# Using the Method of Images to Find the Force on q

• quasar987
In summary, the conversation discusses the use of the method of images to find the force on a charge q. The potential is known to be zero at x=R and at infinity, and obeys a certain Poisson equation between those points. The speaker plans to put a charge q' at x<R to make the potential zero at x=R and at infinity. They also mention the uniqueness theorem and their intuition that they can place q' anywhere to achieve this result. However, they ultimately realize that their intuition may not be sufficient and discuss the need for the same boundary conditions on the entire volume bounded by the surface of the ball and infinity. The conversation ends with a discussion of the speaker's argument and the potential for using an image charge located at
quasar987
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The situation is illustrated in the attachment. I try to find the force on q by the method of images.

I know that the potential is 0 at x=R and at infinity, and I know that The potential obeys a certain Poisson equation between those points. I will try to put a charge q' somewhere at x<R that makes it so V=0 at x=R and at infinity. That way, according to the uniqueness theorem, the potential between x=R and infinity created by the 'sphere-q' system will be the same as the potential created by the 'q'-q' system.

My intuition tells me that I can pretty much put q' anywhere I want and I'll be able to create V=0 at x=R just by adjusting the magnitude of the charge. So I chose to put it at x=0. The potential of this charge configuration is

$$V(x) = \frac{1}{4\pi \epsilon_0}\left(\frac{q}{|x-a|}+\frac{q'}{x}\right)$$

and we have V(R) = 0 iff

$$q' = -\frac{R}{|R-a|}q = -\frac{R}{a-R}q$$

So the force on q is just that exerted by q' at this particular location, i.e.

$$F_{\rightarrow q}=-\frac{1}{4\pi\epsilon_0}\frac{Rq^2}{(a-R)(a)^2}$$

Yes? Or do I have the method of images all wrong?

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quasar987 said:
The situation is illustrated in the attachment. I try to find the force on q by the method of images.

I know that the potential is 0 at x=R and at infinity, and I know that The potential obeys a certain Poisson equation between those points. I will try to put a charge q' somewhere at x<R that makes it so V=0 at x=R and at infinity. That way, according to the uniqueness theorem, the potential between x=R and infinity created by the 'sphere-q' system will be the same as the potential created by the 'q'-q' system.

My intuition tells me that I can pretty much put q' anywhere I want and I'll be able to create V=0 at x=R just by adjusting the magnitude of the charge. So I chose to put it at x=0. The potential of this charge configuration is

$$V(x) = \frac{1}{4\pi \epsilon_0}\left(\frac{q}{|x-a|}-\frac{q'}{x}\right)$$

and we have V(R) = 0 iff

$$q' = \frac{R}{|R-a|}q = \frac{R}{a-R}q$$

So the force on q is just that exerted by q' at this particular location, i.e.

$$F_{\rightarrow q}=\frac{1}{4\pi\epsilon_0}\frac{Rq^2}{(a-R)(a)^2}$$

Yes? Or do I have the method of images all wrong?

Intuition is a great thing, but sometimes it just won't do The charge at x = 0 could be adjusted to make the potential zero at one point on the axis bewteen charges, but not for all r = R, which is what I suspect you need to show. Is that a conducting sphere? The method of images is called that for a good reason. Where might an image of q be formed?

First, I edited a sign error in the original post.

Yes, the ball is a grounded conductor.

So you're basically saying that it's not sufficient to get an image that creates the same boundary conditions along the x axis. I need one that creates the same boundary conditions on the whole volume bounded by the surface of the ball and infinity. This is logical. But I thought it would be sufficient to just have the same boundary condition along the x-axis in view of this argument:

Along the line, V(x,y,z) = V(x,0,0)=V(x). And V(x) obeys a Poisson equation along the x axis. So suppose there exist 2 funtions V1(x) and V2(x) that satisfy this Poisson equation and share the same value at the boundary x=R and x=infinity. Then V3 = V2 - V1 satisfies Laplace equation along the x-axis and V3(R)=V3(infinity)=0. Therefor, V3(x)=0 for all x (since V3 is harmonic). This means that V1 = V2.

In other words, the uniqueness theorem holds.

I suspect that the error is when I said V(x,y,z) = V(x,0,0)=V(x) but I don't see why.

quasar987 said:
First, I edited a sign error in the original post.

Yes, the ball is a grounded conductor.

So you're basically saying that it's not sufficient to get an image that creates the same boundary conditions along the x axis. I need one that creates the same boundary conditions on the whole volume bounded by the surface of the ball and infinity. This is logical. But I thought it would be sufficient to just have the same boundary condition along the x-axis in view of this argument:

Along the line, V(x,y,z) = V(x,0,0)=V(x). And V(x) obeys a Poisson equation along the x axis. So suppose there exist 2 funtions V1(x) and V2(x) that satisfy this Poisson equation and share the same value at the boundary x=R and x=infinity. Then V3 = V2 - V1 satisfies Laplace equation along the x-axis and V3(R)=V3(infinity)=0. Therefor, V3(x)=0 for all x (since V3 is harmonic). This means that V1 = V2.

In other words, the uniqueness theorem holds.

I suspect that the error is when I said V(x,y,z) = V(x,0,0)=V(x) but I don't see why.

I'm not totally following your argument, but I don't think it holds up even on the x axis. Suppose you find the magnitude of an image charge located at x = 0 that will result in zero potential at x = R. This should be pretty easy to do based purely on the 1/r dependence of the potential from a point charge. What will the potential be at x = -R? There is no way the given charge with an image point charge at x = 0 can give you zero potential at both x = R and x = -R. The image charge is going to have to be opposite the sign of the given charge, and located at some x such that 0<x<R. If you find an image charge that works for both x = R and x = -R, I think you have done it.

http://www.phys.ufl.edu/~dorsey/phy6346-00/lectures/lect04.pdf

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Sorry for not having been more explicit with my "argument". It's roughly just a copy/paste of the proof of the uniqueness theorem given by Griffiths.

If you find an image charge that works for both x = R and x = -R, I think you have done it.

I have the solution for the problem we're discussing (though the source of the solution is not entirely thrusthworthy) and this is what they do too. They put the charge where V=0 at both x=R and x=-R. But I don't see why! We just want the potential in the region x>R and the boundary of this region is x=R and x=infinity. Anyway, I'll read that pdf document. It looks great, thx!

quasar987 said:
I have the solution for the problem we're discussing (though the source of the solution is not entirely thrusthworthy) and this is what they do too. They put the charge where V=0 at both x=R and x=-R. But I don't see why! We just want the potential in the region x>R and the boundary of this region is x=R and x=infinity.

Still, it is a multidimensional problem. It would be a totally different problem and a different solution if V = 0 at x = R because there was a plane conductor passing through x = R, or if some other conducting surface passed through that point The boundary conditions of the problem extend off the x-axis, so you cannot solve the problem by looking only along that axis in the region of interest.

## 1. What is the method of images?

The method of images is a mathematical technique used to solve electrostatics problems involving point charges and grounded conductors. It involves creating imaginary charges (images) that satisfy the boundary conditions of the problem, allowing for the determination of the electric field and force on a given charge.

## 2. How does the method of images work?

The method of images works by using the principle of superposition, which states that the total electric field at a point is the sum of the individual electric fields from all charges present. By creating images that cancel out the electric field from the original charge at the boundary of a conductor, the net electric field can be determined and used to find the force on the original charge.

## 3. When is the method of images most useful?

The method of images is most useful when dealing with problems involving grounded conductors and point charges, as it can simplify the calculations and allow for a more intuitive understanding of the problem. It is also useful when dealing with problems involving infinite or semi-infinite geometries, as it allows for the consideration of boundary conditions that would be difficult to solve otherwise.

## 4. Are there any limitations to using the method of images?

Yes, there are limitations to using the method of images. It is only applicable to static electric problems, meaning that it cannot be used for problems involving changing electric fields or moving charges. Additionally, it can only be used for problems with simple geometries and boundary conditions, and may not be accurate for more complex systems.

## 5. How does the method of images relate to other techniques for solving electrostatics problems?

The method of images is one of several techniques used to solve electrostatics problems, including Gauss's law, Coulomb's law, and the method of images. It is often used in conjunction with these other methods, and can provide a more intuitive understanding of the problem and a simpler solution in certain cases.

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