Biomedical Engineering Reference
In-Depth Information
The attractive force at the shearing plane is termed the zeta potential. The entire distance
from the Stern layer to the point where the electric field equilibrates is termed the electric
double layer. All of these potentials and distances are shown in Figure 7.4 .
If there are two charged surfaces that come into close contact with each other, for exam-
ple, a charged protein and an ion in solution, then the attractive force between these two
surfaces can be described by
2
ξ 1 ξ 2
ε l ε
0 Ae 2 d=λ D
F ef ðdÞ 5
ð
7
:
16
Þ
ξ
where
is the surface charge densities for the surface and the molecules within the fluid,
A is the potential area of contact between the surface and the molecules,
ε 0 is the dielectric
constant of vacuum,
λ D is the Debye length.
The Debye length quantifies the electrostatic double layer and is calculated from
ε l is the dielectric constant of the liquid, and
s
ε l ε
0 k B T
e 2 P i ðc i χ
λ D 5
ð
:
Þ
7
17
2
i Þ
10 2 23 J/K), T is the temperature (in Kelvin),
e is the charge of an electron, and c is the concentration of molecule i with a valence of
where k B
is the Boltzmann constant (1.38
3
.
Equation 7.16 is a modification of the electric double-layer formulation, taking into account
that the layer from each of the two surfaces will overlap as they approach. The summation
in Equation 7.17 takes into account the valence and concentration of particular species.
This tends to be difficult to approximate for biological specimens because it is typical that
charged surfaces (e.g., cell membranes) are composed of multiple charged species that can
have a time-varying concentration.
χ
Example
Calculate the attractive force between an endothelial cell and a red blood cell that are 3
μ
m
12 mC/m 2 and that for the red
apart. The surface charge density for the endothelial cell is
2
3 mC/m 2 . Assume that the dielectric constant of the liquid is 75 and that the
blood is flowing at a temperature of 32 C. The summation of all molecular species and valences
on these two cell types is 1.7
blood cell is
2
m 2 3 . Approximate the potential area of contact between the two
cells, using geometric constraints.
μ
Solution
First, calculate the Debye length for this interaction:
t
75 8
1
23 K
12 C 2
Nm 2
:
85 E 2
:
38 E 2
ð
273
32
ÞK
1
λ D 5
8
:
012
μ
m
5
2
m 2 3
ð
1
:
6 E 2
19
ð
1
:
7
μ
Þ
Now calculate the attractive force between the endothelial cell and the red blood cell:
2
12 m m 2
3 m m 2
2
2
Ae 2 3 μ m = 8 : 012 μ m
F ef ðdÞ 5
12 C 2
Nm 2
75 8
:
85 E 2
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