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Line integral ad electric charge

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data
    The force exerted by an electric charge at the origin on a charged particle at point (x,y,z) with position vector r = <x,y,z> is F(r) = Kr / |r|^3
    where K is a constant. Find the work done as the particle moves along a straight line from (2,0,0) to (2,1,5).

    2. Relevant equations

    3. The attempt at a solution
    So if it moves from (2,0,0) to (2,1,5) I can parametrize:
    x = 2
    Y = T
    Z = 5T
    for 0<t<1

    so r(t) = (2, T, 5T)
    r'(t) = (0, 1, 5)
    But how do I put the force function F(r) = Kr / |r|^3 in terms of these parameters so I can do the integral?
     
  2. jcsd
  3. Feb 15, 2009 #2

    AEM

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    The electrostatic force is a conservative force. This means that the work done by the electrostatic force while moving a charge from one point to another is independent of the path and only depends on the end points. The magnitude of the work done by the electrostatic force will be the difference in the potential energy of the system in the two configurations. Using this you can find the answer to your question without computing the line integral along the path requested, but by choosing another simpler path.

    However, if that is not allowed, then you probably should re-express the force law you were given in terms of Cartesian coordinates.
     
  4. Feb 16, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    work= force* distance moved and if the force is variable along the path, it is \int f dx[/itex]. Finally if force is a vector, then it is [itex]\vec{F}\dot\vec{ds}[/itex] where the integral is take along the path the object moves on.

    Here the force is given as [tex]\frac{K\vec{r}}{|\vec{r}|^3}[/itex] which, in Cartesian coordinates, is [tex]\frac{K}{(x^2+ y^2+ z^2)^{3/2}}(x\vec{i}+ y\vec{j}+ z\vec{k}[/tex]. As you say, parametric equations for the path are x= 2, y= t, z= 5t so you can replace x, y, z in the force vector by those. As a vector, the path is given by [tex]\vec{r}= 2\vec{i}+ t\vec{j}+ 5t\vec{j}[/tex] so [itex]d\vec{r}= (\vec{j}+ 5\vec{k})dt[/tex]. Integrate the dot product of those two from t= 0 to t= 1.

    However, AEM is correct. That is a conserative force so you could also find a "potential", an antiderivative, F(x,y,z) such that [tex]\nabla F[/tex] is the force vector, and evaluate that at the two end points. You would have to find F such that
    [tex]\frac{\partial F}{\partial x}= \frac{Kx}{(x^2+ y^2+ z^2)^{3/2}}[/tex]
    [tex]\frac{\partial F}{\partial y}= \frac{Ky}{(x^2+ y^2+ z^2)^{3/2}}[/tex]
    [tex]\frac{\partial F}{\partial z}= \frac{Kz}{(x^2+ y^2+ z^2)^{3/2}}[/tex]
     
  5. Feb 16, 2009 #4
    I'm very happy to read your post, because that's actually how i ended up doing it, though I certainly wasn't sure if it was right. Thanks very much, I really appreciate all your help!
     
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