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I'm having a bit of trouble understanding this: In class we were taught that $$ V(\vec b)-V(\vec a)=-\int_\vec a^\vec b\vec E\cdot d\vec{r} $$ And we were told that when the charge over which we're integrating is finite (i.e, a bounded set), we can take $V($"$\infty$"$)=0$, but that when charges are infinite (unbounded, I suppose), we couldn't.

There are a few things that confuse me: The first is, why can't we set $V=0$ at any point we want? We could do it in the case of gravity, for example. Why not for potential?

The second is, why does having an unbounded charge imply we shouldn't take $V$ to be $0$ at infinity?

Furthermore, given an unbounded charge (an infinitely long charged line, in the $x$-axis), where can we place $V_\vec p=0$?

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You can set $V=0$ wherever you want. All you have to do is try to find a potential funtion such that

  • It satisfies given boundary conditions.

  • Its gradient is the electric field as you know it.

Sure, you can set V to be zero at $\infty$, and if you want, even $\infty$ at $0$

As long as you can find a funtion that gives you those values at those points, and whose gradient is $\frac{\lambda}{2\pi \epsilon_0 r}$.

As always, what is more important is the difference in potential. There is no physical law that says that the potential (or even the force) at $\infty$ must be zero.

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Why can't we set $V = 0$ at an arbitrary point like we did with gravity?

You may indeed set $V = 0$ wherever you like, no matter what your charge distribution looks like, be it finite or infinite. However, it is convenient to set $V=0$ at $\infty$ for finite charge distributions. For finite charge distributions, and for $r$ far from the source, $E\sim r^{-2},$ and as $r\to\infty$, $E\to 0$.

Why does having an unbounded charge imply we shouldn't take $V$ to be $0$ at infinity?

Suppose our charge distribution is infinite in extent. Then, the electric field for this body is defined everywhere, just like the case of a finite distribution, but the electric field does not go to $0$ "fast enough". The electric field in these cases generally goes as $E\sim r^{-1}$, and we get divergent integrals. If we try to compute $V$ using the definition of potential with $\infty$ as our reference point, we find $$ V(p) = -\int_\infty^p E(r)\cdot dr\sim-\int_\infty^p r^{-1}\,dr = -\log p + \lim_{r\to\infty}\log r $$ which we can see diverges. Therefore, it is a good strategy to choose our reference point to be at some finite distance from our charge distribution to avoid the divergence of our potential. In practice there is no such thing as an infinite distribution of charge or a point at infinity, and all that is really important to us is differences in potential between two points.

The fact that the potential function is unique up to a constant means that we can only have meaningful physical results by interpreting potential differences. If it were otherwise, and the constant mattered, we would have an infinite number of meaningful physical results to interpret (an intractable task).

Given an unbounded charge (an infinitely long charged line in the $x$-axis), where can we place $V = 0$?

This goes back to your first question. The answer is that we may set $V = 0$ anywhere we want. This a consequence of the fact that we only consider potential differences as being physically meaningful. So, set $V = 0$ at whichever point $p$ is convenient for you. Perhaps there is a certain symmetry to the problem that calls for $V$ to be $0$ at some particular point. Regardless, keep in mind that we only ever bother with $\Delta V$, so our choice of zero potential is impertinent.

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Let me give you an example, why you cannot always choose V=0 for r = infinity. Consider a infinitely large sheet of charge. Notice that if the shrrt in the x-y plane, r tending to infinity may mean x tending to infinity at a certain z and y, or z tending to infinity at a certain x and y. However, even as x tends to infinity, the potential of the sheet remains unchanged, since it extends upto infinity in the x direction, while in the z direction, it goes to zero as we go to infinity. If you choose the sheet to be at potential 0, then at z tending to infinity, the potential will not be zero. A diagram should help you understand.

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