- #1

josephsanders

- 8

- 3

- 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!

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!

Last edited: