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Doubts about Electric field due to an infinitely long wire

  1. Sep 13, 2015 #1
    About Electric field due to an infinitely long straight uniformly charged wire my book says the assumption that the wire is infinitely long is very important because without this we can not take vector E to be perpendicular to the curved part of the cylindrical gaussian surface.I think it should be parallel in place of perpendicular or it should be flat part rather than curved part.
    I also want to ask there is one more sentence that I am not getting
    the expression of E derived from an infinitely long straight uniformly charged wire i.e Vector E= λ/2πε0r
    is true largely at the central portion of the wire length and not really true at the ends of wire.I want to know why this is so?
     
  2. jcsd
  3. Sep 13, 2015 #2
    No, it means perpendicular to the curved part, although I prefer to think of this as perpendicular to the axis of the cylinder i.e. the wire. This diagram and explanation works for me.
    An infinitely long wire doesn't have ends. This is important because it means that we can ignore what happens at the ends thus providing the symmetry which makes it inevitable that ## \vec E ## is perpendicular to the axis.
     
  4. Sep 13, 2015 #3
    But electric field vector is parallel to area vector of the curved part.
     
  5. Sep 13, 2015 #4
    how?
     
  6. Sep 13, 2015 #5
    Yes, that is because an area vector is perpendicular to the area to which it relates. The statement "the electric field vector is perpendicluar to [an area tangent to ] the curved part" is equivalent to "the electric field vector is parallel to the area vector of [an area tangent to ] the curved part".
     
  7. Sep 13, 2015 #6
    If the cylinder has finite length then in general a point on its surface is closer to one end than the other.
     
  8. Sep 13, 2015 #7
    Except at midpoint,right?
     
  9. Sep 13, 2015 #8
    Exactly, the midpoint is the special case which necessitates the words "in general a point" rather than "every point".
     
  10. Sep 13, 2015 #9
    But how symmetry is relevant ?how does symmetry decide Whether E would be perpendicular or not?
     
  11. Sep 13, 2015 #10
    If and only if the cylinder is infinite, the influence of the charges tending to point E towards one end of the cylinder is equal to the influence of the charges tending to point E towards the other.
     
  12. Sep 13, 2015 #11
    Still not clear.
     
  13. Sep 13, 2015 #12
    Why are we talking about equaling up the effects of charges?I want to know how symmetry decides direction of vector E?
     
  14. Sep 13, 2015 #13

    Dale

    Staff: Mentor

    What other direction could it be? Assume that it pointed 80 degrees instead of perpendicular. How would you know of it is 80 degrees one way or 80 degrees the other way?
     
  15. Sep 14, 2015 #14
    Only if the wire is infinitely long, the field will be a function of only the distance ("r" in cylindrical co-ordinates) from the wire,otherwise as the formula of Electric field at any point says it also depends on amount of source charge.
    This makes different amount of source charge to be at different distance .Thus making field different at most of the points.
    But when we take the wire to be infinitely long, we can picture the field as being a function only of the distance ("r" in cylindrical co-ordinates) from the wire, the farther from the wire, the weaker it is. The loci of all points of any constant field strength we choose/select will be a cylinder with a central axis on the wire.
    Right?
     
  16. Sep 14, 2015 #15

    BvU

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    Good approach ! Both lazy physicists and good physicists always keep their eyes open for symmetries :smile:.

    The case cries out for using cylindrical coordinates ##\rho, \phi, z##

    In this case there are at least four symmetries in evidence

    1. rotational -- same situation when replacing ##\phi## by ##\phi + \alpha ## with a constant ##\alpha\quad \Rightarrow \quad ## no ## \phi ## dependence

    2. translational -- same situation when replacing ##z## by ##z + p ## with a constant ## p \quad \Rightarrow \quad ## no ## z ## dependence

    3. reflection -- same situation when replacing ##\phi## by ##- \phi ## in combination with #1 ## \Rightarrow \quad \phi ## component is zero

    4. reflection -- same situation when replacing ##z## by ## -z ## in combination with #2 ##\Rightarrow \quad z ## component is zero

    In short: Yes. (to your last question)
     
    Last edited: Sep 14, 2015
  17. Sep 14, 2015 #16
    MY POST #14 "Right?'?
     
  18. Sep 14, 2015 #17

    BvU

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    Yes. really.
     
  19. Sep 14, 2015 #18
    But how?
    I thought I was wrong here
     
  20. Sep 14, 2015 #19
    I mean
    pict.png
     
  21. Sep 14, 2015 #20

    BvU

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    My turn:
    What the book tries to bring across here is:

    For a wire with finite length, the expression of E derived from an infinitely ...
    which is probably also the context you left out in your post #1. Thus making the thread a confusion between infinitely long and finite length


    ---

    the answer is Yes. really.
     
  22. Sep 14, 2015 #21
    How?
    When the wire is of infinite length
    G5man.png
    when it has a finite length
    YZ0Yl.png
    I want to understand why this is so?
     
  23. Sep 14, 2015 #22

    BvU

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    Your picture doesn't really show a wire with a finite length. It shows a very thick, very short barrel. For normal folks to call something a wire, the length has to be considerably bigger than the diameter. You can't come close enough to your barrel to let the expression for the electric field approach the infinite wire expression.

    The "infinite wire" approximation is valid as soon as you are so close to the wire that you can't see the ends -- or better: when the viewing angle for the whole wire approaches ##\pi/2##. Check out the worked out problem here : the exact expression is (11) and for L >> ##\rho## (his y) you get his last expression (15),
     
  24. Sep 14, 2015 #23

    ehild

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    The electric field does depend on the amount of surface charge in any case.
    The surface charge can be homogeneous along the finite wire or rod, still the electric field would change along the length. In case of infinite length, the parallel component of the charge elements cancel each other, only the normal component remains and the field lines are perpendicular to the wire .

    finiterod.JPG
     
  25. Sep 14, 2015 #24

    ehild

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    Imagine the charge distributed evenly along the wire. Along the black line between the two charges, the resultant field is perpendicular to the wire as the parallel components cancel.
    wireresfield.JPG
     
  26. Sep 14, 2015 #25
    This is about infinite line,right?
     
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