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Scalar potential

  1. May 19, 2010 #1
    π²³ ∞ ° → ~ µ ρ σ τ ω ∑ … √ ∫ ≤ ≥ ± ∃ · θ φ ψ Ω α β γ δ ∂ ∆ ∇ ε λ Λ Γ ô
    1. The problem statement, all variables and given/known data
    Consider the cylindrical in free space as shown below. The resistor has diameter 2a, length l, is filled with a inhomogeneous material with a z dependent electrical conductivity σ(z) and is capped by 2 thin disks of materials with infinite conductivity. A static voltage φ0 is applied to the 2 disks with a battery.
    Show that the potential inside the battery is given by
    φ (z)= ∫ (C1/σ(z)) dz+ C2

    C1,C2 are constants
    2. Relevant equations



    3. The attempt at a solution

    I am trying to use poisson's equation in cylindrical coordinates.
    But I am not sure, how to get the right hand side

    The potential varies only along Z and is constant along r and φ
    Therefore, poissons eqn will reduce to
    d2(φ)/dz2 = -ρ/ε

    I need σ(z) on the right hand side, but I have ρ/ε. How do I arrive at σ(z)?
     

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  3. May 19, 2010 #2

    kuruman

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    Poisson's Equation is not the way to go. The cylinder is neutral (ρ=0) as current I flows through it.

    Slice the cylinder up like salami and consider a single slice. It has thickness dz, conductivity σ(z), area A and potential difference dV across it. Find an expression for the current I through it. Note that this current is the same for all the slices, i.e. the expression you have is a constant. Call that constant C1. You get my drift?
     
  4. May 20, 2010 #3
    Kuruman, why is ρ=0. Can you explain please.
    Also, if ρ=0 then I can use Laplace equation.
     
  5. May 20, 2010 #4
    Since kuruman not likely to post, I think you should consider his suggestion. Find dV for a crosssection and add them up with the integral.

    There's no reason I can think of to look for pattern fitting to Laplace's equation. φ(z) = ∫(C1/σ(z))dz + C2 is not second order in anything. The voltage is Static, for one. The C1 and C2 in the equation are constants, not capacitance if that's what's causing confusion.
     
  6. May 20, 2010 #5
    oh,

    dv = E.dz

    But E = J/σ(z)

    v = ∫J/σ(z) dz + C2

    Here, C1=J

    This is it?
    I didn't mean to imply C1,C2 as capacitance. I know they are constants. I thought you integrate once, you get C1 and do it again, you get C2.
     
  7. May 20, 2010 #6

    gabbagabbahey

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    I think you're supposed to assume that the current density inside the battery is uniform.That means that at any given time, the amount of charge leaving some region in the conductor will be the same as the amount of current entering the region. Therefor, there will be no charge buildup and you can assume the charge density is zero.[/quote]

    You would need appropriate boundary conditions, which you don't have.
     
  8. May 20, 2010 #7

    kuruman

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    Not really. The resistance of a slice is

    R = dz/σA (A = cross sectional area)

    I = dV/R = σA (dV/dz) = constant

    Take it from here. It's a generalized "resistors in series" problem.
     
    Last edited: May 20, 2010
  9. May 23, 2010 #8
    Ok. I get it now.
    There's another question in the same problem.
    If σ is constant, determine potential φ inside the resistor, current density J, surface and volume charge densities.
    How do I find the surface and volume charge densities?
    I thought they were zero.
    I tried to apply the boundary conditions at the surface(r=a), E field is in the opposite direction of V, in the +z direction. So there's only transverse component. No normal component. Hence no surface charge. Is this right?
    I have no idea how to find the volume charge density. QxVolume of the cylinder?
     
  10. May 23, 2010 #9

    kuruman

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    Did you do the integral to find V(z)? What does Poisson's Equation say?
     
  11. May 25, 2010 #10
    Well, I get C1z+C2=ρ/ε
     
  12. May 25, 2010 #11

    kuruman

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    This is not correct. How exactly did you get the left side of the equation? Please show your work.
     
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