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Let us say we have the following symmetrical apparatus:

enter image description here

Four equal positive charges, all connected to a shaft that can rotate, the connecting rods are insulated, and so does the shaft.

Now suppose that a positive charge, less in magnitude than the four previous ones, pops out in the indicated location below:

enter image description here

Is there any way to tell whether the apparatus would spin without brute forcing the problem with Newtonian mechanics?

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  • $\begingroup$ Why do you describe the Newtonian mechanics as 'brute forcing'? There's only one parameter in the problem (the angle the giant curl-meter makes), it should be pretty easy to say, write down the energy as a function of that angle and find its minimum. $\endgroup$ – jacob1729 Aug 7 at 15:04
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In the example you drew, there will obviously be a clockwise torque, as you indicated with the arrow. You can make the extra charge as close as you like to one of the arms of the curl-meter, so that force can be as large as desired.

The curl-meter is supposed to mock up the curl operator, which is a local thing. Derivatives are local. That means that you need to imagine taking the limit as the curl-meter's size approaches zero.

The OP asks in a comment:

The confusing thing is since the curl of an E-field is null, then so must be its circulation around any loop, so the work done around the circular path by the force generated by the small charge must be zero, and so the curlimter shoud not turn at all, or is it that it rotates clockwise and then gets back to its original position, i.e., it oscillates?

If you rotate the large meter (which isn't really a curl-meter because it's not infinitesimal in size), then the work done is $\int_{\theta_1}^{\theta_2} \tau d\theta$. This work will be zero if you integrate over one full rotation, because this field is conservative. It doesn't matter whether you consider free rotation or not. The work done by the electric field is zero, for example, if you rotate the meter by hand using a crank. The fact that the work integrates to zero over one cycle does not imply that $\tau=0$ for a particular value of $\theta$.

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  • $\begingroup$ The confusing thing is since the curl of an $E$-field is null, then so must be its circulation around any loop, so the work done around the circular path by the force generated by the small charge must be zero, and so the curlimter shoud not turn at all, or is it that it rotates clockwise and then gets back to its original position, i.e., it oscillates? $\endgroup$ – Hilbert Aug 7 at 12:45
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    $\begingroup$ As Ben Crowell says, the curl is a local notion. If your curl-meter shrinks to a size small enough, it will not turn because the field will appear differently (more "homogeneous"). The reason it turns it your picture is because the strong difference in the fields value (due to the small charge) and orientation at the 4 points of your curl-meter. This would not be the case is it is very small. If not done yet, take a look at en.wikipedia.org/wiki/Curl_(mathematics)#Definition $\endgroup$ – David Aug 7 at 14:14

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