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I've started asking myself:

What is the value of the electric field at the surface of a shell with uniform distribution? In particular, is it infinite, as this point is over an (infinitesimal) charge?

After google a few, it seems the correct answer is $E=\frac{k_eQ}{2R^2}$ being R the radius of the shell and Q its total charge. It is not infinite (why?), nor $0$ nor $\frac{k_eQ}{R^2}$.

Still on same target, I've try to compare the expression for total energy in a system of charged particles versus the expression for continuous system. According wiki and other, the expression for particles is:

$$ U = \frac{1}{2} k_e\sum_{i=1}^N q_i \sum_{j=1}^{N(j\ne i)} \frac{q_j}{r_{ij}}. $$

Note the $j \ne i$ in second summatory, that means that, for a single charge, $U=0$, as explained in the comment:

A common question arises concerning the interaction of a point charge with its own electrostatic potential. Since this interaction doesn't act to move the point charge itself, it doesn't contribute to the stored energy of the system.

However, for a continuous system, the expression is:

$$ U = \frac{1}{2} \int_V \rho \Phi dV $$

(not an equivalent to the $i \ne j$ ?)

a) if I describe a particle has $\rho(v)=Q\delta(v)$, then:

$$ U = \frac{1}{2} \int_V \rho \Phi dV = \frac{1}{2} \int_V Q\delta(v) \frac{k_eQ}{r} dV = \frac{k_eQ^2}{r}|_{r=0} = \infty $$

b) skipping Dirac's delta , I try to define a particle as a shell in the limit $R \rightarrow 0$. Inside and in the surface of a shell there are constant potential $\frac{k_eQ}{R}$:

$$ U = \frac{1}{2} \int_V \rho \Phi dV = \frac{1}{2} \int_V \rho \frac{k_eQ}{R} dV = \frac{k_eQ^2}{R} $$

that, again, when $R \rightarrow 0$ then $U \rightarrow \infty$.

(equal result, but longer proof, if I define a particle as a uniformly charged sphere).

Thus, my questions are:

  • why the field in the surface of a charged sphere (uniform or shell) is not infinite?
  • why continuous and discrete expression of total energy doesn't gives me the same result ?
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  • $\begingroup$ "After google a few". Do you have a reference? $\endgroup$
    – my2cts
    Apr 25, 2019 at 18:36
  • $\begingroup$ @my2cts: this is the most complete I've found: ias.ac.in/article/fulltext/reso/023/11/1215-1223 $\endgroup$ Apr 25, 2019 at 18:41
  • $\begingroup$ I suggest to study Wikipedia articles and references therein on electromagnetism instead. Also some good textbooks are accessible online. $\endgroup$
    – my2cts
    Apr 25, 2019 at 18:48
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    $\begingroup$ @my2cts relying on Wikipedia for study of physics is a terrible idea. Textbooks and papers from well known authors are much more reliable source of information. $\endgroup$ Apr 27, 2019 at 11:28
  • $\begingroup$ @ Jan Lalinsky I have good experience with wikipedia articles. They can serve as an introduction to literature. If I find bad ones I edit them. You are right that they can never replace studying textbook, or better, physics in a university program. Note that a decent wikipedia article is a great improvement over the article referred to by the OP for any standard. $\endgroup$
    – my2cts
    Apr 27, 2019 at 11:44

2 Answers 2

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The field at the surface of a uniformly charged shell with charge Q is $\frac{k_eQ}{R^2}$. Inside the shell, as well as inside the conductor just below the charge dustribution the field is zero. I don't see the point of "regularizing" this to some sort of of average as suggested in a different answer.

The answer to your first question is that indeed the field tends to zero at the location of a charge, but for a continuous charge dustribution, the charge itself tends to zero, so a finite answer results.

The answer to your second question is that in the discrete expression self interaction is taken out explicitly by omitting $j=i$.

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Let us first compute the electric field on the surface of a spherical (infinitely thin) shell with a uniform charge distribution. We can use Gauss's law for this, $$\int_{S_r} \vec{E}\cdot d\vec{A} = \frac{Q_{in}}{\varepsilon_0},$$ from which using that the charge density is $\sigma = Q/4\pi R^2$ and taking the Gauss surface at $r=R$, we get (using the spherical symmetry): $$E = \frac{4\pi R^2\sigma}{4\pi r^2\varepsilon_0} \stackrel{r\longrightarrow R}{=} k_e\frac{Q}{R^2}$$ as you stated yourself. However notice that for all things to hold, the limit is taken from above, namely the Gaussian surface is taken from outside and reduced. On the other hand we also know the field inside will be zero, so there is a clear discontinuity in the electric field at $r=R$. So formally the field is not defined there, but we regularize the field by taking the average of the discontinuity jump. In this case, $$E(R):= \frac{\Delta E}{2} = (E_{out}+E_{in})/2 = E_{out}/2 = k_e \frac{Q}{2R^2}$$ again in agreement with the result you found. (For more details on this I suggest Griffith's book on electrodynamics sec. 2.4 and 2.5)

Now to compute the energy stored in a given configuration it is enough to realize that, the energy stored in the electric field of a point charge is already infinite, since it somehow measures how much energy went into building the configuration so you would need to drag let's say some charge $Q$ and accumulate it into a point and Coulomb's law already tells us the force would be infinite.

So when you use the discrete version for computing the energy, one is not taking into consideration the energy needed to build the point charges themselves, reason why you demand $i\neq j$. While in the continuous version of the formula $\Phi$ is the total potential evaluated at the same point as the density, $\rho(\vec{r})\Phi(\vec{r})$, so you expect this quantity to already be infinite if not taken with care, because it is implicitly accounting for the energy of building up each "charge point". Again if you read sec. 2.5 from Griffith's book, you will find some more details.

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  • $\begingroup$ This does not answer the question. It address calculation on a discrete grid, where care must be taken to correctly account for singularities. $\endgroup$
    – my2cts
    Apr 27, 2019 at 11:51
  • $\begingroup$ This answer could be the accepted one if it includes a proof that the energy necessary to contruct a charged shell (or charged wire or sphere, if easier) is finite. $\endgroup$ Apr 27, 2019 at 12:01
  • $\begingroup$ The factor of 1/2 is there because we deal with a closed surface. Other shapes such as half-sphere could have different factors. I see no basis for taking average of the two fields in the general case. $\endgroup$ Apr 27, 2019 at 12:20
  • $\begingroup$ It is true that the factor 1/2 relies on the surface being closed, but the averaging is generic, it is only useful if you actually need to assign a value of the field exactly at an interface, so you will have some non-zero field on both sides over 2 still, as Griffith's puts it "Averaging is really just a device for removing the contribution of the patch itself." $\endgroup$
    – ohneVal
    Apr 29, 2019 at 10:39

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