# Field inside cavity of charged conductor

Can we use gauss law to prove that field in a cavity of charged conductor of any shape is zero?? If so how as the shape may lack symmetry? If not how can we prove the field to be zero? I know that the charge resides on the surface of conductor but I am not satisfied by the reasoning that since charge enclosed is zero field is zero as per my book. Is there a quantitative way to explain it?

• Do you know the proof of gauss law using solid angle? If not I will suggest you to go through it thoroughly. You might check it out over here:scribd.com/document/247926626/Gauss-Theorem-Proof The proof can also be found in Concepts of Physics-Vol2-Gauss Law May 21, 2020 at 4:19

## 2 Answers

If there is no symmetry, Gauss law wont't help you (it's still true, but less useful).

You can prove that field lines on the surface $$S$$ of the conductor must be normal to the surface, otherwise there will be a net force rearranging the charges until this becomes true. This means the surface of the conductor is an equipotential, say $$V_S=V_0$$, which implies $$V=V_0$$ in the entire volume enclosed by $$S$$ due to Laplace's equation. Since $$\mathbf{E}=-\nabla V\implies\mathbf{E}=0$$.

Yes, we can use Gauss Law $$\oint \vec E\cdot d\vec{A}=Q_{net}/\epsilon_{o}$$ to prove that electric field inside the cavity of conductor is zero. Consider a closed surface inside the cavity then the net charge enclosed by Gaussian surface will be zero i.e. $$Q_{net}=0$$
$$\oint \vec E\cdot d\vec{A}=0$$$$\oint |\vec E|dA\cos\theta=0$$ $$\implies |\vec E|=0\quad (\because dA\ne 0, \theta\ne90^\circ )$$