Gauss' Law With an Infinite Cylinder of Charge

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SUMMARY

The discussion focuses on applying Gauss' Law to determine the potential difference between a point S above an infinite cylinder of charge with linear charge density λ and a point on the surface of the cylinder with radius R. The solution involves integrating the electric field E over the surface area of the cylinder, leading to the expression E(s) * 2π * L * S. The confusion arose regarding the variable S, which represents the distance from the point S to the cylinder, and the integration process over the angle φ and the length of the cylinder. The key takeaway is understanding the relationship between the electric field, surface area, and the geometry of the cylinder.

PREREQUISITES
  • Understanding of Gauss' Law
  • Familiarity with electric fields and potential differences
  • Knowledge of integration techniques in physics
  • Basic concepts of cylindrical symmetry in electrostatics
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  • Study the derivation of Gauss' Law in cylindrical coordinates
  • Learn about electric field calculations for different charge distributions
  • Explore the concept of electric potential and its relationship to electric fields
  • Review integration techniques specifically for surface area calculations in electrostatics
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[SOLVED] Gauss' Law With an Infinite Cylinder of Charge

Homework Statement



The test question was to find the potential difference between a point S above a cylinder of charger per length lambda, and a point on the surface of the cylinder having radius R and infinite length.

Homework Equations



Stupid me tried to grind through it during the test. But looking at the prof's solution:

http://www.phys.washington.edu/users/schick/321A/321-07exam1soln.pdf

Here's the test if you need to know the question how he worded it:

http://www.phys.washington.edu/users/schick/321A/321-07exam1.pdf

Problem #1.

I don't get how he got "E(s)*2pi*L*S

I get the 2pi and L come from integrating over phi and the length of the cylinder, but I can't figure out where that S came from.Is it because you'd need a cosine(theta) for the direction that the cylinder is from S? Where cosine is S/distance? Then wouldn't he have to add in a bunch of other stuff? I'm just not seeing something here and I feel embarrassed.
 
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[tex]\int E \cdot dA = EAcos(\Theta)[/tex]

Where [tex]\Theta[/tex] is the angle between the surface of the cylinder and the electric field.

Now, simplify this last expression. What is A? What is [tex]\Theta[/tex]?
 
Ohhhhhh I get it. A is 2pi*Length*radius, and he's just using S as the radius, since it shouldn't matter.

Gotcha, thanks a bunch.
 

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