Parallel-Plate Capacitor

1. Sep 7, 2008

just.karl

When a nerve impulse propagates changes from 7.0 x 10^5 N/C in one direction to 3.0 x 10^5 N/C in the other direction. Approximating the cell membrane as a parallel-plate capacitor, find the magnitude of the change in charge density on the walls on the cell membrane.

I understand the basic idea of a parallel-plate capacitor but I haven't been able to figure out how to set it up or where to go. A hint where to start would be nice, thanks.

2. Sep 7, 2008

Defennder

You have Q=CV for a capacitor. Now, how do you find E for a parallel plate capacitor given V and d, the separation of plates?

You have to assume that the distance between the plates stay the same when the E-field changes.

3. Sep 7, 2008

just.karl

I'm not sure what the equation Q=CV stands for or how it relates to the newton / coulomb charge values I'm given.

4. Sep 7, 2008

just.karl

help????.....

5. Sep 8, 2008

Defennder

Q=CV tells you the relationship between the charge, the capacitance and the voltage difference between the two plates. Since you are told to model it as a parallel plate capacitor, you can safely assume that the capacitance remains constant throughout.

Suppose you have an electric field E in one dimension. How is this related to the potential difference between any two points and the distance between them? This is what you need in order to make use of Q=CV, since E doesn't appear anywhere in Q=CV.

Let Q1 be the old charge on the plates, Q2 be the new charge:
Q1 = CV1
Q2 = CV2
Change in Q = C(V1-V2).

Now one more formula you'll need: The formula for the capacitance of a parallel plate capacitor (assume you're dealing infinite planes of charges here). Use Gauss law to get it.

Once you have this, plug them into the equation above and you're done.

6. Sep 8, 2008

just.karl

This doesn't seem right for how you do the problem. Is all you do is add the two electric fields together to get the magnitude of change and then multiply it by e_0?

7. Sep 8, 2008

Defennder

You can't add the two fields together. They aren't electric fields due to separate sources, they're from the same capacitor. Although the result is mathematically the same, the conceptual understanding underlying your working is mistaken. Anyway I made a mistake, since d (the separation of the plates) cancels out in the equation, you do not have to assume that d remains constant.

8. Sep 25, 2008

just.karl

Sorry for the long delay on the reply. I forgot about it and my exam is tomorrow. So I want to figure this out.

So V=Ed and C=e_oA/d so then Q=Ee_oA and A=$$\Phi$$/E ??? Am I on the right track at all?

9. Sep 25, 2008

Defennder

What is $$\Phi$$ here? It's not V is it?