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In the figure below, direct current (DC) discharge occurs at and after some breakdown voltage, $V_{\text{breakdown}}\,,$ when the plasma pressure $P_{\text{plamsa}}$ is within a range observed to include $\left[0.01,\, 10\right]\,\mathrm{Torr}$, as electrons accelerate under the influence of potential difference to cause further electron emissions and recombinations.

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If there is a potential difference, this implies presence of electric field, implying presence of charge on electrodes.

Question: How is an electric field strong enough to cause emissive collisions of electrons produced when the elecrodes (assume wires in this case) have really low capacitance? The voltage across the electrodes is $2\, \mathrm{kV}$. And distance between electrons is $14\, \mathrm{cm}$.

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For a small capacitance a small charge suffices to create a large voltage, since $V=Q/C$.

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  • $\begingroup$ We know what the potential difference is, according to capacitance, charge will appear on the elecrodes. Which will result in an electric field. I am sorry I am not with providing specifics, but the electrodes are simple thin rods and at a distance of 14 cm (axially). I am not quantifying anything, but is the charge buildup enough, to produce enough electric field, to give enough energy to electrons to cause ionisations. According to my intuition capacitance is too low. But I don't think we can say anything unless I provide quantities. $\endgroup$
    – user539586
    Commented May 22, 2018 at 10:00
  • $\begingroup$ As you know V, the field is roughly V/d. The charge is V/C , precisely the amount to create the voltage difference as this is the function of the power supply. $\endgroup$
    – my2cts
    Commented May 22, 2018 at 13:13

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