From Griffiths:

Two spherical cavities, of radii a and b, are hollowed out from the interior of a (neutral) conducting sphere of radius R. At the center of each cavity a point charge is placed – call these charges $q_a$ and $q_b$.What is the field within each cavity?

By Gauss’s law, fields are $Kq_a/r_a^2,Kq_b/r_b^2$, but thinking in terms of charges, the charge on outer surface cannot affect inside, so contribute nothing to the field inside any cavity. Also does not affects the field in another cavity the charges of one cavity(both present inside and induced). So field inside the cavity must only be due to $q_a$(WLOG), and induced charges $-q_a$.

But the result from gauss's law suggests that even induced charges do not affect the field? Is it due to the fact that charges on surface of a solid charged conducting sphere produce no field inside, but these induced charges are not on outside surface [of the surface of cavity but inside the whole sphere], also if both situations are similiar, then there's difference in that the cavity is like a hollow shell. Is my reasoning correct? Somebody please explain.


Explain without gauss's law, I know that with gauss law you would eventually explain all( I might also ).


2 Answers 2


But the result from gauss's law suggests that even induced charges do not affect the field?

This is because the charges are at the center of the cavity, so the induced charge are symmetrically induced at the cavities(circular) and we know already that the field due to charged hollow sphere is zero at any point inside the sphere, similarly happens here.

Also, the fact that hollow sphere is charged on the outside and the induced charges are now residing inside doesn't matter since I think electric field is unaffected by intervention of a neutral conducting sphere.


The induced charge on the inner surface of the cavities will be $-q_{a}$ and $-q_{b}$. Also there will be a distributed induced charge on the outer surface of the sphere which is $q_{a}+q_{b}$. Now why these charges do not contribute in the field inside the cavity. The reason is to find out the electric field you will choose a Gaussian surface inside the cavity. The charge enclosed by the surface contributes to the electric, nothing else. That is why the induced charges do not affect the electric field inside the cavity.

  • $\begingroup$ I know that, see my question, I'm talking without gauss's law $\endgroup$
    – RE60K
    Aug 16, 2014 at 5:57
  • $\begingroup$ The charges are on the inside wall of the cavity, but they are not enclosed by the surface. $\endgroup$ Aug 16, 2014 at 6:08

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