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I've been taught electric field lines do not exist inside the volume of the conductor. An internal field is created inside its volume which cancels the external field, hence suggesting the net charge inside the conductor is zero.

I am aware this is true when electrostatics is in question. I had an example in mind which stems my confusion:

If we take a conducting shell and a dielectric shell, find electric field due to them on points (i) inside the shell, (ii) outside the shell, (iii) on the surface of the shell, through Gauss's Law,we'll get the same values for both (according to the position of the point).

ie. KQ/r² for (i), KQ/R² for (ii) and 0 for (iii)

This implies there's not much difference (except charge distribution) between the properties of a 'conducting' and a 'non-conducting' surface when we're considering electrostatics. Is that implication correct?

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  • $\begingroup$ What's the difference between (II) and (iii)? $\endgroup$
    – Bob D
    Commented Apr 28 at 18:36
  • $\begingroup$ @BobD consider R to be the radius of the shell. (ii) means r>R and (iii) means r=R. Nothing too significant to the actual question... those are just the conditions considered in my textbook. $\endgroup$
    – Mel
    Commented Apr 28 at 18:39
  • $\begingroup$ So why do you write "we'll get the same values for both"? We know the conducting shell cancels the external field but you apparently just assume the dielectric does that as well! I would think that exactly that is the question here... $\endgroup$
    – Jos Bergervoet
    Commented Apr 28 at 19:35
  • $\begingroup$ @JosBergervoet I forgot to mention, I'm talking about an extremely thin shell, such that its volume is negligible. Therefore, there is no charge enclosed in the Gaussian surface and hence electric field is zero. $\endgroup$
    – Mel
    Commented Apr 29 at 12:28
  • $\begingroup$ @Mei Seeing now your comment, I wonder what your question actually is about. Is it about a metal and a dielectric sphere with equal charges put symmetrically on the shell with no external applied field? That how I understood it for my answer. Or do you want to know what happens in this two spheres when you apply an external electric field. In the latter case, you would have no spherically symmetric charge distribution for the metal case and could not, as suggested, determine the electric field at the considered points from Gauss' law. Please explain. $\endgroup$
    – freecharly
    Commented Apr 29 at 19:39

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As long as you have a spherically symmetric charge distribution and don't look inside the shell wall itself, there should be no difference between electric field of the described conductor and the isolating dielectric cases.

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  • $\begingroup$ In case I consider a thin shell (ie. Volume=negligible) then there certainly is no more difference between the two, right? $\endgroup$
    – Mel
    Commented Apr 29 at 12:30
  • $\begingroup$ @Mei You are right! $\endgroup$
    – freecharly
    Commented Apr 29 at 13:05
  • $\begingroup$ But the question mentions a "conductor placed in an electric field", where it will of course develop a nontrivial charge distribution! The dielectric cannot do that, (unless we let $\varepsilon_\text{r}\rightarrow\infty$ but that is not intended here). Placed in an electric fied they do not behave the same. $\endgroup$
    – Jos Bergervoet
    Commented Apr 29 at 15:57
  • $\begingroup$ @Jos Bergervoet t I assumed in my answer that the question is about spherically symmetrical charges residing on the metal and dielectric shells. This is also implied by the question mentioning Gausses law and the total charge Q to determine the electric field at the indicated locations. I did not interpret it as a problem where an external field is applied, to which your argument would, of course, apply. $\endgroup$
    – freecharly
    Commented Apr 29 at 19:55

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