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E field due to a finite line of charge

  1. Feb 4, 2016 #1
    1. The problem statement, all variables and given/known data

    A finite line of uniform charge is on the line joining (4,0,0) (0,8,0) and has λ= 2μC/m find E at (0,0,0)

    2. Relevant equations

    E = Q / (4 π ε0 r^2)

    dQ = λ dl = λ dx

    3. The attempt at a solution

    I first notice that r will depend on 2 changing values x and y.

    I write an equation for the line.
    y = 8 - 2x

    I use this relation to find r in terms of x
    r = sqrt ( x^2 + (8 - 2x)^2 )

    I set up my integral
    λ / (4 π ε0 r^2) ∫ < -x , 2x - 8> dx / [ x^2 + (8 - 2x)^2]^(3/2) from x=0 to x=4

    when I evaluate I get <-2250 , -2250>

    This does not seem correct to me. Based on the location of the line to the origin I would expect my x component to have a larger magnitude than the y component. I am not sure where I am going wrong.
     
  2. jcsd
  3. Feb 4, 2016 #2

    gneill

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    Staff: Mentor

    I think perhaps you need to think about what dq is in this scenario. Since the line of charge is on an angle, from the point of view of the origin dq will not be simply λ dx.
     
  4. Feb 4, 2016 #3
    My thought was that since I expressed everything in terms of x that dx would resolve the need for dy. Apparently this is not the case. I think then dq =λ dy dx.

    This still does not seem correct it would turn in to ∫ ∫ dy dx which feels very wrong. I am missing some understanding. Can you nudge me?
     
  5. Feb 4, 2016 #4

    gneill

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    Staff: Mentor

    Consider this figure:
    upload_2016-2-4_12-14-15.png (revised image)

    Write an expression for dq in terms of dx and dy. What's dy in terms of dx for your line? (think slope).
     
    Last edited: Feb 4, 2016
  6. Feb 4, 2016 #5
    Ok, I see dq = dy/dx.
    This confuses me where to go from here. If I take the derivative of my expression with respect to x wouldn't the integral just go away? How would I evaluate it then for all dq?
     
  7. Feb 4, 2016 #6

    gneill

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    No, the dq will be λ times the length of the line segment indicated. For your line, for a given dx the length of dy is ...?
     
  8. Feb 4, 2016 #7

    gneill

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    I've revised the image in post #4 slightly.
     
  9. Feb 4, 2016 #8
    I don't see it. The only thing I see doesn't make much sense sqrt (dq^2 - dx^2) ???
     
  10. Feb 4, 2016 #9
    don't answer yet, I didn't see your revision
     
  11. Feb 4, 2016 #10

    gneill

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    What's the length of the line segment containing the charge in terms of dx and dy?

    What is dy in terms of dx for your particular line? (dy = ?dx)

    Substitute for dy and write the length of the line segment in terms of dx only.
     
  12. Feb 4, 2016 #11

    gneill

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    Too late! :smile:
     
  13. Feb 4, 2016 #12
    Is the length of the line segment dl = sqrt(dy^2 + dx^2)???
     
  14. Feb 4, 2016 #13

    gneill

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    Yes. Simple Pythagoras. What's dy in terms of dx for your line?
     
  15. Feb 4, 2016 #14
    sqrt(dl^2 - dx^2) ???
     
  16. Feb 4, 2016 #15

    gneill

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    No, for your line as a whole: What's dy/dx? So dy = (?)dx.
     
  17. Feb 4, 2016 #16
    -2 (from the slope)
     
  18. Feb 4, 2016 #17

    gneill

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    Yes, so replace dy with -2dx in the expression for dl. Then you'll have an expression for dl in terms of dx only.

    So then, what's dq in terms of dx?
     
  19. Feb 4, 2016 #18
    sqrt(5) dx
     
  20. Feb 4, 2016 #19

    gneill

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    Multiply that length by λ to yield the charge element.
     
  21. Feb 4, 2016 #20
    So from that,

    E = λ sqrt(5) / (4 π ε0 ) ∫ < -x , 2x - 8> dx / [ x^2 + (8 - 2x)^2]^(3/2) from x=0 to x=4

    But if I do this I just multiply my answer by a sqrt(5). Something must be wrong with my integral.
     
    Last edited: Feb 4, 2016
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