I have been unable to find a definitive answer to my questions after some extensive browsing online. My question involves the coupling between two adjacent conductors (wires, PCB traces, etc) due to the electric field between the two. Now, from some basic tests on an oscilloscope and function generator I can clearly see that an AC waveform radiated from one antenna wire is picked up an appears as noise on the adjacent wire. The noise level is such that the amplitude increases in proportion to the proximity to the antenna.

It was my understanding that this coupling would occur also with DC current as the "antenna" wire would emit a static electric field which would produce a "static equilibrium" in the receiver wire (with an albeit small current).

I guess the my fundamental question concerns why time varying E field couples voltage in nearby conductors by a static field does not. Am I right in my assumption that the charges within the receiver conductors will redistribute (similar to the Skin Effect) in that charges will orient themselves to oppose the incoming field in a static E field? Apologies for the simplistic nature of this question, however many online resources only provide an analogous explanation to that of capacitors.

If someone could provide a better physical explanation in regards to the results I am experiencing, it would be much appreciated.

  • \$\begingroup\$ This is more commonly called crosstalk: en.wikipedia.org/wiki/Crosstalk \$\endgroup\$ Commented Apr 12, 2016 at 10:13
  • \$\begingroup\$ Crosstalk is a common phenomenon in land line phones. For that reason, the installers are supposed to twist the small phone wires that goes into the residence. This action greatly eliminates the crosstalk. Not all techs do it however. \$\endgroup\$ Commented Apr 12, 2016 at 14:06

1 Answer 1


A static electric field has no ability to deliver any power or energy to something that might be used to measure its presence. Just like a capacitor carries no DC current when a constant DC voltage is applied.

However, static electric fields can be measured by using an electric field mill: -


Click on picture for website information.

This performs commutation of the field to produce enough current to drive a meter but, energy is removed from the field and as such the field is somewhat disrupted and there is of course a fundamental error involved with this type of measurement.

In terms of the "mill" an alternating electric field autocommutates and therefore power can be extracted to drive a meter and make a measurement.

Basically, this means that static crosstalk to other "wires" is of no consequence because no current can be involved and natural impedances present on wires to ground force the crosstalk effects to be zero. Not so with alternating fields because a capacitor conducts current proportional to the rate of change of voltage between its terminals: -

Q = CV and dQ/dt = C dV/dt = current.

  • \$\begingroup\$ Thank you Andy for explanation, especially the field mill. However, I am still having some difficulty with the intuition behind why a static field will no displace some charge and result in a current. For instance, in relation to capacitors current, the cap will only prevent current flow when Vc = Vs. However I am having some trouble relating this to a uniform field in space. Shouldn't the static E field from the antenna displace some charges in the receiver? \$\endgroup\$
    – wubzorz
    Commented Apr 12, 2016 at 10:29
  • \$\begingroup\$ It will certainly displace charges but once displaced the charges can only move (and therefore be current due to I = dQ/dT) when the voltage changes. Fixed displaced charge does not translate to current once the "moving" is done. \$\endgroup\$
    – Andy aka
    Commented Apr 12, 2016 at 10:52
  • \$\begingroup\$ So regardless of whether the field is orthogonal (or any orientation for that matter) to the receiving conductor, those charges will only move to establish an equilibrium upon being initially subject to the E field. After that the charges are stationary, have I got that right? \$\endgroup\$
    – wubzorz
    Commented Apr 12, 2016 at 12:07
  • \$\begingroup\$ yes, the charges will become stationary after an initial settling period. \$\endgroup\$
    – Andy aka
    Commented Apr 12, 2016 at 12:10
  • \$\begingroup\$ Awesome! Appreciate your help Andy, has corrected my understanding of several concepts now. \$\endgroup\$
    – wubzorz
    Commented Apr 12, 2016 at 12:12

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