Feynman shows in his Lectures on physics, Volume 2, Section 5.10 that in a hollow conductor there can be no electric fields inside the cavity. But he also says that no static charge distribution Q inside a hollow conductor can produce electric fields outside the conductor, that shielding also works the other way. But how can that be? The flux through a gaussian surface that surrounds the conductor is not zero but Q/ε.


1 Answer 1


Feynmann is right that no static charge distribution inside a hollow grounded conductor produces any field outside the conductor. As far as Gauss' law goes, because the conductor is grounded, i.e. there must be some small wire connecting it to a good earth, it is impossible to draw an arbitrary Gaussian surface that encloses the conductor which is not punctured by the grounding wire, so Gauss' law doesn't apply in the manner that you have applied it. Essentially, the system is not an electrostatic one, e.g. charges coming up from or going down to earth give the system an electrodynamical aspect that alters the considerations.

However, this does not mean that there is no field outside the conductor if the conductor is not grounded.

For example, consider a hollo spherical conducting shell with a point charge Q, at the center. Now the field lines from the point charge terminate on the inside of the conductor's surface and they are unable to pass through, however, they can and do force the free electrons in the conductor to rearrange themselves into a new equilibrium position, i.e. an equal and opposite charge -Q, is induced on the interior of the conducting surface. This is not, however, without the effect of the exterior of the conducting surface having a charge opposite, to the charge induced on its interior, viz. Q. It is this exterior induced charge that gives rise to the field exterior to the conductor and not the point charge inside.

For the case of the spherical conductor, the field is equivalent to that of the point charge itself, however, for differing conductor geometries, the field will take on characteristics of the conductor geometry, with the overall (absolute) strength determined by the enclosed internal charge.

  • $\begingroup$ Thank you, I see now that I must look more into the details of what's happening with the charges in the conductor. $\endgroup$
    – Leif Dietz
    Commented Feb 27 at 17:15
  • $\begingroup$ @LeifDietz Please see my post again, I made crucial edits that better answer your question, I went to check up on Feynmann's lectures. $\endgroup$ Commented Feb 27 at 17:18
  • $\begingroup$ @LeifDietz Gauss' law doesn't apply in the case of a grounded conductor as given in Feynmann's lecture. $\endgroup$ Commented Feb 27 at 17:19
  • $\begingroup$ @LeifDietz You could apply Gauss' law if you also included all the charges in the earth or ground, however, then your paradox wouldn't exist. $\endgroup$ Commented Feb 27 at 17:21
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    $\begingroup$ @LeifDietz Yes, the flux would be proportional to the sum of the internal charge, plus what ever external charge was added. Of course if the conductor was grounded as in Feynmann's example, any charge added would be leaked away to ground. $\endgroup$ Commented Feb 27 at 17:30

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