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Capacitance of a spherical cell

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A cell membrane is composed of lipid molecules and approximately 10.29 nm thick. If the dielectric constant of the lipid is kappa= 5 , what is the approximate capacitance of a spherical cell that has a diameter of 10.29 micro meters?

    2. Relevant equations
    E=KQ/r^2
    C=Q/deltaV=1/K(1/r1-1/r2)
    r2=D/2




    3. The attempt at a solution
    So I've come up with the final solution for to get the answer is Kappa*C=>Kappa(1/K(1/r1-1/r2). I'm not really sure what I am doing wrong, but I keep coming up with answers like 5.72e-18F, which is incorrect. Any help would be greatly appreciated.
     
    Last edited: Sep 14, 2009
  2. jcsd
  3. Sep 14, 2009 #2

    kuruman

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    You need to show exactly how you came up with that number. Otherwise, we can't figure out why it is incorrect. Do you know the correct answer?
     
  4. Sep 14, 2009 #3
    I don't know the correct answer.
    so what I did was used the aforementioned formula and insterted all the values(after converting them). My guess is I went wrong somewhere with the (1/r1-1/r2). I used the thickness for r1 and the diameter of the sphere/2 for r2. anyway, here is the what I got:
    5(1(8.99e9((1/1.029e-8m)-(1/1.029e-5/2))))=5.73e-18
     
  5. Sep 14, 2009 #4

    kuruman

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    You cannot use the thickness for r1. In the expression for the capacitance r1 is the inner radius and r2 is the outer radius. The thickness is d = r2 - r1. Now note that this thickness is one-thousandth of the diameter, very thin. You need an approximation. What do you get for the difference

    [tex]\frac{1}{r_1}-\frac{1}{r_2}[/tex]

    after subtracting the fractions? Do it with symbols - don't put in numbers yet.
     
    Last edited: Sep 14, 2009
  6. Sep 14, 2009 #5
    so r1=(d/2)-thickness
    that makes a lot more sense and yielded a correct answer!
    thanks a bunch!
     
  7. Sep 14, 2009 #6

    kuruman

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    Not what I had in mind, but with today's calculators, I guess it is possible. For whatever it's worth, this is where I was going

    [tex]
    \frac{1}{r_1}-\frac{1}{r_2}=\frac{r_{2}-r_{1}}{r_{2}r_{1}}\approx\frac{d}{r^{2}}
    [/tex]

    The last approximation is valid because the radii are so close to each other. It is a useful approximation to know.
     
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