I'm understand that grounding sets the potential to $0$, but I'm confused as to which potential it sets to $0$. Let me explain my question:
If I have a sphere and I bring a positive charge near it, a negative charge will form at the surface close to the positive charge. Now, if I ground the sphere, I understand that a negative charge will still remain (ie. there won't be a uniform distribution of charges).
From what I know, if we were to take the potential at a point outside of the sphere (let's say just on the surface of the sphere) $(V_{\text{total}})$, it will be sum the potential due to the accumulation of negative charges induced in the sphere $(V_{\text{conductor}})$ and the potential due to the small positive charge $(V_{\text{external}})$. So $$V_{\text{total}}=V_{\text{conductor}}+V_{\text{external}}~.$$
My question is:
Does grounding set $V_{\text{total}}=0$ or $V_{\text{conductor}}=0$?
Since there are still negative charges on the sphere near the positive charge, I assume $V_{\text{total}}=0$?
Is my reasoning correct on this?
This is unintuitive for me, because the positive charge isn't even connected to the conductor. I initially thought that grounding sets $V_{\text{conductor}}$ to zero instead.
Could someone clarify this and point out any misconceptions I may have? Thanks!