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I am just curious about the nature of an electric spark, how fast are the electrons moving?

Will these electrons slow down when the spark occurs through a dielectric? (e.g., air)

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There are no sparks/arcs in vacuum. Unless you get field emission, which requires very high charge densities i.e. either high voltages or very small structures on the emitting surface, there has to be an initially non-conducting medium that can be ionized in the field between the electrodes. Once that ionization occurs, there will be a cascade of electrons and another cascade of ions moving towards the electrodes. These cascades can settle into an equilibrium if the current trough the medium is limited. The diffusion velocity depends on the species of the charged particles, the gas mixture and other conditions. In the general case this is an enormously complex phenomenon. See e.g. http://doeplasma.eecs.umich.edu/files/PSC_Kolobov4.pdf for a discussion of some of the phenomenology and modern transport theory in these discharges. http://www.dspace.univer.kharkov.ua/bitstream/123456789/1731/2/J_Phys_D_99_Vdr_SF6_CF4.pdf has some measurements of electron drift velocity that can show you how this parameter scales experimentally with gas pressure. http://books.google.com/books?id=Xy8d-OzlDWUC&pg=PA18&lpg=PA18&dq=spark+electron+diffusion+velocity&source=bl&ots=cPK7vLt7L-&sig=qwLpIIGO86ppP-Fwet6kYWKGADI&hl=en&sa=X&ei=ePsqVM6SKIbwiwL_zoGoDw&ved=0CCYQ6AEwAQ#v=onepage&q=spark%20electron%20diffusion%20velocity&f=false gives some formulas for the drift velocity under idealized assumptions in some of these regions.

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The speed of the electron in a vacuum (or close vacuum) depends on the voltage applied to them. The higher the tension, the quicker they are (speed proportional to the square root of the voltage applied).

Dielectric has nothing to do with electrons, but everything to do with light... On top of that, vacuum is a dielectric...

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  • $\begingroup$ If you add a dielectric between two capacitor plates, the capacitor charge is influenced - this has to do with electrons... $\endgroup$ – Steeven Nov 2 '14 at 1:49

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