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Solving Possion's equation: electrostatic potential

  1. Sep 28, 2008 #1
    I have the problem and my work uploaded to this webpage. I believe I have the general idea down, but I'm just afraid that I've done a careless calculation and missed a minus sign somewhere. I feel that my results have a lot of negative signs, making the potential constantly negative. I'm not sure if this is what is expected. Any help is appreciated.

    http://sites.google.com/site/physicsinc/phy
     
  2. jcsd
  3. Sep 28, 2008 #2

    gabbagabbahey

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    You've made an error when applying the boundary conditions:

    [tex]{\varphi}_{I}(R)={\varphi}_{II}(R), \quad {\varphi}_{II}(R+d)={\varphi}_{III}(R+d)[/tex]

    You've for some reason interchanged your constants [itex]C_0[/itex] and [itex]C_1[/itex] with [itex]C_3[/itex] and [itex]C_4[/itex]; you can't do this.

    You have 4 unknown constants to solve for, so you will need 4 boundary conditions. Luckily, [itex]\varphi[/itex] isn't the only thing that is continuous; [itex]\vec{E}(r)=\vec{\nabla} \varphi[/itex] must be continuous as well. This should give you your other two boundary conditions.
     
  4. Sep 28, 2008 #3
    [tex]{\varphi}_{I}(r)=C_1^I[/tex]
    [tex]{\varphi}_{II}(r)=-\frac{4 \pi r^2 \rho }{6} + \frac{C_0^II}{r}+C_1^II[/tex]
    [tex]{\varphi}_{III}(r)=\frac{C_0^{III}}{r}[/tex]

    [tex]{\varphi}_{I}(R)=C_1^I[/tex]
    [tex]{\varphi}_{II}(R)=-\frac{4 \pi R^2 \rho }{6} + \frac{C_0^II}{R}+C_1^II[/tex]
    [tex]{\varphi}_{II}(R+d)=-\frac{4 \pi (R+d)^2 \rho}{6} + \frac{C_0^II}{R+d}+C_1^II[/tex]
    [tex]{\varphi}_{III}(R+d)=\frac{C_0^{III}}{R+d}[/tex]
    [tex]\frac{d {\varphi}_{I}{dr} |_{r=R} = 0 [/tex]
    [tex]\frac{d {\varphi}_{II}{dr} |_{r=R} = \frac{8 \pi R \rho }{6} - \frac{C_0^{II}}{R^2} [/tex]
    [tex]\frac{d {\varphi}_{II}{dr} |_{r=R+d} = \frac{8 \pi (R+d) \rho }{6} - \frac{C_0^{II}}{(R+d)^2} [/tex]
    [tex]\frac{d {\varphi}_{III}{dr} |_{r=R+d} = - \frac{C_0^{III}}{(R+d)^2} [/tex]


    We use the 4 B.C.:

    [tex]{\varphi}_{I}(R) = [tex]{\varphi}_{II}(R) [/tex]
    [tex]{\varphi}_{II}(R+d) = {\varphi}_{III}(R+d) [/tex]
    [tex]\frac{d {\varphi}_{I}{dr} |_{r=R} = \frac{d {\varphi}_II}{dr} |_{r=R} [/tex]
    [tex] \frac{d {\varphi}_{II}{dr} |_{r=R+d} = \frac{d {\varphi}_III}{dr} |_{r=R+d} [/tex]

    I've already caught 2 minus sign errors... but this is how i'm setting up everything to solve for the 4 constants.
     
  5. Sep 28, 2008 #4

    gabbagabbahey

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    Why not write it as:

    [tex]{\varphi}_{I}(r)=C_0[/tex]


    [tex]{\varphi}_{III}(r)=\frac{C_1}{r}[/tex]

    [tex]{\varphi}_{II}(r)=-\frac{4 \pi r^2 \rho }{6} + \frac{C_2}{r}+C_3[/tex]

    to get rid of those cumbersome superscripts and make it a little easier to work with?
     
  6. Sep 28, 2008 #5

    gabbagabbahey

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    You are still missing a couple of negative signs:

    [tex]\frac{d {\varphi}_{II}}{dr} |_{r=R} = -\frac{8 \pi R \rho }{6} - \frac{C_0^{II}}{R^2}[/tex]

    [tex]\frac{d {\varphi}_{II}}{dr} |_{r=R+d} = -\frac{8 \pi (R+d) \rho }{6} - \frac{C_0^{II}}{(R+d)^2} [/tex]

    What do you get for your constants now?
     
  7. Sep 29, 2008 #6
  8. Sep 29, 2008 #7

    gabbagabbahey

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    You are missing a negative sign for C_2, which is affecting your other constants.
     
  9. Sep 29, 2008 #8
    thanks for the help, i'm confident that i do not have any more errors after solving for the e-field and checking the BC and limiting cases
     
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