Electric potential across a cross sectional cylinder

AI Thread Summary
The discussion revolves around calculating the maximum potential difference between the inner cylinder of a van de Graaff generator and the outer pressure vessel, given the breakdown electric field of 1.6*10^7 volts/m. The user attempts to apply Gauss's Law and integrate the electric field to find the potential difference, initially calculating a charge of 0.001335 C for the pressure vessel. They express uncertainty about whether this charge applies to the entire cylinder or just the pressure vessel and seek clarification on the integration process. Another participant suggests treating the problem in two parts: first calculating the potential difference between the shield and the outer cylinder, then between the inner cylinder and the shield, ultimately summing these differences for the total potential. The conversation highlights the complexities of electric fields and potential calculations in cylindrical geometries.
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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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