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I read in theory that we cannot calculate at a particular point due to an infinitely long thin wire with uniform positive linear charge density. Instead we can only calculate the potential difference of two points. I know that electric field due to an infinitely long wire with linear density -- (+x) can be written as 2kx/r. So, if r is at infinity then the point will experience 0 force. And hence won't potential be 0 at that point?

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    $\begingroup$ Zero field doesn’t imply zero potential. Since you know the field, is something preventing you from using it to calculate the potential, at least up to a constant? $\endgroup$
    – Ghoster
    Nov 25, 2023 at 5:43
  • $\begingroup$ infinitely long wire with linear density -x Why are you using $-x$ to mean linear charge density? $x$ is usually a Cartesian coordinate. $\endgroup$
    – Ghoster
    Nov 25, 2023 at 5:49
  • $\begingroup$ it is not minus it's a hyphen $\endgroup$
    – john9
    Nov 25, 2023 at 7:19
  • $\begingroup$ I will use the EF to calculate force on the charge (if it is present) then I will Multiply that force with the displacement from that I will get work done which i will present with a -ve sign to get potential .....in the formula if r is zero force is zero hence work done is zero hence potential is zero. $\endgroup$
    – john9
    Nov 25, 2023 at 7:23

1 Answer 1

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We can definitely calculate the electric potential, as long as we choose a valid (in this case, non-infinity) reference point with respect to which we measure the potential.

The common choice in electrostatics with point charges is to take the potential $V(r)\to0$ at $r\to\infty$. This is not a valid choice in the case of an infinitely long line of charge. You know (and can prove using Gauss's law) that the electric field as a function of distance $r$ of an infinitely long wire with linear charge density $\lambda$ is $$\mathbf{E}=\frac{2k\lambda}{r}\hat{\mathbf{r}}.$$

The electric potential at $r$ with respect to a point infinitely far away will turn out to be infinite: $$V_{\infty}(r)=-\int_\infty^r\mathbf{E}(r')\cdot\mathrm{d}\mathbf{r}'=-\int_\infty^r\mathrm{d}r'\,\frac{2k\lambda}{r'}=-2k\lambda\,[\log r']_{\infty}^r.$$ The potential with respect to any non-infinite distance $a$, however, would lead to a real-valued result like $$V_a(r)=-2k\lambda\,[\log r']^r_a=-2k\lambda\log\frac{r}{a}.$$

Thus, even though the magnitude of the electric field approaches zero as $r\to\infty$, it doesn't approach 0 "fast enough" for its integral to be convergent to any finite value. One way to develop some intuition for this is to consider the fact that there's an infinitely large amount of charge in the distribution. The other infinitely large source of electric field that is often treated in introductory E&M courses, the infinite sheet of charge, has a constant electric field that doesn't approach $0$ at $r\to\infty$, but similarly requires that you define the potential with respect to some "nearby" point.

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  • $\begingroup$ Can you share a graph? E vs P. $\endgroup$
    – john9
    Nov 25, 2023 at 10:21
  • $\begingroup$ @john9 Sure, but what quantity does "P" represent in this case? $\endgroup$
    – Rishi
    Nov 25, 2023 at 17:54
  • $\begingroup$ Potential difference $\endgroup$
    – john9
    Dec 13, 2023 at 17:27

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