I know that the $E$ field inside a conductor is zero. What happens if I put a source of charge inside the conductor?

Say the conductor was spherical centered on the origin and there exists a charge at the origin. Then by Gauss's Law, there exists a non zero field inside the conductor.

Or is it the case that even though this charge is within the sphere, it will redistribute evenly over the surface and hence the E field inside is zero?


1 Answer 1


Firstly Gauss law never predicts that there will be no $\vec E$ inside the conductor , the fact that $\vec E$ is $0$ comes from the fact that we are dealing with electrostatics . What gauss law predicts is that $Q_{inside}$ a conductor will be $0$ under a very special condition/state called electrostatic.

Now what is electrostatics ?

It is a condition achieved , a physical state such that no charge anywhere is moving.

Using this argument to say $\vec E=0$ in a conductor . We reason as follows , we know that any conductor by definition has something has de-localised electrons / charge carriers which are free to flow inside the conductor at least . Now if there's any $\vec E$ inside the conductor , we can surely say then the charges must move . But we are assuming there is electrostatic condition , means no charges are moving . Thus $\vec E$ must be $0$ , because had it not been $0$ we could have never achieved electrostatic condition .

Now the Gauss' law part , as $\vec E$ is 0 , so Flux $\phi=0$ through out gaussian surface by definition of flux and thus it means by Gauss' law that charge contained inside our Gaussian surface is $0$ ,

So we proved that in electrostatic conditions , charge cannot reside inside a conductor using 1. A physical assumption & 2. A law of nature .

Now dealing with your question , assuming electrostatic condition has been achieved , there must be no $\vec E$ inside the conductor , I am assuming not a hollow sphere . Thus net charge inside any gaussian surface you imagine is $0$ .

Now imagining this physically , suppose the whole sphere is made up of very very thin infinitesimal shells , you can assume condition to be like this in every infinitesimal shell (assume these to bespheres).enter image description here

But you know net charge of an isolated system is conserved , so on the outermost surface there will be net charge = the charge you put inside in the first place.

Equal in terms of magnitude and sign .

And it'll also be uniformly distributed . Why ?

Because of the symmetric nature of the sphere , suppose you are the charge , you'll see that all points are adzactly equal to each other in every respect , orientation etc. so you'll distribute same way at each point and also if you calculate , in general if you reason that charge will be more here , I can reason same way for any other point on the sphere , you'll see that only this symmetric distribution here helps us to achieve Electrostatic Conditions .

  • $\begingroup$ That's a nice answer - thank you. So the charge located inside will redistribute (evenly) over the surface of the spherical conductor. And then we conclude no charge so no field? $\endgroup$
    – CAF
    May 13, 2013 at 9:29
  • $\begingroup$ Other way around , no field $\implies$ no charge here in this particular case . First the argument for no field comes . And yes the charge will redistribute evenly over the surface . $\endgroup$
    – user23503
    May 13, 2013 at 9:31
  • $\begingroup$ You said above 'assuming not a hollow sphere'. How would this change things? $\endgroup$
    – CAF
    May 13, 2013 at 9:33
  • $\begingroup$ Basically a hollow sphere is simpler case . Nothing as such in the solution will change , just assume the sphere to be infinite hollow spheres , so , in a way I answered for infinite of them , and taking one hollow sphere is a simple and special case of this . Nothing different here , I emphasised not a hollow sphere , so you can imagine the infinite shell redistribution of charge here . $\endgroup$
    – user23503
    May 13, 2013 at 9:35

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