# Why does the electric potential for an infinite line charge seem to have to be worked out from negative infinity rather than positive infinity?

The electric field of an infinite line charge in the plane perpendicular to the line charge can be given as:

$$E=\frac{1}{2\pi \epsilon r}$$

Where $r$ is the perpendicular distance from the line.

So assuming my integration is correct, the integral of this expression is calculated to give the potential.

$$V=\frac{1}{2\pi \epsilon} \log_e(r) +C$$

If we now set the limits for this integral as $-\infty$ and $r$ we can calculate the logarithm of $-\infty$ as 0 and then just say:

$$V_r=\frac{1}{2\pi \epsilon} \log_e(r) - \frac{1}{2\pi \epsilon}\log_e(-\infty)$$ $$V_r=\frac{1}{2\pi \epsilon} \log_e(r)$$

But what about if we wanted to do the integral from positive infinity (which should be equally valid). You can see that a problem arises with an expression:

$$V_r=\frac{1}{2\pi \epsilon} \log_e(r) - \frac{1}{2\pi \epsilon}\log_e(\infty)$$

Which is not finite. Why is there this discrepancy?

If you want to compute the potential energy, you have to set the zero somewhere else: you cannot choose $V(\pm\infty)=0$ because $V(r)$ doesn't fall off; rather, it increases with distance. The reason is that $V(r)$ is an extensive magnitude, and as such it scales with the volume of the system (cf. the link in the comment section).
• OK, I think I understand. We can define potential at infinity to be 0 for a point charge because the integral of $1/x^2$ is finite, but we can't for $1/x$ which is not finite (harmonic series). Is that correct? So how do we reconcile this with the definition of potential which is the energy per unit charge required to move a positive test charge to a finite point from infinity? – Requiens Feb 12 '17 at 18:14
• @Resquiens 1) yes. 2) you cant reconcile the pictures because an infinite line of charge is an unphysical object. They do not exist in reality. Its no surprise that a mathematical model that includes infinite objects fails at some point. In any case, recall that a potential energy is always defined modulo a constant term, and therefore the position of the zero of $V(r)$ is irrelevant. – AccidentalFourierTransform Feb 12 '17 at 18:16
• @Resquiens 1) when you are given the expression of the potential energy of some object, you can always add or subtract an arbitrary constant, and this doesn't affect the physics. 2) recall that $\int\frac{1}{x}=\log|x|$, so there are no negative values of $x$ inside the logarithm. – AccidentalFourierTransform Feb 12 '17 at 18:36