Electromagnetic Coupling Current passes from $L_2$ to $L_3$ which have no physical contact (except air) between them. Van de Graaff Generator The current still transported through air but unlike previous picture the process is obviously different. Now I know previous image is representing electromagnetic induction and later one electrostatic discharge. But apart from the definition what exactly is happening here internally causing such difference? I am somehow familiar with Maxwell's equations I can handle a little technical answer. Thanks in advance.

  • $\begingroup$ In the first case, $no$ current passes between the left hand and right hand circuits. $\endgroup$ Commented Jan 13, 2018 at 10:06
  • $\begingroup$ One anonymous flagger writes: This question is badly formatted , the images are non in imgur, and the link of the first one gives "The owner of www.rroij.com has configured their website improperly. To protect your information from being stolen, Firefox has not connected to this website." in firefox, so I cannot repair the question by editing. $\endgroup$
    – Qmechanic
    Commented Feb 27, 2019 at 5:50

1 Answer 1


The first diagram represents induction, not conduction. The circuit on the left is an LCR oscillator, with $L_2$ and $L_3$ being inductors (basically just loops of wire). Now, as the current through $L_2$ varies, so does the magnetic field around it. This varying magnetic field induces an alternating current in $L_3$ (as stated by Faraday's Law of Induction), which in turn drives the circuit on the left, without direct contact.

In the second picture, current passes directly through air through a phenomenon is called dielectric breakdown. Normally air is an insulator, but given a high enough electric field (about $2MV/m$ for air), gas molecules get ionised and start conducting. Electrons in the conducting gas constantly leave and recombine with gas molecules, leading to the emission of light. Hence, the path of the current appears as a blue arc.


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