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 $$\Phi_E=\frac{q+q_1}{\epsilon_o}$$.

Note that the flux changes as you suggest (assuming the two charges are equal) but it is not then true that the electric field is twice what you wrote in your first equation. That's because with the second charge being off center, the electric field is no longer spherically symmetric so the flux integral is not easily calculated by assuming E is constant.