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Electric field over finite length rod

  1. Jul 10, 2009 #1
    I have attempted to find solutions for this netwide and can't find one that actually satisfies my questions so I am asking here. I am returning to school at the age of 32 and am trying to pre-learn some material before jumping back into the physics classes I need. A lot of the concepts are fuzzy but I'm trying.

    finiterod.jpg

    1. The problem statement, all variables and given/known data
    A thin nonconducting rod of finite length [tex]l[/tex] carries a total charge [tex]q[/tex], spread uniformly along it. Show that E at point P (y units above the rod) on the perpendicular bisector is given by

    [tex]E=(\frac{q}{2y\pi\epsilon}) (\frac{1}{\sqrt{l^2+4y^2}})[/tex]


    2. Relevant equations
    [tex]E=\frac{kQ}{d^2}[/tex]
    [tex]\lambda=\frac{Q}{l}[/tex]


    3. The attempt at a solution
    I've attempted this for 3 days and I can't figure it out.

    I started by drawing the diagram and looking at the resultant forces acting on P from the rod beneath it.

    Step 1.)
    If [tex]E=\frac{kQ}{d^2}[/tex] , then [tex]dE=\frac{kdQ}{d^2}[/tex]

    Step 2.)
    The resulting field will be perpendicular to the rod's length. The field will be pointing upward because any dx segments evaluated from the right of the bisector will be canceled by the -dx segment on the left of the bisector.

    Each dx segment of the length of the rod would produce a field at P of [tex]dE=\frac{kdQ}{d^2}[/tex]

    The components of this resulting E vector are [tex]dE[/tex]y=[tex]dEcos\theta[/tex] and [tex]dE[/tex]x=[tex]dEsin\theta[/tex]

    Step 3.)
    The resulting upward E-field is the sum of all the dx segments along the rod.

    [tex]dE[/tex]y=[tex]\int{}{dEy}[/tex]

    [tex]dE=\int_0^l{cos\theta dE}[/tex]

    Step 4.)
    if [tex]dE=\frac{kdQ}{r^2}[/tex], we substitute in what we know.

    [tex]dQ=\lambda dx[/tex] because Q charge over L length is the charge density along a length dx of the rod.

    [tex]dE=\frac{k\lambda dx}{x^2+y^2}[/tex]

    [tex]E=\int_0^l{cos\theta \frac{k\lambda}{x^2+y^2}dx}[/tex]

    [tex]E=k\lambda\int_0^l{cos\theta \frac{1}{x^2+y^2}dx}[/tex]

    Step 5.)
    This is where I fall apart. I have an equation, which is hopefully right, describing the Electric field at point P. If this isn't right so far, then I am completely clueless to where I made the mistake. Since [tex]\theta[/tex] and x are not independent, I have to change one into the other. I can use [tex]x=ytan\theta[/tex] and [tex]dx=ysec^2\theta d\theta[/tex] I believe. (I looked up that derivative as I can't remember how it's derived.) and substitute these into the equation above but from there, it gets really really messy and I can't get through it.

    Have I done everything right so far? If so, where do I go? If not, where has the logic failed me and am I destined to fail electricity & magnetism in the fall semester?

    Thanks.
     
  2. jcsd
  3. Jul 10, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi exitwound! Welcome to PF!:smile:
    Now just use Pythagoras! :wink:
     
  4. Jul 10, 2009 #3
    I can't figure out what you mean. Where do I apply ol' Pythagoras?
     
  5. Jul 10, 2009 #4

    rl.bhat

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    cos theta = y/r
     
  6. Jul 10, 2009 #5
    I've attempted this route too, and it still gets too messy for me to solve.

    [tex]E=k\lambda\int_0^l{\frac{y}{\sqrt{x^2+y^2}}\frac{1}{x^2+y^2}dx}[/tex]

    [tex]E=ky\lambda\int_0^l{\frac{1}{\sqrt{x^2+y^2}}\frac{1}{x^2+y^2}dx}[/tex]

    [tex]E=ky\lambda\int_0^l{\frac{1}{(x^2+y^2)^\frac{3}{2}}dx}[/tex]

    The antiderivative of [tex]\frac{1}{(x^2+y^2)^\frac{3}{2}}[/tex] is [tex]\frac{-2}{(x^2+y^2)^\frac{1}{2}}[/tex] , correct?

    [tex]E=ky\lambda (\frac{-2}{(l^2+y^2)^\frac{1}{2}} - \frac{-2}{(y^2)^\frac{1}{2}})[/tex]

    [tex]E=ky\lambda (\frac{-2}{(l^2+y^2)^\frac{1}{2}} - \frac{-2}{(y^2)^\frac{1}{2}})[/tex]

    [tex]E=-2ky\lambda (\frac{(y^2)^\frac{1}{2} - (l^2+y^2)^\frac{1}{2}}{(l^2+y^2)^\frac{1}{2}(y^2)^\frac{1}{2}}[/tex]

    [tex]E=-2ky\lambda (\frac{y - \sqrt{l^2+y^2}}{y\sqrt{l^2+y^2}})[/tex]

    At this point I have no idea if I'm still on track or if I've done something wrong.
     
  7. Jul 10, 2009 #6

    rl.bhat

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    The antiderivative of LaTeX Code: \\frac{1}{(x^2+y^2)^\\frac{3}{2}} is LaTeX Code: \\frac{-2}{(x^2+y^2)^\\frac{1}{2}} , correct?
    No.
    Substitute x = ytan theta and solve.
     
  8. Jul 10, 2009 #7
    As mentioned above, I tired that, and I can't solve it. I don't know what I'm doing wrong.
     
  9. Jul 10, 2009 #8

    rl.bhat

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    x = ytantheta
    dx = y*sec^2theta*d(theta)
    x^2 + y^2 = y^2tan^2theta + y^2
    = y^2(tan^2thets + 1)
    = y^2sec^2theta
    Substitute this in the integration and solve for E.
     
  10. Jul 11, 2009 #9
    So...

    [tex]\frac{k\lambda}{y} \int{\frac{sec^2\theta}{tan^2 \theta +1}d \theta}[/tex]

    [tex]\frac{k\lambda}{y} \int{\frac{1}{u+1}du}[/tex] where [tex]u=tan \theta[/tex] and [tex]du=sec^2 \theta d\theta[/tex]

    ...so I'm no longer integrating over [tex]x[/tex] but over [tex]\theta[/tex]. But how do I find out the limits if the length is L? Would it be twice the integral from 0--> L ?

    And I don't know where to go from here.

    Sorry if I'm being lame. I took calculus 10 years ago and i'm trying to relearn.
     
  11. Jul 11, 2009 #10

    LowlyPion

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    If I may, it really seems a bit simpler than I think the discussion has taken it.

    I think you're looking good to this point with a small modification in the limits to exploit the symmetry of the configuration:
    [tex] E=ky\lambda\int_\frac{-L}{2}^\frac{L}{2}{\frac{1}{(x^2+y^2)^\frac{3}{2} }dx} [/tex]

    But taking the integral looks to me to yield

    E = k*λ/y*(x/(x2 + y2)1/2) | with x from -L/2 to L/2

    which by inspection looks to yield the desired result.
     
  12. Jul 11, 2009 #11

    tiny-tim

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    Hi exitwound! :smile:

    i] you left out the ()3/2

    ii] you've made two susbtitutions, and almost got back where you started :rolleyes: … u = x/y
     
  13. Jul 11, 2009 #12
    I can't figure out the calculations done to get that X on top.

    [tex]E=k\lambda y \frac{x}{(x^2 + y^2)^\frac{1}{2}} |_\frac{-l}{2}^\frac{l}{2}[/tex]

    The chain rule kicks my toosh. I really can't figure this out. It doesn't relate to any homework assignment and I've done all I can without ripping my hair out. Is there any way the last few steps can be written out so I can figure out how to get through it and possbily attempt another problem?
     
  14. Jul 12, 2009 #13

    rl.bhat

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    E = k*lambda/y*Intg(sex^2theta/(tan^2theta + 1 )^3/2
    1 + tan^2theta = sec^2theta.
    So integration gives Int.(cos*theta*d*theta) = sin*theta.
    Electric field due to full length of rod = 2k*lambda*sin(theta ) form zero to theta.
    Sin theta = x/(x^2 + y^2)^1/2
    E = k*2x*lambda/y*(x^2 + y^2)^1/2
    Put x = l/2 and simplify to get the result.
     
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