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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?

Qmechanic
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Resquiens
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1 Answers1

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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).

AccidentalFourierTransform
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