Electric potential across a cross sectional cylinder

In summary, the problem involves finding the highest potential difference that can be maintained between a cylindrical high voltage terminal and a pressure vessel, given the radii of the terminal, intershield, and pressure vessel. Using Gauss's Law and the equation E=q/(2*pi*r*ε), the electric field at the surface of the pressure vessel is calculated to be 1.6*107 volts/m. By finding the charge density and electric field as a function of r, the potential difference between the shield and outer cylinders is determined. Similarly, the potential difference between the inner and shield cylinders is also calculated. The sum of these potential differences is the highest potential difference that can be maintained between the inner and outer cylinders.
  • #1
Londonfish
3
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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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  • #3
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?
 

FAQ: Electric potential across a cross sectional cylinder

What is electric potential?

Electric potential is the amount of electric potential energy per unit charge at a point in an electric field.

What is a cross sectional cylinder?

A cross sectional cylinder is a 2-dimensional representation of a cylinder, showing the shape of the cross section when the cylinder is cut perpendicular to its axis.

How is electric potential calculated across a cross sectional cylinder?

To calculate electric potential across a cross sectional cylinder, you need to know the charge distribution and the distance from the point of interest to the cylinder. Then, you can use the equation V = kQ/r, where V is the potential, k is the Coulomb's constant, Q is the charge, and r is the distance.

What factors affect the electric potential across a cross sectional cylinder?

The electric potential across a cross sectional cylinder is affected by the charge distribution, the distance from the point of interest to the cylinder, and the material properties of the cylinder.

How is electric potential related to electric field?

The electric field is the gradient of electric potential, meaning that it is the rate of change of potential with respect to distance. In other words, electric potential is the potential energy per unit charge at a point in an electric field, while electric field is the force per unit charge at a point in an electric field.

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