that would mean 2r⋅R'cosθ = 2rR'cos^2θ right?

 

[anuttarasammyak]

I see you mean by r⋅R' inner product of vectors
\mathbf{r} \cdot \mathbf{R'}
Then mathematics says that side length of triangle with other side lengths r,R' with angle $\theta$ between is
|\mathbf{r}-\mathbf{R'}|=\sqrt{r^2+R'^2-2rR'\cos\theta}
On the x-y plane where $\theta=\pm \pi/2$, $\cos \theta=0$ which gives zero potential there as you expect.

 

[SammyS]

I see you mean by r⋅R' inner product of vectors
\mathbf{r} \cdot \mathbf{R'}
Then mathematics says that side length of triangle with other side lengths r,R' with angle $\theta$ between is
|\mathbf{r}-\mathbf{R'}|=\sqrt{r^2+R'^2-2rR'\cos\theta}
On the x-y plane where $\theta=\pm \pi/2$, $\cos \theta=0$ which gives zero potential there as you expect.

@josephsanders ,

In terms of the angle, $\theta$, as you are using it as a spherical coordinate, the angle between $\mathbf{r}$ and $\mathbf{R'}$ is $\pi - \theta$. The angle $\theta$ is between the positive z-axis and $\mathbf{r}$, where as $\mathbf{R'}$ lies along the negative z-axis.

Since $\cos (\theta) = -\cos(\pi - \theta)$ the result you need is

$\displaystyle \quad \quad |\mathbf{r}-\mathbf{R'}|=\sqrt{r^2+R'^2+2rR'\cos\theta \ }$