Are Human Cell Walls Capacitors?

[doggydan42]

Homework Statement

Some cell walls in the human body have a layer of negative charge on the inside surface. Suppose that the
surface charge densities are $±0.50 \times 10^{-3} C/m^2$ , the cell wall is $5.0 \times 10^{-9} m$ thick, and the cell wall material has a dielectric constant of $\kappa = 5.4$ .
(a) Find the magnitude of the electric field in the wall between two charge layers.
(b) Find the potential difference between the inside and the outside of the cell. Which is at higher potential?
(c) A typical cell in the human body has volume $10^{−16} m^3$. Estimate the total electrical field energy stored in the wall of a cell of this size when assuming that the cell is spherical. (Hint: Calculate the volume of the cell wall.)

Homework Equations

$$U=\frac{1}{2} V^2C = \frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}QV = \frac{1}{2}\epsilon_0 E^2Ad
\\ C = \kappa C_0 = \kappa \epsilon_0\frac{A}{d}\text{(for parallel plates)}
\\ C = \frac{Q}{V}= 4\pi \epsilon_0\frac{R_1R_2}{R_2-R_1}$$

The Attempt at a Solution

For part a), I used the following capacitance equation and two equations from U to solve for E:
$$\frac{1}{2} V^2C=\frac{1}{2}\epsilon_0 E^2Ad$$
By plugging in the equation for C, I solved for: $E = \sqrt{\kappa}\frac{\sigma}{\epsilon_0}$ for a charge density $\sigma$. The result was $E = 1.31 \times 10^8 \frac{N}{C}$
I understand I could do something with the equation: $\frac{1}{2} V^2C = \frac{1}{2}\epsilon_0 E^2Ad$, but I was not sure what V would represent.

Thank you in Advance

 

[kuruman]

... but I was not sure what V would represent.

V represents the potential difference between the cell walls. That's the answer to (b).

$U=\frac{1}{2} V^2C = \frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}QV = \frac{1}{2}\epsilon_0 E^2Ad$.

The first three equations are valid for any capacitor, but $U=\frac{1}{2}\epsilon_0 E^2Ad$ is valid for a parallel plate capacitor. Here you have a spherical capacitor. However, you can approximate $U=\frac{1}{2}\epsilon_0 E^2\times(Volume~between~cell~walls)$ if you can justify that the electric field changes linearly when you move radially from one wall to the other. This would be the case if the wall separation is much smaller than either radius. Use the hint.

 

[doggydan42]

V represents the potential difference between the cell walls. That's the answer to (b).

The first three equations are valid for any capacitor, but $U=\frac{1}{2}\epsilon_0 E^2Ad$ is valid for a parallel plate capacitor. Here you have a spherical capacitor. However, you can approximate $U=\frac{1}{2}\epsilon_0 E^2\times(Volume~between~cell~walls)$ if you can justify that the electric field changes linearly when you move radially from one wall to the other. This would be the case if the wall separation is much smaller than either radius. Use the hint.

So for question b, would the potential difference be outer - inner, or inner - outer?

Thank you

 

[kuruman]

Actually it's the "voltage" which means the absolute value of the difference whichever way you take it.

 

[doggydan42]

So would the higher potential be between the plates, because there is no electric field outside the plates, since they have opposite charges?

Actually it's the "voltage" which means the absolute value of the difference whichever way you take it.

Also, why would you have to justify that the electric field changes linearly from wall to wall?

 

[kuruman]

So would the higher potential be between the plates, because there is no electric field outside the plates, since they have opposite charges?

Yes, although use of Gauss's Law gives a better explanation why the potential is zero outside the shells.

Also, why would you have to justify that the electric field changes linearly from wall to wall?

Sorry, I misspoke. I meant to say you have to justify that the voltage changes linearly between the walls which means that the electric field is approximated as uniform. You can see that the E field is not uniform if you draw field lines between the concentric spheres. The lines become farther apart as you move towards the outer shell which means that the magnitude of the field becomes weaker, therefore is not constant.

 

[doggydan42]

Sorry, I misspoke. I meant to say you have to justify that the voltage changes linearly between the walls which means that the electric field is approximated as uniform. You can see that the E field is not uniform if you draw field lines between the concentric spheres. The lines become farther apart as you move towards the outer shell which means that the magnitude of the field becomes weaker, therefore is not constant.

So why would you have to justify that the voltage changes linearly to be able to use:
$U=\frac{1}{2}\epsilon_0 E^2\times(Volume~between~cell~walls)$? Also, if you cannot use that equation since the voltage does not change linearly, then what would you use? From the equations given to us, the closest one to that was $\frac{1}{2}\epsilon_0 E^2 \times Ad$, though that is for parallel plates and $V=Ad$.

Thank you

 

[kuruman]

So why would you have to justify that the voltage changes linearly to be able to use ...

When the voltage changes linearly, the electric field has constant magnitude. You must justify that to a good approximation either the voltage changes linearly or that the magnitude of the E-field is constant. Look at the equation $U=\frac{1}{2}\epsilon_0 E^2\times Volume$ What value would you use for $E$ if it is different at different points between the cells? Hints: How does the separation between cells compare with the radius of either cell? Why is that relevant?

 

[doggydan42]

Oh, so the magnitude would be a good approximation, since the separation between cells is much smaller than the either radii. I found for a lower case v being volume:

$$v_1 = \frac{4}{3}\pi r_1^3, ~v_0 = \frac{4}{3}\pi (r_1-d)^3
\\v_{walls} = v_w = v_1-v_0 = \frac{4}{3}\pi(r_1^3-(r_1-d)^3$$

Since $v_1 = 10^{-16} m^3$, which is the volume with radius $r_1$, where $r_0$ is the inner radius, and $r_1$ is the outer radius bounding the cell, so

$$r_1 = \sqrt[3]{\frac{3v_1}{4\pi}} = 2.88 \times 10^{-6} m$$
And so using that radius and the thickness, d, to solve for $v_0$, I found: $v_w = 5.2 \times 10^{-19} m^3$
So $r_1 >> d \Rightarrow 2.88 \times 10^{-6} m >> 5\times 10^{-9} m$

With that, would I be able to use the equation for the energy with the volume, and set that equal to the equation that uses capacitance and potential difference?

Thank you.

 

[kuruman]

That's the idea. Good work. What are you going to use for $A$ and $d$ in $Ad$ for the volume?

 

[doggydan42]

That's the idea. Good work. What are you going to use for $A$ and $d$ in $Ad$ for the volume?

Would I not just be able to use the volume between walls, since it is an approximation? If I had to use A and d, then I would use $A = 4\pi r_0^2$, so the inner surface area, and multiply that by the thickness d.

Thank you

 

[kuruman]

You got it.

 

[doggydan42]

So I can just use the volume between walls, or do I need to use $Ad$?

Thank you

 

[kuruman]

So I can just use the volume between walls, or do I need to use $Ad$?

If $A$ and $d$ are what you said in post #11, isn't $Ad$ the volume between walls to within the approximation? Calculate $Ad$ as you indicated in post #11 and then as the difference between the volume of the outer and inner spheres and see how different the results are.

 

[doggydan42]

I see. Thank you.