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It is well known that the method of image charges describes how if you have a charge near a conducting plate one can obtain a correct expression for the potential and E field by assuming that (a) the charge is reflected in all nearby conducting surfaces with opposite sign, (b) any image charges in metal plates are themselves reflected, and (c) the E-field always is normal to the conductor.

This works because of the uniqueness and existence theorems of Laplace's equation with the boundary condition of the metal conductor -- and the motion of image charges explains lots of wonderful things, e.g. [Graham]-Smith Purcell radiation can be explained by image charges alone.

My question is this: imagine I have three semi-infinite conducting plates meeting at a vertex 120º apart, with a charge $+Q$ placed on one of the other axes of symmetry.

The image charges I'll get will be placed in the points of an equilateral triangle, and according to the "rules" above they'll be apparently inconsistent: each reflected charge should be both $-Q$ and $+Q$ simultaneously, as can be seen by either going clockwise or anticlockwise around from the solid "real" charge below:

Three planes and an image charge

What does the picture of image charges look like, and why? Doesn't it look like the distribution of images is inconsistent with itself? How you reconcile the "classical" image charge explanation that is often given in first-year textbooks with this one. (Likewise, what happens if the angle is anything that would give you an odd number of charges, such as a pentagon rather than a triangle?)

To know what the "right answer" is, I performed some EM simulations by numerically solving Laplace with a charge of +10 C on a 5 mm sphere with d=30 mm, surrounded by either (a) three infinite equipotentials:

enter image description here

This qualitatively is consistent with having two $-Q$ spheres below:

enter image description here

In contrast, the $±Q$ situation seems decidedly wrong (as you'd expect):

enter image description here

As is the $+2Q$ situation:

enter image description here

Why? How do I explain this "simply"?

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  • $\begingroup$ Why do you think that it is inconsistent to have two negative charges? $\endgroup$ – FGSUZ Jan 30 '19 at 23:59
  • $\begingroup$ @FGSUZ -- Because the image in the third metal plane of either of them produces a positive charge on top of the original negative image charge? $\endgroup$ – Landak Jan 31 '19 at 0:26
  • $\begingroup$ Not sure if I understand, but you do not have to cancel all charges. You only have to compensate the "real region" of the problem. The charges you place beyond the planes are virtual charges, imaginary ones, and they do not have to be compensated. $\endgroup$ – FGSUZ Jan 31 '19 at 0:31
  • $\begingroup$ Even the first situation does not seem to me in agreement with the 3 equipotential plans. The field created by two charges + Q and -Q will be orthogonal to the plane of antisymmetry of the charges. That's what we expect. But the field created by the third will be contained in this plane. As a result, the total field cannot be normal to the equipotentials. Qualitatively, we may have this impression because the third is farther than the other two and creates a weaker field ? $\endgroup$ – Vincent Fraticelli Jan 31 '19 at 19:54
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There's nothing "inconsistent", it's just that the method of image charges doesn't work in this situation, in the sense that there is no simple arrangement of image point charges that will enforce the desired boundary conditions. In fact, the method of images almost never works, it only works for a small number of highly symmetric boundary conditions, so this isn't really surprising.

Furthermore, contrary to what the other answer says, you can't fix the situation by moving the charge. If you do this, the method of images will fail in an even worse way: reflecting in the planes will force an image charge to appear in the physical region (i.e. the sector where the real charge is), while the whole point of image charges is that you can satisfy boundary conditions by putting charges in unphysical regions alone.

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  • $\begingroup$ The method of images can work in non symmetrical cases by applying the principle of superposition several times in a row and by decomposing the highly non symmetrical original problem into a series/superposition of symmetrical problems. A very simple example is the case of three coplanar charges placed in (0,0), (a,0), and (0,b). There's no azimuthal symmetry, no rotational one either, no translational one, etc. Yet the method works. $\endgroup$ – AccidentalBismuthTransform Apr 28 '19 at 17:39
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    $\begingroup$ @thermomagneticcondensedboson The method of images works if the boundary conditions are highly symmetrical, such as if you have a conducting sphere or a conducting infinite plane. It doesn't have much to do with what the physical charges are. In this case, the boundary conditions just aren't nice enough. $\endgroup$ – knzhou Apr 28 '19 at 17:49
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    $\begingroup$ I see! I hadn't thought about that. It would be nice if you included this info in your answer. Because as it is, if you take the example I described (or replace the centered charge by a conducting sphere), there is no symmetry and yet the method works. But if we consider the surface of the conducting sphere a boundary condition then yeah it is highly symmetric although not the potential/E field. $\endgroup$ – AccidentalBismuthTransform Apr 28 '19 at 18:42
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Original image plus labels

First of all, the real charge is screened by the conductive plates. Its presence affects only the top-right space, i.e. the space between the plates "e" and "d". Thus, the first simplification is to remove the plate "f". Only two plates, "e" and "d".

However, the problem is not yet solved. If the real charge A is +Q, the image charges B and C would be -Q. Consider the charge B. It correctly adds its electric field to the electric field generated by the real charge A, so that the sum of the two electric fields A and B is perpendicular to the plate "e". In the same way, C adds its electric field to the electric field of A so that the sum of A and C is perpendicular to "d". But on "d", we also have the electric field generated by B, which makes the total not perpendicular to "d" (and the same holds on "e" for the detrimental effect of C).

What we forgot is that we must also take the reflection of B by "d". How is it possible? We must imagine that "d" is prolonged to the left. But we are unlucky: prolonging "d" passes exactly through B, and prolonging "e" passes exactly through C.

So the "paradox" is simply that we want to make the reflection of a charge which stays on the reflective surface. This is not difficult to solve: just, displace the charge by $\epsilon$, and send $\epsilon\to 0$.

Summarizing: first, displace A from its unlucky position, just a bit, $\epsilon$. Make the (infinite number) of reflections, of the real charge and of all the image charges. Make the infinite sum. Send $\epsilon\to 0$. The result could be correct if i) the procedure converges; ii) all the image charges are beyond the plates with respect to the real charge A. But in general, you cannot say.

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  • $\begingroup$ Much of this argument relies on having the initial test charge lie symmetrically at the angle bisector. That type of restriction is generally not desirable in image-charge methods. $\endgroup$ – Emilio Pisanty Apr 28 '19 at 17:33
  • $\begingroup$ @EmilioPisanty : the comment is not clear. Are you claiming that my answer requires that the charge is in a given position? Actually, it is the opposite, i.e. I'm saying that it is better to start with the charge displaced from that special position. On the other hand, I'm not claiming that the special position cannot be treated; I'm just suggesting to start from a more generic situation, just to analyzing a more clear situation. $\endgroup$ – Doriano Brogioli Apr 28 '19 at 17:39
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    $\begingroup$ As I stated in my answer, this just doesn't work. If you displace the original charge, you will eventually reflect image charges into the physical region, in which case the method breaks down completely. $\endgroup$ – knzhou Apr 28 '19 at 17:51
  • $\begingroup$ Exactly as I wrote at the end of my answer. Thank you for pointing this out again. $\endgroup$ – Doriano Brogioli Apr 28 '19 at 18:05

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