# Does grounding set the potential on a surface of a sphere to $~0~$ or the potential due to the charges of the sphere to $~0~$?

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!

There cannot be a potential difference inside a static conductor. For if there was, there'd be an electric field and charges would redistribute until that force vanishes.

This applies to the situation without earthing where the potential on the sphere is uniform throughout the sphere.

But it also applies when the conductor is grounded. A connection to "earth" is a conducting path to a large enough reservoir of charges such that it's always at a certain potential (say arbitrarily 0). And hence there cannot be a potential difference between earth and the sphere. Thus grounding sets $$V_{conductor}=0$$.

In the presence of the influencing charge, balancing charges are brought in from earth to neutralize any potential difference between conductor and earth due to influence.

One my phrase it the following way: The electrostatic potential $$\phi(x)$$ at any point $$x$$ in space is the sum of the potentials generated by the external free charge $$\phi_{ext}$$ and the induced charges on the conducting surface $$\phi_{ind}$$ (this is what OP calls $$V_{conductor}$$ which I think is misleading terminology): $$\phi(x) = \phi_{ext}(x) + \phi_{ind}(x)$$

Since the surface is grounded, we know its potential to be zero $$\phi(x\text{ on the surface}) = 0$$ and thus $$\phi_{ext}(x\text{ on the surface}) = - \phi_{ind}(x\text{ on the surface})$$

• So if I measured the $V_{total}$ at the surface of the sphere closest to the positive charge, it wouldn't be zero, due to the potential by the positive charge? How about the negative charges acclumated in the conductor near the positive charge? – Yip Jung Hon Aug 4 at 9:11
• @YipJungHon, what is $V_{total}$? The (conductive) sphere is an equipotential surface isn't it? If you measure the potential of the (grounded) sphere anywhere on the surface of the sphere, you will measure 0V. The mobile charge is distributed just so that this holds. – Alfred Centauri Aug 4 at 11:58
• I denoted $V_{total}$ as the sum of potential due to the charges on the conductor $(V_{conductor})$ and the potential due to the positive charge $(V_{external})$. I see what you mean. Does this mean that at the surface of the conducting sphere where there is a positive charge placed near it, $V_{conductor}$= $-V_{external}$ in order to make $V_{total}=0?$ – Yip Jung Hon Aug 4 at 12:33
• @YipJungHon Essentially yes, see the edit. I have named the potentials a little differently, because I think calling the potential due to the induced charges V_conductor is misleading. – Nephente Aug 4 at 14:05