The usual argument about the electric field being zero within the empty cavity of a conductor is as follows.

Consider any two different points on the inner surface of the conductor. Let the first point have a positive charge while the second have a negative charge. If we form a loop containing the electric field line joining these charges, we get a non-zero result for the loop integral $\oint\vec E\cdot d\vec l$, if we assume a non-zero electric field within the cavity. Thus we are forced to conclude that the electric field is indeed zero within the cavity for such a system.

However, if we now place a charge $q$ within the cavity then by using the same arguments above we can create a loop around involving the electric field between some point on the surface and $q$, which in this case is not on the surface. If we again pass this loop from the conductor we will get a non-zero integral, leading to the result that the field within this system's cavity is zero too.

But I believe that my above argument is wrong, that is there is some contribution of induced charges due to the presence of $q$ that I am neglecting. However, I am not able to catch an error in reasoning.

  • $\begingroup$ The field within the (ideal) conductor material is zero, because there are mobile charges, which would keep moving till they cancel out the field. This is why the external field does not penetrate the cavity. The charges in the cavity however do create field, which does not have to be zero: the EM101 problem of a spherical shell around a point charge is an example (use Gauss law). $\endgroup$
    – Roger V.
    Commented Jul 1, 2022 at 7:03

1 Answer 1


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Source: https://www.fiberoptics4sale.com/blogs/electromagnetic-optics/a-plain-explanation-of-maxwells-equations

As seen in the equation above, In a steady state system, Where the magnetic fields don't change over time, you can never have a non-zero value for the closed loop LHS integral.


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