Biological Physics: Energy, Information, Life

7.4. A repulsive interlude[[Student version, January 17, 2003]] 235

Your Turn 7e

Now combine the Poisson equation with the Boltzmann distribution to get

d^2 V

dx^2

=− 4 πBc 0 e−V. Poisson–Boltzmann equation (7.23)

The payoff for introducing the abbreviationsV and (^) Bis that now Equation 7.23 is less cluttered,
and we can verify at a glance that its dimensions work: Both d^2 /dx^2 and (^) Bc 0 have unitsm−^2.
Like any differential equation, Equation 7.23 has not one, but a wholefamilyof solutions; to get
aunique solution we need to specify additional information, namely some “boundary conditions”
on the unknown functionV(x). For example, it you throw a rock upward Newton’s Law says that
its heightz(t)obeys the equationd
(^2) z
dt^2 =−g.But this equation won’t tell us how high the rock will
go! We also need to specify how hard you threw the rock, or more precisely its speed and location
when it left your hand, at time zero. Similarly, we should not expect Equation 7.23 to specify the
full solution, because it doesn’t mention the surface charge density. Instead the equation has a
family of solutions; we must choose the one corresponding to the given value ofσq.
Tosee howσqenters the problem, we now apply the surface form of the Gauss Law (Equa-
tion 7.18), which gives−ddVx

∣∣

∣

surface

=−σεq,or

dV

dx

∣∣

∣∣

surface

=4πB

σq

e

. (when the allowed region isx>0) (7.24)

When using this formula, remember thatσqis a positive number; the surface has charge density
−σq.

Example How does one remember the correct sign in this formula?

Solution: Notice that the potential goes down as we approach a negative object.

Thus, approaching counterions feel their potentialeV decrease as they approach the

surface, which they like, so they’re attracted. Ifxis the distance from a negatively

charged surface, thenV will be decreasing as we approach it, or increasing as we

leave: dV/dx>0. That shows that the signs are correct in Equation 7.24.

Solution of the P–B equation Wehave reduced the problem of finding the counterion distri-
bution outside a surface to solving Equation 7.23. This is a differential equation, so we’ll need to
impose some conditions to determine a unique solution. Furthermore, the equation itself contains
an unknown constantc 0 ,which requires another condition to fix its value. The conditions are:

• The boundary condition at the surface (Equation 7.24),


• An analogous condition dV/dx=0at infinity, since no charge is located there, and


• The convention thatV(0) = 0.
  It’s usually not easy to solve nonlinear differential equations like Equation 7.23. Still, in some
  special situations we do get lucky. We need a function whose second derivative equals its exponen-
  tial. We recall that the logarithm of a power ofxhas the property that both its derivative, and its
  exponential, are powers ofx.Wedon’t wantV(x)=lnx,since that’s divergent (equal to infinity) at
  the surface. Nevertheless a slight modification gives something promising:V(x)=?Bln(1 + (x/x 0 )).