Charge Density and Neuronal Cells-Help

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Homework Help Overview

The discussion revolves around calculating the charge density of neuronal cell membranes, modeled as a parallel plate capacitor. The problem involves understanding the relationship between electric field strength and charge density, as well as the geometry of the cell membrane.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss using the electric field equation to solve for charge density, with some expressing uncertainty about how to determine the area needed for calculations. Others point out the approximation of the cell membrane as a parallel plate capacitor versus its actual cylindrical shape.

Discussion Status

There are multiple interpretations of the problem, with some participants providing guidance on the equations to use. However, there is no explicit consensus on the correct approach, as some participants have reported discrepancies in their calculations.

Contextual Notes

Participants note the challenge of applying the parallel plate capacitor model to a structure that is more cylindrical in nature, raising questions about the validity of the assumptions made in the problem setup.

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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!
 
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bump for answer :smile:
 
Well using the first equation rearrange for charge density because that's 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.
 
That is what i did. I solved for charge density by the equation derived in my textbook 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??
 
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.
 
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.
 

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