# Is the electric field within a cavity of a conductor affected only by other charges within the cavity?

The following exercise below confuses me: I don't understand why the answer is (A). I seem to be realizing empirically that the electric field at points inside a cavity in a conductor such as this is only affected by charge inside the cavity like $$+q_c$$ rather than charges outside of it like $$+q$$. However, I can't justify this physically.

If my assumption is correct, why is this? If it's not, what's the proper thinking here? If our point of observation wasn't along a line between the two charges, how would this change things?

• Gauss law might be able to help if you are already able to prove spherical symmetry of field – Aditya Garg May 1 '19 at 17:37
• Is this a n exercise from a text book? – my2cts May 1 '19 at 22:07
• No, it is not - to my knowledge. – sangstar May 2 '19 at 9:17

Consider this situation: there is no cavity and no $$q_C$$, only a copper sphere and an external point charge $$q$$.
You know that in this situation the charges in the conductor rearrange so that the field in the conductor becomes zero. This is done by means of a non-uniform surface charge density $$\sigma$$, whose generated field in the inner of the conductor cancels with $$q$$'s field. This charge is located only on the surface of the sphere and its total is zero (we' re supposing a neutral conductor).
Finally, the effect of adding $$q_C$$ in the cavity can be simply figured out by means of the principle of superposition.
Because of superposition the effect of both charges can be discussed separately. The correct answer is indeed A. The presence of the conductor does not affect the contribution of $$q_C$$ to the field, although it induces a charge $$-q_C$$ on the inner surface and $$+q_C$$ on the outer. The outer charge, $$q$$, induces a charge distribution $$\rho$$ on the outer surface of the conductor such that inside the conductor the field is cancelled. The surface integral of $$\rho$$ sums vanishes. Its value can be found from Gauss's law, $$\vec \nabla \cdot \vec E = \rho / \epsilon_0$$. Inside the conductor the field is zero, no charge distribution is induced on the inner surface and no field contribution exists in the inner cavity.