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Line of charge

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data

    [​IMG]

    A thin rod of length L with total charge Q. Find an expression for the electric field E at distance x from the end of the rod. Give your answer in component form.


    2. Relevant equations

    Well this isn't a infinite line of charge so this formula should work:

    E = [tex]\frac{1}{4\pi\epsilon_{0}}[/tex][tex]\frac{Q}{r\sqrt{r^{2} + (\frac{L}{2})^{2}}}[/tex]


    3. The attempt at a solution

    I think I will have to integrate because I cannot cancel components because there is no symmetry in the diagram all electric fields will point in the positive x, negative y direction.
     
  2. jcsd
  3. Jan 18, 2009 #2

    Doc Al

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    Staff: Mentor

    :confused: Where did this formula come from?

    Yes, you'll have to integrate. Hint: Break the line charge into small elements of length dy.
     
  4. Jan 18, 2009 #3
    I got that formula from and example in my physics book it is the result after integrating this formula:

    (Ei)x = Eicos[tex]\theta[/tex]i = [tex]\frac{1}{4\pi\epsilon_{0}}[/tex][tex]\frac{\Delta Q}{r_{i}^{2}}[/tex]cos[tex]\theta[/tex]i

    I think that formula is for a point that bisects the rod only.
     
  5. Jan 18, 2009 #4

    Doc Al

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    Staff: Mentor

    Set up the integral and derive your own formula.
     
  6. Jan 18, 2009 #5
    Ok here is what I got for the x component:

    (Ei)x = Eicos[tex]\theta[/tex]i = [tex]\frac{1}{4\pi\epsilon_{0}}[/tex][tex]\frac{\Delta Q}{r_{i}^{2}}[/tex]cos[tex]\theta[/tex]i

    r = [tex]\sqrt{y_{i}^{2} + x^{2}}[/tex]

    cos[tex]\theta[/tex]i = [tex]\frac{x}{\sqrt{y_{i}^{2} + x^{2}}}[/tex]

    (Ei)x = [tex]\frac{1}{4\pi\epsilon_{0}}[/tex][tex]\frac{\Delta Q}{y_{i}^{2} + x^{2}}[/tex][tex]\frac{x}{\sqrt{y_{i}^{2} + x^{2}}}[/tex]

    = [tex]\frac{1}{4\pi\epsilon_{0}}[/tex][tex]\frac{x\Delta Q}{(y_{i}^{2} + x^{2})^{3/2}}[/tex]

    Ex = [tex]\sum^{N}_{i = 1}[/tex](Ei)x = [tex]\frac{1}{4\pi\epsilon_{0}}[/tex][tex]\sum^{N}_{i = 1}[/tex][tex]\frac{x\Delta Q}{(y_{i}^{2} + x^{2})^{3/2}}[/tex]

    [tex]\Delta Q[/tex] = [tex]\lambda\Delta y[/tex] = ([tex]\frac{Q}{L}[/tex])[tex]\Delta y[/tex]

    Ex = [tex]\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}[/tex][tex]\sum^{N}_{i = 1}[/tex][tex]\frac{x\Delta y}{(y_{i}^{2} + x^{2})^{3/2}}[/tex]

    = [tex]\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}[/tex][tex]\int_{0}^{L}[/tex][tex]\frac{xdy}{(y^{2} + x^{2})^{3/2}}[/tex]

    = [tex]\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}[/tex][tex]\frac{y}{x\sqrt{y^{2} + x^{2}}}[/tex][tex]\left|_{0}^{L}[/tex]

    = [tex]\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}[/tex][tex]\left[\frac{L}{x\sqrt{L^{2} + x^{2}}} - \frac{0}{x\sqrt{0^{2} + x^{2}}}\right][/tex]

    = [tex]\frac{1}{4\pi\epsilon_{0}}[/tex][tex]\frac{Q}{x\sqrt{L^{2} + x^{2}}}[/tex]
     
  7. Jan 18, 2009 #6

    Doc Al

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    Staff: Mentor

    Excellent!
     
  8. Jan 18, 2009 #7
    Ok here is what I got for the y component, I am not sure if this one is right:

    (Ei)y = Eisin[tex]\theta[/tex]i = [tex]\frac{1}{4\pi\epsilon_{0}}[/tex][tex]\frac{\Delta Q}{r_{i}^{2}}[/tex]sin[tex]\theta[/tex]i

    r = [tex]\sqrt{y_{i}^{2} + x^{2}}[/tex]

    sin[tex]\theta[/tex]i = [tex]\frac{y_{i}}{\sqrt{y_{i}^{2} + x^{2}}}[/tex]

    (Ei)y = [tex]\frac{1}{4\pi\epsilon_{0}}[/tex][tex]\frac{\Delta Q}{y_{i}^{2} + x^{2}}[/tex][tex]\frac{y_{i}}{\sqrt{y_{i}^{2} + x^{2}}}[/tex]

    = [tex]\frac{1}{4\pi\epsilon_{0}}[/tex][tex]\frac{y_{i}\Delta Q}{(y_{i}^{2} + x^{2})^{3/2}}[/tex]

    Ey = [tex]\sum^{N}_{i = 1}[/tex](Ei)y = [tex]\frac{1}{4\pi\epsilon_{0}}[/tex][tex]\sum^{N}_{i = 1}[/tex][tex]\frac{y_{i}\Delta Q}{(y_{i}^{2} + x^{2})^{3/2}}[/tex]

    [tex]\Delta Q[/tex] = [tex]\lambda\Delta y[/tex] = ([tex]\frac{Q}{L}[/tex])[tex]\Delta y[/tex]

    Ey = [tex]\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}[/tex][tex]\sum^{N}_{i = 1}[/tex][tex]\frac{y_{i}\Delta y}{(y_{i}^{2} + x^{2})^{3/2}}[/tex]

    = [tex]\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}[/tex][tex]\int_{0}^{L}[/tex][tex]\frac{ydy}{(y^{2} + x^{2})^{3/2}}[/tex]

    = [tex]\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}[/tex][tex]\frac{y}{y\sqrt{y^{2} + x^{2}}}[/tex][tex]\left|_{0}^{L}[/tex]

    = [tex]\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}[/tex][tex]\left[\frac{L}{L\sqrt{L^{2} + x^{2}}} - \frac{0}{0\sqrt{0^{2} + x^{2}}}\right][/tex]

    = [tex]\frac{1}{4\pi\epsilon_{0}}[/tex][tex]\frac{Q}{L\sqrt{L^{2} + x^{2}}}[/tex]
     
  9. Jan 18, 2009 #8

    Doc Al

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    Staff: Mentor

    Good.

    Cancel those y's. (And be careful with the sign of the integral.)
     
  10. Jan 18, 2009 #9
    Ok, I made the changes:

    = -[tex]\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}[/tex][tex]\frac{1}{\sqrt{y^{2} + x^{2}}}[/tex][tex]\left|_{0}^{L}[/tex]

    = -[tex]\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}[/tex][tex]\left[\frac{1}{\sqrt{L^{2} + x^{2}}} - \frac{1}{\sqrt{0^{2} + x^{2}}}\right][/tex]

    = -[tex]\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}[/tex][tex]\left[\frac{1}{\sqrt{L^{2} + x^{2}}} - \frac{1}{\sqrt{x^{2}}}\right][/tex]
     
  11. Jan 18, 2009 #10

    Doc Al

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    Staff: Mentor

    That looks good. Move that outside minus sign inside. (Switch the order of subtraction in the brackets.)
     
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