# Electric field outside conductor with cavity

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?

• Can you elaborate your problem, please? – user36790 Sep 6 '15 at 16:05
• I believe your teacher meant a spherical conductor with cavity... – John Dvorak Sep 6 '15 at 16:06
• @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. – Quaternion Sep 7 '15 at 7:39
• @JanDvorak, actually he/she insisted the conductor's shape was arbitrary. – Quaternion Sep 7 '15 at 7:43