In the context of this page, I have more or less understood the proof and I have written a summary below:

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  1. Consider any Gaussian surface inside a large surface which contains the cavity. Flux through this surface is zero as the electric field at every point on it is zero. Therefore either zero charges enclosed or an equal amount of positive and negative charge
  2. To prove that it is the case of zero charge, inside the large surface consider a line integral around some curve $\Gamma$ which has part of it in the conductor (but not in the cavity) and part in the cavity with the part in the cavity along an electric field line(refer here). Note that the line integral along the path of the conductor is zero.
  3. However, if there is a equal opposite and negative charge there must be some quantity of line integral inside the cavity, however, it is well known that a loop integral is zero in the electrostatic case because it is characterized by a lack of curl. Hence no charge!

But this one point in Feynman's explanation I can not understand:

Now imagine a loop Γ that crosses the cavity along a line of force from some positive charge to some negative charge, and returns to its starting point via the conductor (as in Fig. 5–12). The integral along such a line of force from the positive to the negative charges would not be zero.

if you have a line integral passing through a charge, isn't the line integral actually running through a singularity of the electric field? I think this should mess things up. ( Charges are like singularities of the electric field as the coulombs law expression blows up as you get very close to it)


1 Answer 1


There is no singularity of the electric field on a smooth metal surface, a smooth surface distribution of charges produces a finite field. This is not true any longer if you have sharp edges and some such but Feynman's proof will still hold if you approach the edges by a sequence of smooth surfaces and take the limit with some "handwaving"...

  • $\begingroup$ Isn't there some kind of blow up the part where the loop pierces the metal? $\endgroup$
    – Babu
    Commented Nov 11, 2020 at 6:25
  • $\begingroup$ think of a simple capacitor in the middle of the plates with nearly uniform charge distribution on the surface, why would the field not be finite and smooth there? $\endgroup$
    – hyportnex
    Commented Nov 11, 2020 at 11:07
  • $\begingroup$ Good point but difficult to wrap my head around it $\endgroup$
    – Babu
    Commented Nov 11, 2020 at 11:42
  • $\begingroup$ a point charge has a field with singularity, when it is distributed uniformly over a sphere of finite radius $R$ there is no singularity, in fact, the field is identical to that of the point charge for $r>R$. The field singularity is caused by the finite amount of charge being squeezed in an infinitesimal surface or infinitesimal volume. A finite surface density of charges has an infinitesimal charge over an infinitesimal surface. $\endgroup$
    – hyportnex
    Commented Nov 11, 2020 at 11:47
  • 1
    $\begingroup$ Just keep on with Feynman; I also like Schwartz, Jackson and Landau-Lifshitz $\endgroup$
    – hyportnex
    Commented Nov 11, 2020 at 12:11

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