Charge Density and Neuronal Cells-Help!

1. Jan 10, 2007

deenuh20

Charge Density and Neuronal Cells--Help!

The cell membrane if a typical nerve cell consists of an inner and outer wall separated by a distance of 0.10 µm. The electric field within the cell membrane is 7.0 x 105 N/C. Approximating the cell membrane as a parallel plate capacitor, determine the magnitude of the charge density on the inner and outer cell walls.

E=charge density/Eo
Q=charge density*Area

I tried doing this problem by using E=Charge density/Eo and just solving for charge density. I'm not really quite sure how to find the area to use in the equation (I know its cylindrical, but I'm not sure how I can find the area with just the distance)

Thank you!

2. Jan 11, 2007

deenuh20

3. Jan 11, 2007

Kurdt

Staff Emeritus
Well using the first equation rearrange for charge density because thats all that is asked for. You know the E-field and epsilon nought is a constant. It is however a very crude approximation as the cell is more cylindrical than anything else and the gap is not a vacuum.

4. Jan 11, 2007

deenuh20

That is what i did. I solved for charge density by the equation derived in my text book for cylindrical approximations and got 1.239x10^-5 C/m^2 when using the equation, Charge Density = E*2Eo. However, when I type this answer in (it's online homework), it keeps saying it is incorrect, which is making me wonder if I am missing something??

5. Jan 11, 2007

Kurdt

Staff Emeritus
The question states treating it as a parallel plate capacitor not a cylindrical one. For a parallel plate the charge density will be half the value you have obtained.

6. Jan 11, 2007

deenuh20

Nevermind, I got it, FINALLY!

I had to solve for Q using E=k*(Q/r^2) then find Area and solve for charge density by Q/A.