# Electric field outside of a conductor with charge $q$

Suppose that I have a spherical conductor (radius $$R$$) with a cavity at the center and a charge $$q$$ inside it. I know that the electric field outside of the conductor will be $$\vec{E}=\frac{q}{4\pi\epsilon_0}\frac{1}{r^2}\hat{r}$$

My question is what happens when there is another charge of the same magnitude at distance $$R+a$$, where $$a, from the center of the conductor. Does the flux of the electric field outside of the conductor (for example, at $$r=2R$$) remain $$\Phi_E=\frac{q}{\epsilon_0}$$ or does it become $$\Phi_E=\frac{2q}{e_0}$$ due to the other charge?

• At $r=2R$, if we assume a sperical Gaussian surface, we can see that the net enclosed charge in this surface is $q+q=2q$ and thus the flux is $\Phi_E=\frac{2q}{\epsilon_o}$
– Manu
Sep 16, 2021 at 13:58
• @Manu You should write this comment as an answer Sep 16, 2021 at 14:17
• You are assuming that the second charge is equal to the first. Sep 16, 2021 at 14:17
• Yes, if it is not the case then $\Phi_E=\frac{q+q_1}{\epsilon_o}$.
– Manu
Sep 16, 2021 at 14:51

At $$r=2R$$, if we consider a sperical Gaussian surface, we can see that the net enclosed charge in this surface is $$q+q=2q$$ and thus the flux is $$\Phi_E=\frac{2q}{\epsilon_o}$$.
If tje outside charge is not same as $$q$$ (say $$q_1$$), then the flux becomes $$\Phi_E=\frac{q+q_1}{\epsilon_o}$$.