Solving for Flux Using Enclosed Charge

  • #1

Homework Statement



https://www.smartphysics.com/Content/Media/Images/EM/03/h3_lineD.png [Broken]

Where, λ1 = -3.2 μC/cm, λ2 = 9.6 μC/cm, a = 8.5 cm and h = 11.4 cm. a/2 = 4.25 cm.

What is the total flux that passes through the surface of the cylinder?

Homework Equations



E = 2kλ/r
[itex]\Phi[/itex] = E*A = qenclosed/[itex]\epsilon[/itex]o

The Attempt at a Solution



The first thing I did was solve for the total electric field of the two line charges, which is 338823529.411765 N/C. After this, the next step would be to simply multiply by the surface area of the cylinder. However, this doesn't work because the electric field is not constant throughout the surface of the cylinder.

My next step was to try and solve it using the enclosed charge, but since I have only been given the charge densities of each line, I have no clue how to figure out enclosed charge. Is there a specific method as to how you can do this?

Any help is greatly appreciated.
 
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Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,909

Homework Statement



https://www.smartphysics.com/Content/Media/Images/EM/03/h3_lineD.png [Broken]

Where, λ1 = -3.2 μC/cm, λ2 = 9.6 μC/cm, a = 8.5 cm and h = 11.4 cm. a/2 = 4.25 cm.

What is the total flux that passes through the surface of the cylinder?

Homework Equations



E = 2kλ/r
[itex]\Phi[/itex] = E*A = qenclosed/[itex]\epsilon[/itex]o

The Attempt at a Solution



The first thing I did was solve for the total electric field of the two line charges, which is 338823529.411765 N/C. After this, the next step would be to simply multiply by the surface area of the cylinder. However, this doesn't work because the electric field is not constant throughout the surface of the cylinder.

My next step was to try and solve it using the enclosed charge, but since I have only been given the charge densities of each line, I have no clue how to figure out enclosed charge. Is there a specific method as to how you can do this?
Any help is greatly appreciated.
As you have noted, the electric field is not constant on the surface, so it is not AE. But the total flux is always equal to the enclosed charge divided by ε0: you need to calculate that only from the λ-s and length h.
 
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  • #3
As you have noted, the electric field is not constant on the surface, so it is not AE. But the total flux is always equal to the enclosed charge divided by ε0: you need to calculate that only from the λ-s and length h.
Alright, thank you. :) I think I got it now.

I do have one more question though. Is the method used to find the enclosed charge different depending upon what type of field you are dealing with?
 
  • #4
ehild
Homework Helper
15,543
1,909
I do not get you. It is valid for the relation between electric flux and enclosed charge.

ehild
 

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