# Why is the electric potential continuous when we aproach an infinite uniformly charged sheet?

Supose we have an infinite uniformly charged sheet on the plane z=0, and at $z>0$ $\to\vec{E}=\frac{\sigma}{2\epsilon_{0}}$ and at $z<0$ $\to\vec{E}=-\frac{\sigma}{2\epsilon_{0}}$. Therefore, calculating the electric potential $V=-\int\vec{E}.d\vec{z}$, we will have something like this: $V=\frac{\sigma}{2\epsilon_{0}}z$ for $z>0$ and $V=-\frac{\sigma}{2\epsilon_{0}}z$ for $z<0$. . And taking limits and proceding mathematically, we see that the potential is continuous but not differentiable at $z=0$, but here's what it's causing me a headache (and please correct me if I'm wrong):

We define electric potential as the amount of work an external agent has to exert on an unit charge to move it from infinity (taking of course $V(\infty)=0$) to a certain position in the z direction (that's the only direction that matters as we're talking of an infinite sheet). And I may be confusing myself with the point charge model but it is not intuitive to me that the closer we get to the infinite plate (starting from $z=+\infty$) the potential aproaches 0. I would expect to make a tremendous amount of work in order to get that unit charge as close as possible to the plate, and not simply become very easy all of the sudden.

• Is it valid to take $V\to 0$ as $z\to\infty$ here? – probably_someone Mar 27 '18 at 15:54
• Indeed you cannot have $V(\infty)=0$ since there are charges at $\infty$ in the setup of your system. – ZeroTheHero Mar 27 '18 at 16:09

$$V=-\int\vec{E}.d\vec{z}$$

is correct, but remember that there is a constant of integration. So the potential is:

$$V(z)=\frac{\sigma}{2\epsilon_{0}}z + V_0$$

for some constant $V_0$. That means it is incorrect to say the potential is zero when $z=0$. We cannot assign an absolute value to the potential. We can only calculate potential differences.

As you say, it's common to take the potential to be zero at infinity but this only makes sense when the electric field tends to zero at infinity. for example it makes sense for a point charge because the field decreases as $r^{-2}$. The problem with the infinite flat sheet is that the field strength is independent of distance. It does not tend to zero at infinity so the potential at infinity is not a well defined quantity and we cannot usefully set it to zero.

Given the symmetry of the sheet it is an obvious choice to set the potential to zero at the sheet i.e. to choose $V_0=0$. But all this does is to define our potential function $V(z)$ as equal to the energy needed to move a unit charge from the sheet to a distance $z$.

• I'm really sorry about the next question but this is killing me. How can we define $V_0$=0 if we can only talk about potential differences? It kind of made sense to the point charge to set $V(\infty)$=0 as there was no electric field but now at the plate it's confusing – Bidon Mar 27 '18 at 16:40
• @Duartejfs we are defining our potential function $V(z)$ as the energy need to move a unit charge from the sheet (i.e. $z=0$) to the distance $z$ i.e. the potential difference between the sheet and the distance $z$. This automatically means the potential is zero at the sheet i.e. $V(0)=0$. – John Rennie Mar 27 '18 at 16:53
• I think I got it. This whole thing of settting $V(\infty)=0$ or $V(0)=0)$ is only a matter of "Where do I want to start to measure?", and therefore it's pretty obvious this whole idea of the closer we get to the plate the closer to 0 is the potential, which is a consequence of the independence of the distance on the electric field of the infinite plate, my intuition is now beig challenged by the idea of an electric field independent of the distance from the souce, but that's another question, anyway, thank you! – Bidon Mar 27 '18 at 17:22
• @Duartejfs Yes, exactly :-) – John Rennie Mar 27 '18 at 17:30