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Electric Field outside a cylinder

  1. Aug 9, 2007 #1
    The problem statement, all variables and given/known data
    Immediately outside a very long cylindrical wire of radius Ri = 0.5 mm, the electric field (in air) is E = 40 kV/m,
    away from the wire’s surface. Use clearly labelled diagrams, explain any principle/law used, for all parts.
    a) What is the charge per unit length along the wire?

    The attempt at a solution

    I'm not sure which formula I should apply, can somebody direct me? :confused:
     
  2. jcsd
  3. Aug 9, 2007 #2

    learningphysics

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    There's an important law that relates the charge to electric flux (which is field strength*area)... can you think of the law? That's what you need to use here.
     
  4. Aug 9, 2007 #3
    Gausses?..............
     
  5. Aug 9, 2007 #4

    learningphysics

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    Yes. :smile: That's what you need to use here. Take an arbitrary length x of the cylinder... Then use gauss' law over that cylinder over that length... You should be able to solve for charge per unit length...
     
  6. Aug 9, 2007 #5
    I'm still confused. Which formula exactly should I be using?
     
  7. Aug 9, 2007 #6

    learningphysics

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    It's Gauss' law... what does Gauss' law say?
     
  8. Aug 9, 2007 #7
    Ok, so I should be using

    Φ = EA = Q/εo
    = 40*2*pi*0.0005*x = Q/εo

    Therefore Q/x=1.112*10^-12 C/m

    Correct?
     
  9. Aug 9, 2007 #8

    learningphysics

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    Yes, looks correct to me.
     
  10. Aug 9, 2007 #9
    Cool!
    How about if I wanted to find the field at a radius of 2mm from the centre of the wire?

    Using Gauss again, I know all values but x (length of wire), Q, and E (which is what I want to find).

    If I use the Q/x value I just found out from the previous question, I get 1.24*10^-25 V/m.

    Sounds like quite a far out number, could it be right?
     
  11. Aug 9, 2007 #10

    learningphysics

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    Oops... before for E, you needed to use 40,000... not 40 (you have to convert to volts). So that would give Q/x=1.112*10^-9 C/m

    For this part, I get 10,000V/m or 10kv/m.
     
  12. Aug 9, 2007 #11
    Good spot with the kV/m!

    I got 10kv/m for part 2 as well, just a silly arithmatic error!

    Thanks ALOT!
     
  13. Aug 9, 2007 #12

    learningphysics

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    You're welcome.
     
  14. Aug 9, 2007 #13
    Managed to do one part by myself (good acheivement by my standards) before being getting stuck again.

    How is it possible to calculate the potential diff betweem r=2 and r=0.5?
    As the length of the wire is supposedly the same and the E for r=0.5 is 40000 and the E for r=2 is 10000, is it simply 40000-10000 = 30000V?
     
  15. Aug 9, 2007 #14

    learningphysics

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    No. The potential diff is found using an integral. Your text should have that integral.
     
  16. Aug 9, 2007 #15
    I think this is the one?

    [​IMG]

    where f = 2
    and i = 0.5

    I'm not sure how to do the actual mathematics part of it and there is no example in the book.

    edit: my understanding is that the test charge is moving parallel to E and hence angle will be 0degrees.
     
    Last edited: Aug 9, 2007
  17. Aug 9, 2007 #16

    learningphysics

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    Yes, you take a path radially outward from the wire... and E is parallel to that path. cos 0=1. So it is just -integral Edr
     
  18. Aug 9, 2007 #17
    So basically...
    next steps

    =-∫Edr (with terminals 0.5 and 2)

    =-∫Er (with terminals 0.5 and 2)

    = -([E*0.5]-[E*2])

    Do I use the corresponding E values for each radial distance?

    Am I even on the right track?
     
  19. Aug 9, 2007 #18

    learningphysics

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    No, you need a formula for E in terms of r... ie you need E(r)... then you'd calculate:

    =-∫E(r)dr (with terminals 0.5 and 2)
     
  20. Aug 9, 2007 #19
    Well in my textbook I have found

    E=(λ)/(2pi*εo*r)

    where lambda is Q/x.

    Is this equation still suitable even though lambda has been brought in?
     
  21. Aug 9, 2007 #20

    learningphysics

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    Not sure... do exactly what you did to calculate the field at 2mm... except instead of 2mm just leave the radius as r. So solve for E, and you'll have an expression in terms of r.
     
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