# Curl of the $E$ field of a moving charge

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

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.

• Maxwell-Faraday equation gives the curl Commented May 21, 2023 at 19:25
• The Maxwell-Faraday equation applies to time-varying magnetic fields, which do not exist here. Commented May 21, 2023 at 19:36
• 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. Commented May 21, 2023 at 19:45
• I see. Thank you! Commented May 21, 2023 at 19:57

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 $$\nabla\times{\bf E}= \frac{3q({\bf r\times v)(r\cdot v)}} {\gamma^2[r^2-({\bf v\times r})^2]^{\frac{5}{2}}}$$.
• Thanks for you answer. What formula is this? Do the units of $r$ and $v\times r$ differ? Commented May 24, 2023 at 20:39
• Ok, I guess then $v \times r$ should be $\frac{v}{c} \times r$. Commented May 26, 2023 at 21:22