Recently I heard a lecturer claim that the electric field outside a solid conductor containing a cavity with a point charge $q$ at a point $\mathbf{r}$ is the same as the electric field due to the point charge: that is, $\mathbf{E(\mathbf{r})} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{\mathbf{r}}$ outside the conductor.

I know that for a spherical conductor with an arbitrary cavity containing a point charge $q$ the field outside is $\mathbf{E(\mathbf{r})} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{\mathbf{r}}$, where $\mathbf{r}$ is measured from the center of the sphere. But I don't think the lecturer's statement is true - for example, if the shape of the conductor is arbitrary and the field outside is radial, as claimed, then the field cannot in general be perpendicular to the surface of the conductor as it should.

Is my objection to the lecturer's claim valid?

  • $\begingroup$ Can you elaborate your problem, please? $\endgroup$
    – user36790
    Commented Sep 6, 2015 at 16:05
  • 1
    $\begingroup$ I believe your teacher meant a spherical conductor with cavity... $\endgroup$ Commented Sep 6, 2015 at 16:06
  • $\begingroup$ @user36790, Thanks for your comment - I've tried to clarify the question, but I don't see much to elaborate on. It's quite a simple question (I'm only asking because a lecturer said . . .), and in fact I've already accepted an answer. $\endgroup$
    – Quaternion
    Commented Sep 7, 2015 at 7:39
  • $\begingroup$ @JanDvorak, actually he/she insisted the conductor's shape was arbitrary. $\endgroup$
    – Quaternion
    Commented Sep 7, 2015 at 7:43

1 Answer 1


Close to the conductor the field depends strongly on its shape. An arbitrary conductor won't produce an electric field comparable to the field of a point charge.

But in the far field every non-zero charge distribution will result in an electric field, which behave like a equivalent point charge. By experience, the far field starts around 100 times the extent of the conductor.

This can by shown by multipole expansion of the charge distribution. You find, that "higher" multipole moments like the quadrupole moment will decline faster the the monopole moment (total charge).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.