The cathode drop in an electric arc
- Creators
- Mackeown, S. S.
Abstract
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 5×10^(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 2×10^(-5) centimeters, so that the original assumption is justified.
Additional Information
©1929 The American Physical Society. Received 14 March 1929.
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Additional details
- Eprint ID
- 2545
- DOI
- 10.1103/PhysRev.34.611
- Resolver ID
- CaltechAUTHORS:MACKpr29.915
- Created
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2006-04-10Created from EPrint's datestamp field
- Updated
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2022-10-05Created from EPrint's last_modified field