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A rather basic physics question. How does the circuit below behave, if the sphere is solid copper, and proportionally much larger than the input and output conductor diameters? The wires are connected to an AC source and ground. I assume the inverse square law would apply, since the wave spreads out in the spherical conductor. Will there be a voltage loss for the signal proportional to it?

enter image description here

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    $\begingroup$ It is just a conductor, with some resistance (and capacitance). Why would an inverse square law be of much use? $\endgroup$
    – Jon Custer
    Jan 17, 2022 at 19:34
  • $\begingroup$ I was thinking since the current must be spreading out. Normally when analyzing electromagnetic radiation, it loses amplitude because it spreads out, and same for any other wave form like water ripples. And even though electric current is not the same as electromagnetic radiation waveforms, it still spreads out in the conductor if it is a sphere or cone etc. $\endgroup$
    – user53289
    Jan 17, 2022 at 21:17

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The inverse square law applies when the sphere's outer surface is the important surface for the consideration. Here the important thing is the surface at any given distance that the current passes through. So, in the picture, the total current passing each black line will be the same. Each black line cuts the sphere making a circular disk. The areas of these will not be proportional to the square of distance from either wire. This will be the case no matter where on the sphere the connections attach.

There are also a lot of very interesting effects. For example the skin effect can be important for AC current.

enter image description here

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  • $\begingroup$ Thanks. If I understand right, the current does spread out (per inverse square law?) but then it converges again, so the sum current is the same. Does the diameter of the sphere (the area of the sections in your image) also contributed to reduced resistance? $\endgroup$
    – user53289
    Jan 17, 2022 at 21:11

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