Flux of an infinite line of charge through a cylinder

AI Thread Summary
The discussion centers on calculating the total electric flux through a cylindrical surface due to two infinite line charges with different charge densities. The user attempts to apply Gauss' law but struggles with the non-uniform electric field and the orientation of the field relative to the surface area. They calculate the total charge contained within the cylinder by multiplying the charge densities by the height of the cylinder, aiming to find the net flux using the formula Flux = (Qcontained/ε₀). Despite their calculations, they express confusion over why their approach appears incorrect, indicating a need for clarification on the application of Gauss' law in this scenario. The thread highlights the complexities involved in applying theoretical concepts to practical problems in electromagnetism.
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Homework Statement



http://smartphysics.com/Content/Media/Images/EM/03/h3_lineD.png

charge density 1 = .00029 C/m
charge density 2 = -.00087 C/m
h = .116 m
a = .094 m

a) What is the total flux Φ that now passes through the cylindrical surface of height h=.116 m? Enter a positive number if the net flux leaves the cylinder and a negative number if the net flux enters the cylinder.

b) The initial infinite line of charge is now moved so that it is parallel to the y-axis at x = -4.7cm (i.e. the two lines are equidistant from the center of the cylinder). What is the new value for Ex(P), the x-component of the electric field at point P?

c) What is the total flux Φ that now passes through the cylindrical surface? Enter a positive number if the net flux leaves the cylinder and a negative number if the net flux enters the cylinder.

Homework Equations



gauss' law

The Attempt at a Solution



a) How can we use Gauss' law? The magnitude of the field isn't uniform so it doesn't simplify.

b) This is easy enough, included as context for the next part.

c) The net flux should be the same as in part a) since the total charge flowing through the surface is the same.
 
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I think I made some headway. For the first question the electric field is not constant over the suface nor is it everywhere perpendicular to dA so we can't use the flux = ∫E*DA half of Gauss' law. We can use the right side, i.e. flux = Qcontained/eo. Multiply each charge density by the height of the cylinder to get the total charge through each piece of rod contained by the cylinder. Then we can add them to get the net charge through the cylinder and lastly divide by the permittivity of free space constant to get the net flux, which will be negative.

Charge density1 = +.00029 C/m

Charge density2 = -.00087 C/m.

Q1 = (.00029)*(.116)
Q2 = (-.00087)*(.116)

Flux = (Q1+Q2)/eo

But apparently this is incorrect.
 
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This problem is killing me. I just don't see how my approach could be incorrect. Anyone?
 
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