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Jan 20, 2010 18:09

A neuron is spherical with diameter of 20microns. What is its predicted capacitance. Assume that specific membrane capacitance is 1microFarad/cm^2 ( Read more... )

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k_shaker January 21 2010, 05:12:14 UTC
umm somethings fishy there with your conversion

20 um = 0.002 cm = 2 * 10^-3 cm

since you want a radius

1 * 10^-3 cm and then the square for surface area

1 * 10 ^-6 cm^2

4pi is 12.56

SA = 1.256 * 10^-5 cm^2

hmm somethings up with your formula because your units aren't canceling
out. it comes out to cm^4. You probably should be multiplying membrane capacitance by the surface area in order to get a capacitance measurement for the whole cell.

You should get something like 12.56 pF, pico being 1 * 10 ^-12.

I'm sure you've fixed your problem by now, if not, hope that helps.

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ucla_bruin January 21 2010, 07:10:56 UTC
OMG, you are my hero. You're right, my conversion was off. It was such a dumb mistake. Thank you for catching that!!!!

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ucla_bruin January 21 2010, 07:25:39 UTC
Gawd, could you check the work for this one too?

You record from a neuron whose resting potential is -75mV. You inject a hyperpolarizing current of 0.2nA and observe that the membrane potential settles to -87mV. The trajectory of the membrane potential has time constant 150ms (Time Constant = resistance*capacitance). Estimate the cell diameter, assuming it's a sphere. Remember that specific membrane capacitance is 1uF/cm^2.

Rearranging Ohms law I found the resistance of the cell to be

R = 12*10^-3/2*10^-10 = 6*10^7 ohms

Plugging that into the time constant equation, I found the capacitance:

C = .15s/6*10^7 --> C = 2.5*10^-9F

If 1uF/cm^2 = C/SA then --> 1uF/cm^2 = C/4(pi)r^2

Solving for r I get 14*10^-5m and doubling that I get 28*10^-5m as the diameter.

Does that look about right??

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ucla_bruin January 21 2010, 07:35:44 UTC
ugh that last part should be 1.41E-4m for the radius, so doubling that to get the diameter would be 242um

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