Electric potential across a cross sectional cylinder

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SUMMARY

The discussion focuses on calculating the highest potential difference between the inner cylinder and the outer pressure vessel of a van de Graaff generator, given specific dimensions and an electric field threshold of 1.6 x 107 volts/m. Using Gauss's Law, the potential difference is derived through integration, resulting in a calculated value of approximately 11.28 megavolts. The conversation also explores the relationship between charge density and electric field strength, emphasizing the need to consider both the inner and outer cylinders to determine the total potential difference accurately.

PREREQUISITES
  • Understanding of Gauss's Law and its application in electrostatics
  • Familiarity with electric field calculations in cylindrical geometries
  • Knowledge of potential energy and voltage concepts in physics
  • Basic calculus skills for integration and logarithmic functions
NEXT STEPS
  • Study the derivation of electric fields in cylindrical coordinates
  • Learn about charge density and its impact on electric fields
  • Explore advanced applications of Gauss's Law in complex geometries
  • Investigate potential difference calculations in multi-layered cylindrical systems
USEFUL FOR

Physics students, electrical engineers, and anyone involved in high-voltage systems or electrostatics will benefit from this discussion, particularly those studying van de Graaff generators and electric field interactions in cylindrical structures.

Londonfish
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Homework Statement


A cross-section of a cylindrical high voltage terminal (inner cylinder) of a van de Graaff generator, surrounded by an 'intershield' (middle cylinder) and a pressure vessel (outer cylinder). The gas in the pressure vessel breaks down in electric fields greater than 1.6*107volts/m. If the radii of the terminal, intershield and pressure vessal are 1.5m, 2.5m, 4m respectively, what is the highest potential difference that can be maintained between the terminal and the pressure vessel? (hint: the intershield must be maintained at a potential such that breakdown is about to occur on its own outer surface as well on the surface of the terminal.)


Homework Equations



rterminal=1.5m
rintershield=2.5m
rpressure vessel=4m
Emax, p-vessel= 1.6*107volts/m

∫E*ds= q/ε (Gauss's Law)
I believe E= q/ (2*pi*r*ε);where r= radius

V(rA) - V(rB) = -∫ q/ (2*pi*r*ε) dr ; V= potential energy, integrate from A to B


The Attempt at a Solution



Emax,p-vessel= q/ (2*pi*r*ε)
q=Emax*2*pi*ε*(4-2.5)
q=0.001335 C.
Would this q work for the entire cylinder if I changed the radius or is it just for the pressure vessel because that's where the electric field is?

Then,
V(rA) - V(rB) = -∫ q/ (2*pi*r*ε) dr
= -q/ (2*pi*ε)∫1/r dr
= -q/ (2*pi*ε)(ln(B)-ln)A) ; Where A,B would be the radius
plugging in q and using ri=2.5m and rpv=4m
= (-2.4*107)*(-0.47)
= 1.128*107 volts or 11.28 megavolts

I don't really know where to go from here, or even if this step is true. Is there any easier way I could be doing this?
 
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Are you treating this as two problems, first ignore the inner cylinder. You know the maximum electric field at the surface of the shield cylinder. As the electric field goes as 1/r you can work backwards and determine the charge density and thus the electric field as a function of r, with that you can determine the potential difference between the shield and the outer cylinder.

Now you do the same for the inner cylinder, you will get a potential difference between the inner and shield cylinders? The sum of those potential differences is the potential difference between the inner and outer cylinders?
 

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