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In the framework of classical electrodynamics, at distances much greater than a conductor's dimension, the field ought to approach that of a point charge located at the conductor. But where at?

For a highly symmetric conductor, we ought to be able to deduce some information about the "point charge" location. Yet, consider an arbitrarily shaped conductive body. Is the point located at the centroid?

Edit: I am not questioning why we approximate a source as a point charge at far distances. This question regards geometrical convergence in a physical scenario. If the field lines converge to that of a point charge, they emanate from a point. Are there known relations between the point and the body itself?

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  • $\begingroup$ If you have a square sheet of metal of side 1cm. You can consider the point to be anywhere near the metal if you are a mile away. The point is NOT the centroid. The centroid is a pure geometric quantity. The charge density can vary over the geometry. You usually get the location of the point when you apply the approximations to the exact formula. $\endgroup$ – Yashas Jan 26 '17 at 6:50
  • $\begingroup$ @YashasSamaga That makes sense. I suppose we could chase the point location down from the formula expansion. That would provide some case by case analysis. I put centroid out there as an example, or bait I suppose, but I'm curious as to whether there are any known relationships that stem from the geometry of the conductor. $\endgroup$ – zahbaz Jan 26 '17 at 7:30
  • $\begingroup$ There is no fixed point but you get the location after you apply the approximation. For example, consider a dipole, if you have used $r$ as the distance from the point where you wish to find the field strength to the midpoint of the dipole. Then you are going to assume that all the charge is concentrated at the midpoint of the dipole from a far away place. $\endgroup$ – Yashas Jan 26 '17 at 7:36
  • $\begingroup$ How do you get ​ "If the field lines converge to that of a point charge, they emanate from a point." ? ​ ​ ​ ​ $\endgroup$ – user21968 Jan 26 '17 at 7:36
  • $\begingroup$ @RickyDemer It was poor phrasing. Maybe this is better: "If the field lines of a source converge to those of a point charge, we must be able to locate the position of said point." $\endgroup$ – zahbaz Jan 26 '17 at 7:46
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Based on some of the back-and-forth I see, I think you're asking the wrong question. I think the question you want to ask is "Given a charge distribution $\rho(\mathbf{r})$, where should I place a point source so that the exact potential $\phi(\mathbf{r}) = \int \rho(\mathbf{r}')/|\mathbf{r}-\mathbf{r}'| dv'$ is most closely approximated by the potential from the point source?"

The answer is that you want to choose $\mathbf{r}_0$ such that

$\int (\mathbf{r}'-\mathbf{r}_0) \rho(\mathbf{r}') dv' = 0$

If the charge distribution is uniform, then the answer is at the centroid. The reason this is the right point is it makes the dipole moment of the difference between exact and approximate solutions go to zero. So the error in the potential is $\mathcal{O}(1/r^3)$, whereas with any other choice the error would include the dipole term, and therefore be $\mathcal{O}(1/r^2)$. (Properly setting the magnitude of the point charge accounts for the monopole term of $\mathcal{O}(1/r)$.)

Further clarification:

The choice of $\mathbf{r}_0$ that satisfies the dipole constraint above is

$\mathbf{r}_0 = \frac{\int \mathbf{r}' \rho(\mathbf{r}') dv'}{\int \rho(\mathbf{r}') dv'}$

and can be thought of a as a "center-of-charge" similar to a center-of-mass.

The multipole expansion of the potential $\phi(\mathbf{r})$ contains terms of increasing order in $1/r$

  • Monopole terms decay with $\mathcal{O}(1/r)$. Any charge distributions with the same total charge within a local region have the same monopole moment. That's why a point charge with the same total charge works as an approximation, and it doesn't matter where it is, as long as it's close to the same region. With this approximation, the error between the exact potential and the approximation will be $\mathcal{O}(1/r^2)$. If $r$ is big enough, then like everyone else says, it works fine and it doesn't matter where $\mathbf{r}_0$ is.
  • However, if we want, we can be even more accurate with a judicious choice of the location of the point charge. Dipole terms decay with $\mathcal{O}(1/r^2)$. Since the point source clearly has no dipole moment, picking the point $\mathbf{r}_0$ so that the exact potential has no dipole moment about $\mathbf{r}_0$ removes $\mathcal{O}(1/r^2)$ dependence from the error. This leaves only $\mathcal{O}(1/r^3)$ and higher error terms.
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  • $\begingroup$ I believe "the charge is constant" must be "the charge distribution is uniform". $\endgroup$ – Yashas Jan 26 '17 at 9:06
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    $\begingroup$ "If the charge distribution is uniform, then the answer is at the centroid". Is this really correct? The centroid is the volume centre of the conductor but the charge on the conductor has to reside on the surface. Also, it is impossible to make the charge uniform on an arbitrarily shaped conductor because it has to satisfy the surface equipotential condition. $\endgroup$ – Raziman T V Jan 26 '17 at 9:27
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    $\begingroup$ @Raziman The point is mathematically correct, but it says nothing about how that charge got there or whether it's physically achievable. It also doesn't assume that there's even a conductor. The reason to include this sentence is to give an intuitive geometric meaning to the equation. $\endgroup$ – LedHead Jan 26 '17 at 9:31
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The answer is that it doesn't matter.

The distance at which fields resemble that from a point charge is also the distance at which it does not matter where that point is located within the structure. The change of field due to switching the origin within the conductor will be comparable to the corrections to the point charge approximation, both arising from same order multipole terms.

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    $\begingroup$ If it did matter it would not be field of a point charge! $\endgroup$ – ZeroTheHero Jan 26 '17 at 7:03
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    $\begingroup$ Even physically. You can make the "field lines emanating from a point" approximation only when you are so far away that the near field effects are suppressed and it doesn't matter where that point is situated on the structure that has now shrunk to a point. $\endgroup$ – Raziman T V Jan 26 '17 at 7:12
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    $\begingroup$ But this is the mathematical treatment! Even if you go very far away there will be "non point charge" terms in the fields. They only reduce in magnitude but never die completely. So you are making an approximation when you say that the field reduces to that of the point charge. And you can prove that this approximation is of the same order as the error you get when you switch your coordinate origin in the structure. $\endgroup$ – Raziman T V Jan 26 '17 at 7:30
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    $\begingroup$ It does answer your question: the answer is that no matter where you choose that point to be, the result is the same after taking the far field approximation. This is a mathematical result. (Edit: Looks like I was too late) $\endgroup$ – Noiralef Jan 26 '17 at 7:32
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    $\begingroup$ @zahbaz But there is no point charge. It doesn't exist in reality. If you draw all the field "lines", you'll find that they don't all cross a single point in space. We just approximate the differences to a single point where the difference doesn't really matter to your calculation - but the charge is still spread over the conductor, we just ignore the insignificant differences. No matter what point you pick, you'll get results with the same accuracy within the required precision - that's the whole point of approximation. $\endgroup$ – Luaan Jan 26 '17 at 8:13
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The multipole expansion is a useful approximation to low order (mono-, di, - quadrupole) if the diameter of the charge distribution $d$ is much smaller than the distance at which you observe the field or potential $r$.

That observation distance is with respect to an origin, which you conveniently place somewhere inside the charge distribution. Emphasis on somewhere, because wiggling with the origin at most a distance $d$ won't change much $$ |r|^{-1} \approx |r+d|^{-1} \text{for $d\ll r$}$$ You may place the orgin, aka the source of the monopole field, anywhere inside$^1$ the charge distribution. It is not a point dictated by the theory. One would normally choose it in such a way to make higher moments vanish.

One can always make the dipole moment vanish by placing the origin at the center of charge.


$^1$ You may place it outside the (convex cone around) the charges, but then you will never get higher moments to vanish.

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You are implicitly assuming that sufficiently far away from a bounded charge distribution, the electric field becomes perfectly radial, i.e. every electric field line would meet at a single point if you just projected it in a straight line back to the neighborhood of the charge distribution. But this isn't the case; if you could measure the direction of the faraway electric fields so precisely that you could extrapolate them back the source region with a precision better than the spatial distribution of the source charge, you would find that they don't exactly intersect at the same point. The electric fields at different faraway points would point to slightly different parts of the charge distribution. As you get farther and farther away, the electric fields become closer and closer to being radial, so you need to know their directions more and more accurately to tell which specific part of the charge distribution they're aligned with. Measuring the fields with this extra accuracy corresponds to measuring the dipole rather than monopole contribution (which falls off faster). This dipole data in turn gives you more information about the details of the charge configuration than just its total charge, which is what you get from the purely radial part of the field.

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I'd like to add another example and clarify one of my statements in the comments I made to the OP's question.

If you have a square sheet of metal of side 1cm. You can consider the point to be anywhere near the metal if you are a mile away. The point is NOT the centroid. The centroid is a purely geometric quantity. The charge density can vary over the geometry. You usually get the location of the point when you apply the approximations to the exact formula.

There is no fixed point but you get the location after you apply the approximation. For example, consider a dipole, if you have used r as the distance from the point where you wish to find the field strength to the midpoint of the dipole. Then you are going to assume that all the charge is concentrated at the midpoint of the dipole from a far away place.

enter image description here

Your point of interest P is so far away from the dipole that the distance between the point P and the dipole's midpoint is essentially the same as the distance between the point P and negative charge of the dipole.

The change of the location of point charge does not cause any significant change in the answer.

You can assume that the point charge is at the midpoint of the dipole, or it is at the positive charge or it is at the negative charge. Any point you choose, your answer will yield pretty much the same number. Manier times, the uncertainty in our measuring instruments is much bigger that we cannot notice the difference by changing the location of the point. Of course, you cannot go way to off with your approximation.

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protected by Qmechanic Jan 27 '17 at 16:26

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