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Electric Potential of a Cylinder using Poisson's Equation!

  1. Apr 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider a homogeneously charged, infinitely long, straight wire of finite radius R. Determine the potential [itex]\phi[/itex](r) of the wire for r ≤ R and for r ≥ R. You must use the Poisson-Equation!
    2. Relevant equations
    Δφ(r) = -ρ(r)/ε ⇔ [itex]\frac{1}{r}[/itex][itex]\frac{∂}{∂r}[/itex](r[itex]\frac{∂φ}{∂r}[/itex]) + [itex]\frac{1}{r^{2}}[/itex][itex]\frac{∂^{2}φ}{∂θ^{2}}[/itex] + [itex]\frac{∂^{2}φ}{∂z^{2}}[/itex] = -ρ[itex]_{0}[/itex]/ε

    3. The attempt at a solution
    Ok first of all, I'm really frustrated, cause this problem is a two-minutes work using the Gauss's Law and I'm stuck for an hour:mad:
    This problem should be really trivial, cause the partial DE could be reduced due to symmetry (the electric field must have only radial dependency) to the following ordinary DE :
    [itex]\frac{1}{r}[/itex][itex]\frac{d}{dr}[/itex](r[itex]\frac{dφ}{dr}[/itex]) = -ρ[itex]_{0}[/itex]/ε , with the solution : φ(r) = -[itex]\frac{ρ [itex]_{0}[/itex]r^{2}}{4ε}[/itex] + C[itex]_{1}[/itex]ln(r) + C[itex]_{2}[/itex]
    The problem is I've no idea how I could calculate C[itex]_{1}[/itex] and C[itex]_{2}[/itex] for the two areas r ≤ R and r ≥ R ! I've thought of integrating with upper and lower bounds instead of indefinitely but to no avail...
     
    Last edited: Apr 19, 2012
  2. jcsd
  3. Apr 19, 2012 #2
    What boundary conditions are you using? First separate the solution into a solution of the two regions. Then, you know that the potential is continuous going between the two regions.
     
  4. Apr 19, 2012 #3
    My problem is purely of mathematical nature. At what stage of the calculation should I separate the regions (after I've found the general solution of the DE?) and how I'm supposed to? Using Dirac's function maybe? But then again that is already done during the derivation of Poisson's law so it wouldn't help to go back. As already written, I've tried to put boundaries to the to integrals (e.g. from r' to R for the first and from R to r' for the second region or something like that) but it didn't help too much or at least that's what I understood...

    There are no specific boundary conditions given! Just homogenity of charge til r=R.

    Continuity is a good point! though I've never used that before, it could still lead me somewhere...
     
  5. Apr 19, 2012 #4
    What I mean is that your general solution now needs to be split into two different regions. For example, for r<R then the natural log can't be there because it's undefined at 0 in that region. Continuity will be helpful too.

    I don't know, if you know what you're doing (and after this problem you'll know what you're doing) you may be able to argue that the Poisson solution takes just about as long as the Gaussian formulation. Additionally, Poisson solutions hold in more situations than Gaussian -- I've never seen in infinite charged cylinder in real life.
     
  6. Apr 19, 2012 #5
    Ok I've solved that, thanks for the tips...

    As a matter of fact indefinite integration was the only possible approach to this problem. The trick in order to "separate" the regions is simply to consider ρ=ρ[itex]_{0}[/itex] on the inside and ρ=0 on the outside. Two of the emerging constants can be eliminated/calculated by the continuity of the potential and by the mere fact that in the centre of the sphere the potential must be finite. The other two constants cannot be found without any boundary conditions (e.g. φ(R)=0)
     
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