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Infinite Lines of Charge

  1. Jan 17, 2009 #1
    1. The problem statement, all variables and given/known data

    [​IMG]
    The figure is a cross section of two infinite lines of charge that extend out of the page. Both have linear charge density [tex]\lambda[/tex]. Find an expression for the electric field strength E at the heigth y above the midpoint between the lines.

    2. Relevant equations

    Well E = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\frac{2\lambda}{r}[/tex] for an infinite line of charge.

    3. The attempt at a solution

    I am not quite sure what this question is asking. I think I have to integrate that formula to get an expression at y but I am not sure.
     
  2. jcsd
  3. Jan 17, 2009 #2

    Doc Al

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    Staff: Mentor

    No need to integrate--use superposition. Use that formula to find the electric field from each line charge at the point in question. (What would "r" be? What's the direction of each field contribution?) Then just add the two vectors to find the total field at that point.
     
  4. Jan 17, 2009 #3
    I placed the fields like this.
    [​IMG]

    I said:

    r = [tex]\sqrt{y^{2} + (\frac{d}{2})^{2}}[/tex]

    E1 = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex]

    E2 = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex]

    ENET = E1 + E2
     
  5. Jan 17, 2009 #4

    Doc Al

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    Looks good, but you're not done.
    Actually find the resultant (in terms of the given variables).
     
  6. Jan 17, 2009 #5
    Like this?

    ENET = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex] + [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex] = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\left[\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}} + \frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}\right][/tex]
     
  7. Jan 17, 2009 #6

    Doc Al

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    Staff: Mentor

    No. You must add them like vectors. (Find the x and y components.)
     
  8. Jan 17, 2009 #7
    Ok, I think I got it.

    (E1)x = (E1)cos[tex]\theta[/tex]
    (E2)x = (E2)cos[tex]\theta[/tex]

    (E1)y = (E1)sin[tex]\theta[/tex]
    (E2)y = (E2)sin[tex]\theta[/tex]

    (E1)x = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex][tex]\frac{\frac{d}{2}}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex]

    (E2)x = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex][tex]\frac{\frac{d}{2}}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex]

    (E1)y = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex][tex]\frac{y}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex]

    (E2)y = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex][tex]\frac{y}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex]

    then add the x components and the y components:

    ENET = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\left[\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}\frac{\frac{d}{2}}{\sqrt{y^{2} + (\frac{d}{2})^{2}}} + \frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}\frac{\frac{d}{2}}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}\right][/tex] i , [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\left[\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}\frac{y}{\sqrt{y^{2} + (\frac{d}{2})^{2}}} + \frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}\frac{y}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}\right][/tex] j
     
  9. Jan 17, 2009 #8

    Doc Al

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    Staff: Mentor

    Careful with the signs of these components. (You should be able to look at the diagram and immediately have an idea of which way the total field will point.)

    Correct the signs as needed and redo. Be sure to simplify your final answer as much as possible.
     
  10. Jan 17, 2009 #9
    Ok, I think I finally have it. I fixed the signs.

    (E1)x = -(E1)cos[tex]\theta[/tex]
    (E2)x = (E2)cos[tex]\theta[/tex]

    (E1)y = (E1)sin[tex]\theta[/tex]
    (E2)y = (E2)sin[tex]\theta[/tex]

    (E1)x = -[tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex][tex]\frac{\frac{d}{2}}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex]

    (E2)x = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex][tex]\frac{\frac{d}{2}}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex]

    (E1)y = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex][tex]\frac{y}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex]

    (E2)y = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex][tex]\frac{y}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}[/tex]

    then add the x components and the y components:

    the x components cancel as they are equal but opposite.

    ENET = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\left[\frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}\frac{y}{\sqrt{y^{2} + (\frac{d}{2})^{2}}} + \frac{2\lambda}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}\frac{y}{\sqrt{y^{2} + (\frac{d}{2})^{2}}}\right][/tex] j

    = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\left[\frac{y2\lambda}{y^{2} + (\frac{d}{2})^{2}} + \frac{y2\lambda}{y^{2} + (\frac{d}{2})^{2}}\right][/tex] j

    = [tex]\frac{1}{4\Pi\epsilon_{0}}[/tex][tex]\left[\frac{y4\lambda}{y^{2} + (\frac{d}{2})^{2}}\right][/tex] j
     
  11. Jan 17, 2009 #10

    Doc Al

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    Staff: Mentor

    Looks good! (Cancel those 4s. :wink:)
     
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