How do I calculate Volts per cm (V/cm) with the following given?

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SUMMARY

The calculation of Volts per centimeter (V/cm) for two coaxial tubes with a 20kV voltage difference involves understanding the geometry of the tubes rather than simply calculating area. The inner tube has an outer diameter (OD) of 0.5 inches and the outer tube has an OD of 1.00 inch, both with a length of 4 inches. The correct approach is to treat the coaxial tubes as a capacitor and use the formula for electric field strength, which is the voltage divided by the distance between the tubes. The distance is determined by the difference in their radii.

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  • Understanding of electric fields and capacitance
  • Familiarity with coaxial cable geometry
  • Knowledge of voltage and its relationship to electric field strength
  • Basic skills in unit conversion (inches to centimeters)
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  • Learn how to calculate capacitance for coaxial cylinders
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Homework Statement



How do I calculate Volts per cm (V/cm) with the following given?
Two coaxial tubes with 20kV placed across them.
Inner tube (actutally solid rod) has an OD of .5" and a length of 4"
Outer tube has an OD of 1.00" and a length of 4"

Homework Equations





The Attempt at a Solution


I was going to calculate the Area of the inner and outer tubes and convert them to cm, then divide the total area by the applied voltage? Somehow I don't think that is right?
 
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HMS-776 said:

Homework Statement



How do I calculate Volts per cm (V/cm) with the following given?
Two coaxial tubes with 20kV placed across them.
Inner tube (actutally solid rod) has an OD of .5" and a length of 4"
Outer tube has an OD of 1.00" and a length of 4"

Homework Equations





The Attempt at a Solution


I was going to calculate the Area of the inner and outer tubes and convert them to cm, then divide the total area by the applied voltage? Somehow I don't think that is right?

I'm not understanding the problem statement. Why would you have two coax cables with a net voltage between them. Usually the shields of each coax will be at ground potential, or somebody is about to lose their job...
 
They are just coaxial tubes (one inside the other) which form a capacitor. Sorry if I explained that wrong.
 

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