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Electrostatic Potential from a uniform line of charge?

  1. Aug 4, 2010 #1
    1. The problem statement, all variables and given/known data
    a) What is the value of an electrostatic potential V, a distance r from a point charge Q?

    b) A uniform line of charge, of linear charge density ([tex]\lambda[/tex]), extends along the x-axis from x = 0, to x = a.

    i) By considering the contributions from the infitisimal elements of the charge, show that the electrostatic potential at a point on the x-axis with x = b, where b>a, is given by:

    V(b) = [tex]\stackrel{\lambda}{4pie0}[/tex]ln([tex]\frac{b}{(b-a)}[/tex]

    ii) Explain how the electrical field E can be derived from the distribution of the electrostatic potential.

    iii) By considering how the potential varies with b, find the electrostatic field along the x-axis, again for b>a.

    iv) Demonstrate that the value for E you have determined is consistant with that expected from an equivalent point charge, in the limit b>>a.


    3. The attempt at a solution

    a) for this is simply have: v =Q/4pie0r.

    b) I know that I am dealing with a continuous charge distribution problem here. But have no idea how I can show that... any advice would be appreciated. Thanks
     
  2. jcsd
  3. Aug 4, 2010 #2

    ehild

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    Consider the line as a series of point-like charges, λdL each, where dL is the length of the charged line element. You know what is the potential of a point charge at a given distance. Integrate from L=0 to L=a all this contributions to the potential at the point x=b.

    ehild
     
  4. Aug 4, 2010 #3
    Since you have uniform linear distribution of charge, you know that linear density of charge is

    [tex]\lambda=\frac{Q}{a}=\frac{dq}{dx},[/tex]

    where Q is total charge.

    In order to find potential, you need to know electric field. Contribution of infinitesimal element is

    [tex]d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{dq}{x^{2}}\hat{x}[/tex]

    (I assumed that charge is positive!)

    On integration

    [tex]\vec{E}=\frac{\lambda}{4\pi\epsilon_{0}}\int_{x-a}^{x}\frac{dx}{x^{2}}\hat{x}[/tex]

    [tex]\vec{E}=\frac{\lambda}{4\pi\epsilon_{0}}(\frac{1}{x-a}-\frac{1}{x})\hat{x}.[/tex]

    Then use the formula for potential and integrate again.

    [tex]d\vec{U}=-\vec{E}\cdot d\vec{x}[/tex]
     
  5. Aug 4, 2010 #4
    OK so when I integrate over the limits from: b to 0. I get the correct answer so thanks!!!
     
  6. Aug 5, 2010 #5
    Ok I am working through the rest of this question! However for part ii I have said we need to use the del operator on in the form E = -delV

    However for the third part I think I just have to do the partial derivative in reference to b. However we are sort of just working back from the derivation... is this wrong? thanks
     
  7. Aug 6, 2010 #6

    ehild

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    the charge density is 0 for x>a.
    [tex]
    \vec{E}=\frac{\lambda}{4\pi\epsilon_{0}}\int_{x-a}^{x}\frac{dx}{x^{2}}\hat{x}
    [/tex]
    equals to 0.

    ehild
     
  8. Aug 6, 2010 #7
    You can also solve this problem by finding the potential first.

    [tex]dU=\frac{\lambda}{4\pi\epsilon_{0}}\frac{dx}{x}[/tex]

    [tex]U=\frac{\lambda}{4\pi\epsilon_{0}}\int_{b-a}^{b}\frac{dx}{x}[/tex]

    [tex]U=\frac{\lambda}{4\pi\epsilon_{0}}\ln\frac{b}{b-a}[/tex]

    Then use the formula

    [tex]\vec{E}=-\nabla U[/tex]

    to find electric field. It's maybe a bit harder, but you should get the same results.

    For the iv) part you have

    [tex]
    E=\frac{\lambda}{4\pi\epsilon_{0}}(\frac{1}{b-a}-\frac{1}{b})=\frac{\lambda}{4\pi\epsilon_{0}}\frac{a}{b(b-a)}.
    [/tex]

    For b>>a

    [tex]
    E=\frac{1}{4\pi\epsilon_{0}}\frac{\lambda a}{b^{2}}=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{b^{2}}.
    [/tex]
     
  9. Aug 6, 2010 #8

    ehild

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    Explain, please, what is x in your equations?

    ehild
     
  10. Aug 6, 2010 #9
    If you are referring to equation

    [tex]
    U=\frac{\lambda}{4\pi\epsilon_{0}}\int_{b-a}^{b}\frac{dx}{x}
    [/tex]

    x is the distance between infinitesimal charge and the point on x-axis.

    I believe that this equation is confusing.

    [tex]
    \vec{E}=\frac{\lambda}{4\pi\epsilon_{0}}\int_{x-a}^{x}\frac{dx}{x^{2}}\hat{x}
    [/tex]

    This is better

    [tex]
    \vec{E}=\frac{\lambda}{4\pi\epsilon_{0}}\int_{x-a}^{x}\frac{dr}{r^{2}}\hat{x}=\frac{\lambda}{4\pi\epsilon_{0}}(\frac{1}{ x-a}-\frac{1}{x})\hat{x}
    [/tex]

    x is a point on x-axis. I was trying to express the electric field in terms of x (any point on axis), so I can calculate the potential later by using the formula

    [tex]
    dU=-\vec{E}\cdot d\vec{s},
    [/tex]

    where s would be a vector pointing in direction where potential is smaller.

    I hope it's better now.
     
  11. Aug 6, 2010 #10

    ehild

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    Yes, the notation x is used for the x coordinate in the text of the problem. So the distance of a charge dq from a point at x is better be denoted with r.
    Then the potential of the whole line charge at x is

    [tex]

    U=\frac{\lambda}{4\pi\epsilon_{0}}\int_{x-a}^{x}\frac{dr}{r}
    =\frac{\lambda}{4\pi\epsilon_{0}}\ln\frac{x}{x-a}

    [/tex]



    The integration is for the length of the line.

    The lectric field is the negative gradient of U, and its x components is

    [tex]E_x=-dU/dx=-\frac{\lambda}{4\pi\epsilon_{0}}\frac{x-a}{x}\frac{-a}{(x-a)^2}=\frac{\lambda}{4\pi\epsilon_{0}}\frac{a}{x(x-a)}[/tex]


    for every x>a.


    ehild
     
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