The Cathode Drop in an Electric Arc

By making the assumption that the total cathode drop occurs in a distance less than one mean free path from the cathode, Poisson's equation can be solved. If the experimental values of 4000 amperes per square centimeter, and 10 volts are used for the current density at the cathode, and the cathode drop in a mercury arc, values for the electric field existing at the surface of the cathode can be determined for varying percentages of the current carried by positive ions. If 5% of the current at the cathode is carried by positive ions, the field existing at the surface of the cathode exceeds 5\ifmmode\times\else\texttimes\fi{}${10}^{5}$ volts per cm. This is probably sufficient to furnish the necessary electron current by "field" currents produced by this high field. The whole cathode drop occurs within a distance of approximately 2\ifmmode\times\else\texttimes\fi{}${10}^{\ensuremath{-}5}$ centimeters, so that the original assumption is justified.

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