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A parallel plate electrode system is separated by a distance of 0.6 cm. When a voltage of 11 kV is applied 500 A steady current flows between the electrodes. At constant E field conditions, the distance is reduced to 0.12cm. The new current value is 50 A. Calculate the Townsend ionization coefficient * a =19.1 ionizing collisions / cm a =4.79 ionizing collisions. cm a =4.79 ionizing collisions / cm a =3.8 ionizing collisions. cm

A parallel plate electrode system is separated by a distance of 0.6 cm. When a voltage of 11 kV is applied 500 A steady current flows between the electrodes. At constant E field conditions, the distance is reduced to 0.12cm. The new current value is 50 A. Calculate the Townsend ionization coefficient * a =19.1 ionizing collisions / cm a =4.79 ionizing collisions. cm a =4.79 ionizing collisions / cm a =3.8 ionizing collisions. cm

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A parallel plate electrode system is separated by a distance of O.6 cm. When a voltage of 11 kV is applied 500 A steady current flows between the electrodes. At constant E field conditions, the distance is reduced to 0.12cm. The new current value is 50 A. Calculate the Townsend ionization coefficient * a =19.1 ionizing collisions / cm a =4.79 ionizing collisions. cm a =4.79 ionizing collisions / cm a =3.8 ionizing collisions. cm
Transcribed Image Text:A parallel plate electrode system is separated by a distance of O.6 cm. When a voltage of 11 kV is applied 500 A steady current flows between the electrodes. At constant E field conditions, the distance is reduced to 0.12cm. The new current value is 50 A. Calculate the Townsend ionization coefficient * a =19.1 ionizing collisions / cm a =4.79 ionizing collisions. cm a =4.79 ionizing collisions / cm a =3.8 ionizing collisions. cm
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