250 Chapter 7. Entropic forces at work[[Student version, January 17, 2003]]
7.4.2′ The formulas above are special cases of the general Gauss law, which states that
∫
E·dS=q/ε.
In this formula the integral is over any closed surface. The symbol dSrepresents a directed area
element of the surface; it is defined asˆndS,where dSis the element’s area andnˆis the outward-
pointing vector perpendicular to the element.qis the total charge enclosed by the surface. Applying
this formula to the two small boxes shown in Figure 7.7 on page 232 yields Equations 7.20 and 7.18,
respectively.
7.4.3′The solution Equation 7.25 has a disturbing feature: The potential goes to infinity far from
the surface! It’s true that physical quantities like the electric field and concentration profile are well
behaved (see Your Turn 7f), but still this pathology hints that we have missed something. For one
thing, no macromolecule is really an infinite plane. But a more important and interesting omission
from our analysis is the fact that any real solution has at least some coions; the concentrationc∞
of salt in the surrounding water is never exactly zero.
Rather than introducing the unknown parameterc 0 and then going back to set it, this time we’ll
choose the constant inV(x)sothatV→ 0 farfrom the surface; then the Boltzmann distribution
reads
c+(x)=c∞e−eV(x)/kBT and c−(x)=c∞e−(−e)V(x)/kBT
for the counterions and coions respectively. The corresponding Poisson–Boltzmann equation is
d^2 V
dx^2
=−^12 λ−D^2[
e−V−eV]
, (7.33)
where againV=eV /kBTandλDis defined as
λD≡(8π Bc∞)−^1 /^2. Debye screening length (7.34)In a solution of table salt, withc=0. 1 M,the screening length is about one nanometer.
The solutions to Equation 7.33 are not elementary functions (they’re called “elliptic functions”),
but once again we get lucky for the case of an isolated surface.
Your Turn 7i
Check that
V(x)=−2ln
1+e−(x+x∗)/λD
1 −e−(x+x∗)/λD(7.35)
solves the equation. In this formulax∗is any constant. [Hint: It saves some writing if you
define a new variable,ζ=e−(x+x∗)/λDand rephrase the Poisson–Boltzmann equation in terms
ofζ,notx.]Before we can use Equation 7.35, we still need to impose the surface boundary condition. Equa-
tion 7.24 fixesx∗,via
ex∗/λD= e
2 π BλDσq(
1+
√
1+(2π BλDσq/e)^2)
. (7.36)
Your Turn 7j
Suppose we only want the answer at distances less than some fixedxmax. Show that at low
enough salt concentration (big enoughλD)the solution Equation 7.35 becomes a constant plus
our earlier result, Equation 7.25. How big mustλDbe?