Method of images and spherical coordinates

In summary, the conversation discusses finding the potential of a point charge above an infinite xy-plane. The solution involves placing an image charge at a specific position and using spherical coordinates. There is confusion about the last term in the distance formula and whether it should be written as 2r⋅R'cosθ or 2rR'cosθ. It is clarified that in spherical coordinates, the angle between the two vectors is actually π-θ, which changes the sign of the cosine term. The final formula is given as |r-R'| = √(r^2+R'^2+2rR'cosθ), which confirms that the potential is only zero on the xy-plane.
  • #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!
 
Last edited:
Physics news on Phys.org
  • #2
The difference in the two formula you show is 2r⋅R'cosθ or 2rR'cosθ. They have a same value.
 
  • #3
anuttarasammyak said:
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?
 
  • #4
I see you mean by r⋅R' inner product of vectors
[tex] \mathbf{r} \cdot \mathbf{R'} [/tex]
Then mathematics says that side length of triangle with other side lengths r,R' with angle ##\theta## between is
[tex]|\mathbf{r}-\mathbf{R'}|=\sqrt{r^2+R'^2-2rR'\cos\theta}[/tex]
On the x-y plane where ##\theta=\pm \pi/2##, ##\cos \theta=0## which gives zero potential there as you expect.
 
Last edited:
  • Like
Likes josephsanders
  • #5
anuttarasammyak said:
I see you mean by r⋅R' inner product of vectors
[tex] \mathbf{r} \cdot \mathbf{R'} [/tex]
Then mathematics says that side length of triangle with other side lengths r,R' with angle ##\theta## between is
[tex]|\mathbf{r}-\mathbf{R'}|=\sqrt{r^2+R'^2-2rR'\cos\theta}[/tex]
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 \ }##
 

Related to Method of images and spherical coordinates

1. What is the method of images?

The method of images is a mathematical technique used to solve electrostatic problems in which a point charge is placed near a conducting surface. It involves creating a mirror image charge on the opposite side of the surface, which results in a cancellation of the electric field lines and satisfies the boundary conditions.

2. How is the method of images applied in spherical coordinates?

In spherical coordinates, the method of images involves creating a mirror image charge at a specific distance and angle from the original charge. This distance and angle are determined by the location of the original charge and the geometry of the conducting surface. The resulting electric field is then calculated using the spherical coordinate system.

3. What are the advantages of using the method of images in spherical coordinates?

The method of images in spherical coordinates allows for a simpler and more intuitive solution to electrostatic problems involving spherical conductors. It also allows for the calculation of the electric field at any point in space, rather than just at the surface of the conductor.

4. Are there any limitations to the method of images in spherical coordinates?

One limitation of the method of images in spherical coordinates is that it can only be applied to problems involving spherical conductors. It also assumes that the conductors are perfect and have no imperfections or irregularities.

5. Can the method of images be used to solve problems involving multiple charges?

Yes, the method of images can be extended to solve problems involving multiple charges in spherical coordinates. In this case, multiple mirror image charges are created to satisfy the boundary conditions and the resulting electric field is calculated by summing the contributions from each charge.

Similar threads

Replies
8
Views
1K
Replies
4
Views
985
Replies
7
Views
2K
Replies
4
Views
691
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
404
Replies
14
Views
2K
  • Advanced Physics Homework Help
Replies
10
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
774
Replies
32
Views
2K
Back
Top