How Does Membrane Thickness Affect the Electric Field in a Nerve Cell?

In summary, the cell membrane in a nerve cell has a thickness of 0.12 micrometers. This thickness affects the electric field within the membrane, which is constant.
  • #1
la3
4
0
The cell membrane in a nerve cell has a thickness of 0.12 micrometers.
(a) Approximating the cell membrane as a parallel plate capacitor with a surface charge density of 5.9 x 10^(-6) C/m2, find the electric field within the membrane.

(b) If the thickness of the membrane were doubled, would your answer to (a) increase, decrease, or stay the same. Explain.


Can anyone help me get started on this? I think one of the formulas to use would be something like:
E = sigma / permittivity of free space
where E= electric field, sigma= charge density, and permittivity of free space= 8.85 x 10^-12

I am unsure how the membrane thickness comes into play.
 
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  • #2
Sounds like you're on the right track. It's up to you to determine if membrane thickness affects the field within. :wink:
 
  • #3
I cannot see how membrane thickness matters since it would only affect A (area), but A cancels from both sides of the equation.
However, since it is included in both parts (a) and (b) I feel that thickness probably matters...
 
  • #4
la3 said:
I cannot see how membrane thickness matters since it would only affect A (area), but A cancels from both sides of the equation.
If you mean plate area, then thickness--the distance between the plates--has nothing to do with it.
However, since it is included in both parts (a) and (b) I feel that thickness probably matters...
They want to know if you know what matters and what doesn't.
 
  • #5
Well I am going to have to say that membrane thickness does not matter since the electric field is constant within a parallel plate capacitor since both plates run exactly parallel to each other.
The voltage (perpendicular to the electric field lines) will change as you move from one plate to another (ie. as equipotential lines), but the electric field remains constant.

So, for (a):
E = sigma / permittivity of free space
E = (5.9 x 10^(-6) C/m2) / (8.85 x 10^(-12) C2/Nm2)
E = 6.7 x 10^5 N/C

And for (b):
The answer to (a) would stay the same for above reasons.

How does this sound?
 
  • #6
Perfecto! :approve:
 
  • #7
Haha... great! Thanks for your help!
I had a hard time visualizing the problem on my own.
 

Related to How Does Membrane Thickness Affect the Electric Field in a Nerve Cell?

1. What is a parallel plate capacitor?

A parallel plate capacitor is a device used to store electrical energy by creating an electric field between two metal plates separated by a dielectric material. It is one of the simplest forms of capacitors and is commonly used in electronic circuits.

2. How does a parallel plate capacitor work?

A parallel plate capacitor works by using two conductive plates, usually made of metal, to store opposite charges. When a voltage is applied to the plates, one plate becomes positively charged and the other becomes negatively charged. The electric field created between the plates stores the energy in the form of electrical potential energy.

3. What factors affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is affected by the distance between the plates, the area of the plates, and the type of material used as the dielectric between the plates. The capacitance increases with a larger plate area and decreases with a larger distance between the plates. The type of dielectric material also plays a role, with materials such as air having a lower capacitance than materials like glass or plastic.

4. How is the capacitance of a parallel plate capacitor calculated?

The capacitance of a parallel plate capacitor can be calculated using the equation C = ε0A/d, where C is the capacitance in farads, ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. This equation assumes a vacuum as the dielectric material. For other dielectric materials, the equation is modified to include the dielectric constant (k) of the material: C = kε0A/d.

5. How can the capacitance of a parallel plate capacitor be increased?

The capacitance of a parallel plate capacitor can be increased by increasing the surface area of the plates, decreasing the distance between the plates, or using a material with a higher dielectric constant between the plates. Additionally, connecting multiple capacitors in parallel can increase the overall capacitance.

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