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Integrating the Cartesian form of Coulomb's law

  1. Jan 18, 2009 #1
    Hi, I want to calculate the potential energy between two opposite charges (a dipole) and I know how to integrate Coulomb’s law in the polar form, i.e. in terms of “r”

    [tex]\[
    U=\int\mathbf{F}d\mathbf{r}=-\frac{e^{2}}{4\pi\epsilon_{0}}\int\frac{1}{\mathbf{r^{2}}}d\mathbf{r}=\frac{e^{2}}{4\pi\epsilon_{0}}\frac{1}{r}\][/tex]

    But I want to know how to integrate it when it’s in the Cartesian form i.e.

    [tex]\[
    F(x,y,z)=\frac{qQ}{4\pi\epsilon_{0}}\frac{1}{[x^{2}+y^{2}+z^{2}]^{\frac{3}{2}}}\left(\begin{array}{c}
    x\\
    y\\
    z\end{array}\right)\][/tex]

    Please I need this urgently, and I’m more interested on how to do the integral because I need this for a much more complicated problem that involves moving charges but if I can’t do it for the stationary charges then I can’t do it for that problem. Please I don’t want anybody to suggest integrating the force in the polar form and then changing the variables to Cartesian because as I said I need to know how to do the integral.
     
  2. jcsd
  3. Jan 19, 2009 #2

    jtbell

    User Avatar

    Staff: Mentor

    Exactly where are you stuck? Can you expand the dot product [itex]\vec F \cdot d \vec r[/itex] in Cartesian coordinates?
     
  4. Jan 19, 2009 #3
    Thanks jtbell, good hint, I stupidly forgot that it's a dot product.
    But I have another problem now and this is what I got
    [tex]\[
    U=\int\mathbf{F}.\mathbf{dr}\]
    and in the cartesian form we have \[
    \mathbf{r}=x\mathbf{\hat{x}}+y\mathbf{\hat{y}}+z\mathbf{\hat{z}}\][/tex]
    [tex]\[
    \mathbf{\hat{r}}=\frac{x\mathbf{\hat{x}}+y\mathbf{\hat{y}}+z\mathbf{\hat{z}}}{[x^{2}+y^{2}+z^{2}]^{1/2}}\][/tex]
    [tex]\[
    d\mathbf{r}=dx\mathbf{\hat{x}}+dy\mathbf{\hat{y}}+dz\mathbf{\hat{z}}\]
    Therefore \[
    U=-\frac{e^{2}}{4\pi\epsilon_{0}}\int\frac{1}{[x^{2}+y^{2}+z^{2}]^{3/2}}(x\mathbf{\hat{x}}+y\mathbf{\hat{y}}+z\mathbf{\hat{z}}).(dx\mathbf{\hat{x}}+dy\mathbf{\hat{y}}+dz\mathbf{\hat{z}})\][/tex]
    [tex]\[
    U=-\frac{e^{2}}{4\pi\epsilon_{0}}\{\int\frac{xdx}{[x^{2}+y^{2}+z^{2}]^{3/2}}+\int\frac{ydy}{[x^{2}+y^{2}+z^{2}]^{3/2}}+\int\frac{zdz}{[x^{2}+y^{2}+z^{2}]^{3/2}}\}\][/tex]
    [tex]\[
    U=-\frac{e^{2}}{4\pi\epsilon_{0}}\{-\frac{1}{[x^{2}+y^{2}+z^{2}]^{1/2}}-\frac{1}{[x^{2}+y^{2}+z^{2}]^{1/2}}-\frac{1}{[x^{2}+y^{2}+z^{2}]^{1/2}}\}\][/tex]
    [tex]\[
    U=\frac{e^{2}}{4\pi\epsilon_{0}}\frac{3}{[x^{2}+y^{2}+z^{2}]^{1/2}}=\frac{3e^{2}}{4\pi\epsilon_{0}}\frac{1}{r}\][/tex]

    So there is an extra 3 in comparison with polar form shown in my first post.

    Where did I go wrong?
     
  5. Jan 19, 2009 #4

    jtbell

    User Avatar

    Staff: Mentor

    This is a path integral:

    [tex]U = - \int_{{\vec r}_0}^{\vec r_{final}}{\vec F \cdot d \vec r}[/tex]

    so you have to pick a starting point [itex]{\vec r}_0[/itex] at infinity, then set up a path from that point to the point where you want the potential energy. (Imagine holding one charge fixed at the origin and moving the other charge in from infinity along that path.) Then integrate along that path.

    You're using Cartesian coordinates, so path integrals are easier to do if they're made of segments along or parallel to the x, y and z axes.

    For example, you could start at [itex](x, y, z) = (+\infty, 0, 0)[/itex] and set up a path to [itex](x_{final}, y_{final}, z_{final})[/itex] with three straight-line segments. You pick the segments!

    Simply integrating the x, y and z integrals from [itex]\infty[/itex] to [itex]x_{final}[/itex], [itex]\infty[/itex] to [itex]y_{final}[/itex], and [itex]\infty[/itex] to [itex]z_{final}[/itex] respectively, doesn't define a proper path. I've been groping unsuccesfully for a way to put the reason why into words...
     
  6. Feb 19, 2011 #5
    not so much as a thank you? :surprised
     
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