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Figure 5.15 from Purcell: "Electricity & Magnetism" (3rd edition) shows the electric field of a uniformly moving charge.

Figure of field of a uniformly moving charge.

It reads:

The field in Fig. 5.15 is a field that no stationary charge distribution, whatever its form, could produce. For in this field the line integral of $E$ is in general not zero around a closed path. Consider, for example, the closed path ABCD in Fig. 5.15. The circular arcs contribute nothing to the line integral, being perpendicular to the field; on the radial sections, the field is stronger along BC than along DA, so the circulation of $E$ on this path is not zero.

In other words, this is a non conservative field. According to Helmholtz's theorem a non conservative field has a solenoidal field component, i.e. a curl. I wonder how the curl and the curl-free (irrotational) components of this field look like graphically and algebraically.

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    $\begingroup$ Maxwell-Faraday equation gives the curl $\endgroup$
    – Ghoster
    Commented May 21, 2023 at 19:25
  • $\begingroup$ The Maxwell-Faraday equation applies to time-varying magnetic fields, which do not exist here. $\endgroup$ Commented May 21, 2023 at 19:36
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    $\begingroup$ A moving point charge has a time-varying magnetic field. You are incorrect. The Maxwell-Faraday equation applies always, as do all of Maxwell’s equations. $\endgroup$
    – Ghoster
    Commented May 21, 2023 at 19:45
  • $\begingroup$ I see. Thank you! $\endgroup$ Commented May 21, 2023 at 19:57

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The electric field is given (in Gaussian units) by ${\bf E} = \frac{q{\bf r}} {\gamma^2[r^2-({\bf v\times r})^2]^{\frac{3}{2}}}$. Just take it's curl. The result is $\nabla\times{\bf E}= \frac{3q({\bf r\times v)(r\cdot v)}} {\gamma^2[r^2-({\bf v\times r})^2]^{\frac{5}{2}}}$.

Check my algebra.

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  • $\begingroup$ Thanks for you answer. What formula is this? Do the units of $r$ and $v\times r$ differ? $\endgroup$ Commented May 24, 2023 at 20:39
  • $\begingroup$ The first equation is Eq. {11.154) and Jackson's textbook or (15.51) in my textbook. The units are the usual units, r in cm and v in cm/sec. $\endgroup$ Commented May 26, 2023 at 1:28
  • $\begingroup$ Ok, I guess then $v \times r$ should be $\frac{v}{c} \times r$. $\endgroup$ Commented May 26, 2023 at 21:22
  • $\begingroup$ I am so used to setting c=1 that I left it out there. $\endgroup$ Commented Jun 4, 2023 at 14:30

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