Flux of an infinite line of charge through a cylinder

In summary, the problem involves finding the total flux passing through a cylindrical surface of height h = 0.116 m, with two charge densities of 0.00029 C/m and -0.00087 C/m. Using Gauss' law, the net flux can be calculated by multiplying each charge density by the height of the cylinder and then adding them together, and finally dividing by the permittivity of free space constant. However, this approach may not be correct and further clarification or assistance may be needed.
  • #1
kindofsour
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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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  • #2
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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  • #3
This problem is killing me. I just don't see how my approach could be incorrect. Anyone?
 

FAQ: Flux of an infinite line of charge through a cylinder

What is the flux of an infinite line of charge through a cylinder?

The flux of an infinite line of charge through a cylinder is the measure of electric field passing through the surface of the cylinder. It is represented by the symbol Φ.

How is the flux of an infinite line of charge through a cylinder calculated?

The flux through a cylinder can be calculated by using the formula Φ = ∫ E ⋅ dA, where E is the electric field and dA is the area element of the cylinder's surface.

Is the flux of an infinite line of charge through a cylinder affected by the size of the cylinder?

Yes, the flux through a cylinder is directly proportional to the size of the cylinder. As the size of the cylinder increases, the surface area also increases, resulting in a higher flux value.

What happens to the flux of an infinite line of charge through a cylinder if the cylinder is parallel to the line of charge?

If the cylinder is parallel to the line of charge, the electric field passing through the cylinder's surface will be zero. Therefore, the flux through the cylinder will also be zero.

Can the flux of an infinite line of charge through a cylinder be negative?

Yes, the flux through a cylinder is a scalar quantity and can be positive, negative, or zero. A negative flux value indicates that the electric field and the area vector are in opposite directions.

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