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Homework Help: Resistance, Current, two cylinders

  1. Oct 6, 2011 #1
    1. The problem statement, all variables and given/known data

    An oceanographer is studying how the ion concentration in seawater depends on depth. She makes a measurement by lowering into the water a pair of concentric metallic cylinders at the end of a cable and taking data to determine the resistance between these electrodes as a function of depth. The water between the two cylinders forms a cylindrical shell of inner radius [itex]r_a[/itex], outer radius [itex]r_b[/itex], and length L much larger than [itex]r_b[/itex],. The scientist applies a potential difference ΔV between the inner and outer surfaces, producing an outward radial current I. Let ρ represent the resistivity of the water.

    2. Relevant equations

    ΔV = El
    R = ρ l / a

    3. The attempt at a solution

    The potential difference is maintained across concentric metallic cylinders. We know for a wire of length l with a potential difference maintained across is ΔV = El. In this case, an outward radial current I is produced. In other words, current flows inbetween the cylinders. Thus, charges flow from [itex]r_a[/itex] to [itex]r_b[/itex]. Hence, length l is [itex]r_a[/itex] - [itex]r_b[/itex]. Finally, ΔV = E([itex]r_a[/itex] - [itex]r_b[/itex]).

    In the equation R = ρ l / A, A is cross-sectional area of the conductor (saltwater is good conductor) which is pi(b^2-a^2).

    So, I learn that current density equation is only valid if the cross sectional area A is perpendicular to the current density. So, an area perpendicular to the current are cylinders.

    So A = 2 pi r L, for some radius r.
    Then, I need integrate from [itex]r_a[/itex] to [itex]r_b[/itex] with respect to r.

    So, I write

    dA = 2pi r L dr ?

    The reasoning for length is still valid.

    So, ∫dR = ∫[itex]^{r_b}_{r_a}[/itex]ρ ( ([itex]r_a[/itex] - [itex]r_b[/itex]) / (2pi r L dr)

    R = [itex]\frac{ρ(r_a - r_b) }{2\pi L }[/itex]∫[itex]^{r_b}_{r_a}[/itex] [itex]\frac{1}{ r dr}[/itex]

    R = [itex]\frac{ρ(r_a - r_b) }{2\pi L }[/itex] ln[itex]\frac{r_a}{ r_b}[/itex].

    Please help
  2. jcsd
  3. Oct 6, 2011 #2


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    Staff: Mentor

    Your reasoning looks good up until you write the integral. Why do you include the (ra - rb) term in the integral? Does it have some physical significance to the differential resistance element?
  4. Oct 7, 2011 #3
    I have trouble with integrals.
  5. Oct 7, 2011 #4


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    Staff: Mentor

    The integral is a sum of differential elements. In this case you're trying to compute a total resistance, so you should expect to be adding differential elements of resistance, or dR.

    The resistivity formula you have for a slab of material is R = ρl/A, where l is the 'thickness' of the slab and A the cross sectional area perpendicular to l. You've identified A as being 2πrL for a given radius r. The thickness of a differential element is dr, and so that is identified with l. You should, then, be able to directly 'transcribe' the resistivity formula into a formula for dR.
  6. Oct 7, 2011 #5

    I wrote formula for dR without recognizing dr.
  7. Oct 7, 2011 #6

    [itex]\rho = \frac{1}{\sigma} [/itex]

    [itex]\rho = \frac{E}{J}[/itex]

    [itex]\rho = \frac{AE}{I} [/itex] where A = 2 [itex]\pi[/itex] rdr

    [itex]\rho = \frac{A \Delta V L}{I} [/itex] where L = r

    Does [itex]\rho[/itex] depend on geometry?
    Why do I need to integrate?
  8. Oct 8, 2011 #7
    ρ=AΔVL / I where L = r

    should be


    then from part a, replacing A and r,

    ρ=(2pi L)ΔV/(ln(r_b / r_a) I
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