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Power draw of a spherical conductor surrounded by an insulator & a conducting shell

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data

    A spherical conductor of radius a is surrounded by a spherical conducting shell of radius b, and the gap is filled with an insulating material of resistivity ρ. A thin wire connects the inner surface of the shell to the surface of the conductive sphere, and a potential of V is applied to the outer surface of the conducting shell.

    I. Determine the current drawn from the voltage source.
    II. Integrate the power density (σE2) over the insulator volume v and compare to the power drawn from the voltage source.

    2. Relevant equations

    dR = ρ dr / 4πr2
    i = V / R
    E = ρ J = -[itex]\partial[/itex]V/[itex]\partial[/itex]r
    Pdissipative = I2R
    P/v = σE2
    V = -∫E dot dr

    3. The attempt at a solution

    Part I was fairly straightforward. I found R by integrating from b to a (because the current travels inward from b) and got

    R = ρ (b - a) / 4πab

    which yielded

    i = 4πVab / ρ(b - a)

    Part II is giving me fits, however. The power drawn from the voltage source is just I*V. To get the power from the power density σE2 I would need to find an expression for E. Since E is in the same direction as the radial displacement vector r,

    V = -∫E dr = -Er, evaluated over the limits of integration (from b to a, in this case).

    I have a feeling that this is the route I need to take, but I'm not sure where to take it from there. I also considered using

    E = ρJ = ρ(i/A)

    but am also not sure where to take that... I figure I would need to find a function J(r) since the current density is r-dependent. (Current is constant over an r-dependent geometry.)
    Anyone who can help me with this, or give a hint as to the right direction, would be doing me a huge favor.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 6, 2011 #2
    Re: Power draw of a spherical conductor surrounded by an insulator & a conducting she

    Bump for the morning crowd.
     
  4. Oct 7, 2011 #3
    Re: Power draw of a spherical conductor surrounded by an insulator & a conducting she

    Okay, so I think I've made some progress, and I'd really appreciate if someone could check my thought process here. Some notes I've made and thoughts I've had since I posted the thread:
    • I'll also use U to represent volume, since V is already taken and there's no potential-energy function at play here.
    • [itex]J = di/dU[/itex]
    • [itex]E = \rho J[/itex]

    Since I found earlier that [itex]dR = dr/4\pi r^{2}[/itex], we have

    [itex]i = V/R = 4\pi V ab/\rho(b-a)[/itex]

    and

    [itex]di/dr = 4\pi\sigma Vr[/itex]

    So now we can use [itex]J = di/dU = \frac{4\pi\sigma Vr dr}{4\pi r^{2} dr} = \sigma V/r[/itex]

    and thus,

    [itex]E = J/\sigma = \sigma V/\sigma r = V/r[/itex] [itex]\Rightarrow[/itex] [itex]E^{2} = V^{2}/r^{2}[/itex]

    which we sub into the equation

    [itex]dP/dU = \sigma E^{2} \Rightarrow dP = \sigma E^{2} dU = V^{2}/r^{2} * \sigma * 4 \pi r^{2} dr = 4 \pi\sigma V^{2} dr[/itex]

    Integrating dP with respect to r, we conclude that

    [itex] P = \int ^{b}_{a} 4\pi\sigma V^{2} dr = 4\pi\sigma V^{2}r |^{b}_{a}[/itex]

    When evaluated, this yields

    [itex]P = 4\pi\sigma V^{2}(b - a)[/itex]

    as our answer for the integral of the power-density function over the insulator volume.

    To compare to the power that is dissipated as heat by the insulator, we simply need to take a straightforward approach:

    [itex] dP = V di \Rightarrow P = \int^{b}_{a} V di = \int^{b}_{a} 4\pi\sigma V^{2} dr = 4\pi\sigma Vr |^{b}_{a}[/itex]

    When evaluated, this yields

    [itex]P = 4\pi\sigma V^{2}(b - a)[/itex]

    demonstrating that the power expenditure is equivalent using either method of calculation.

    So what do you folks think? Did I do it right? Any glaring errors I should know about? :)
     
    Last edited: Oct 7, 2011
  5. Oct 8, 2011 #4

    Redbelly98

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    Re: Power draw of a spherical conductor surrounded by an insulator & a conducting she

    I agree with your result for i, but disagree with your power result. For starters, the "straightforward result" would simply use P=iV, using the expression you already got for i. (No integration required.)

    I think you went awry in coming up with an expression like

    [tex]\frac{di}{dr} = 4 \pi \sigma Vr[/tex]

    as if i were a function that changes with r. But i is actually a constant expression.

    I like that approach better. A would be the surface area of a sphere of radius r that the current is flowing through. That would make E2=____?, then you can integrate the power density over the entire volume between r=a and r=b.
     
  6. Oct 8, 2011 #5
    Re: Power draw of a spherical conductor surrounded by an insulator & a conducting she

    Oh. Duh. Thanks, I have this habit of assuming everything is supposed to be a screwball and whatnot because my professor has a proclivity for that kind of thing. And so I overcomplicate things sometimes. :(
     
  7. Oct 8, 2011 #6
    Re: Power draw of a spherical conductor surrounded by an insulator & a conducting she

    So for clarity's sake, whatever i is, I treat it as a constant instead of trying to find a differential charge, and set A = the area at an arbitrary radial distance r and just end up dividing i/4pi r^2 to get J?
     
  8. Oct 8, 2011 #7

    Redbelly98

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    Re: Power draw of a spherical conductor surrounded by an insulator & a conducting she

    Yes, correct.
     
  9. Oct 8, 2011 #8
    Re: Power draw of a spherical conductor surrounded by an insulator & a conducting she

    Thanks so much for that. So from there I would just take E (= [itex]\rho[/itex]J), square it, multiply by sigma, and evaluate the integral

    P = ∫[itex]\sigma[/itex]E2*4[itex]\pi[/itex]r2dr

    leaving the a's and b's in the current function as a's and b's instead of being silly and trying to find them as functions of r.

    Right?
     
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