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When I have a force field $\boldsymbol{\vec{F}}$ it tells me how the momentum of the particle will change given in a position on space.

When we talk about an electric field $\boldsymbol{\vec{E}} = \boldsymbol{\vec{F}}/q$ it seems to be a generalization of this concept since now it's independent of the other charge, now I can tell how would, the behavior of any charge under the influence of this electric field, be like. When I talk about an electric potential $V$, it seems to be again a generalization of work, since $\int\boldsymbol{\vec{E}}\cdot dr = \Delta V$. And now it makes sense why we always talk about the difference of potential because what we can measure in nature is the variation of energy.

When we say that something has an electric potential $V$ normally we put the point of reference on infinity to make $V(x) - V(\infty) = V(x) - 0$, but we still talk about the difference of potential $\Delta V$.

If I'm thinking correctly about it, my question is:

When I have a charge at distance $d$ from a conductive grounded plane like the image below, by induction, the plane will have a charge equal to -q.

grounded plane whit a charge at distance d

When I say that the conductor is grounded and has a potential equal to zero, is the same as saying that when a take any other charge at infinity distance of the plane and I bring it to this plane the work done should be zero? this seems to me to be a little counterintuitive

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2 Answers 2

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No, it means that you are measuring the electric potential relative to the potential of the plane rather than the potential at infinity.

Note that the potential at infinity, which you call $V(\infty)$, is only meaningful to talk about if $\lim_{|\mathbf r|\rightarrow \infty} V(\mathbf r)$ is well-defined. If you have an infinitely large, uniformly charged plane, for example, then this limit is not well-defined because it $V(\mathbf r)$ diverges to infinity if you walk in the direction perpendicular to the plane while remaining constant if you walk in some direction parallel to the plane.

Even if the aforementioned limit exists, it's often convenient (e.g. in circuits) to measure potential relative to some point which isn't at spatial infinity. If I have a basic circuit with a single battery in it, it makes more sense to measure the potential relative to the anode of the battery than relative to some point out beyond Alpha Centauri.

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  • $\begingroup$ Just to comment that, this answer is mostly correct and it's the answer I would have given, but I would have highlighted the “localized” criterion as sufficient but (I’m pretty sure?) not necessary. Indeed in this case it happens to be true that the potential at infinity is also zero relative to the plane, even though the charge induced on the plane is technically an infinite distribution with the same spatial domain as the uniformly charged sheet. I also might have made some statement about boundary conditions—these kinds of statements assume, say, no EM waves traveling in from infinity etc. $\endgroup$
    – CR Drost
    Commented Jun 24, 2021 at 19:16
  • $\begingroup$ @CRDrost Yes, that's a good point. I have weakened that criterion. I'll have to think about the precise statement I'd make about the boundary conditions. $\endgroup$
    – J. Murray
    Commented Jun 24, 2021 at 19:31
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If we postulate some charge $q$ at a distance from ground, this does not mean that ground has the opposite charge; that opposite charge could be anywhere.

We define ground as the reference from which all other potentials are measured. It has an undefined charge, however any charge it might accumulate is finite and it therefore must always have zero potential. So we can treat it as a reservoir having infinite capacity.

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