252 Chapter 7. Entropic forces at work[[Student version, January 17, 2003]]
Tofix the constant of integration we have noted that the electric field is zero at the midplane, and
c+(0) =c 0 there.
Next we need the free energy density per unit area in the gap. You found the free energy density
of an inhomogeneous ideal gas (or solution) in Your Turn 6k on page 209. The free energy for our
problem is this quantity, plus the electrostatic energy^10 of the two negatively charged plates at
x=±D:
F/(kBT·area) =−1
2
σq
e(V(D)+V(−D)) +
∫D
−Ddx[
c+ln
c+
c∗+
1
2
c+V]
.
In this formulac∗is a constant whose value will drop out of our final answer (see Your Turn 6k).
Wesimplify our expression by first noting that ln(c+/c∗)=ln(c 0 /c∗)−V,sothe terms in square
brackets arec+ln(c 0 /c∗)−^12 c+V. The first of these terms is a constant timesc+,soits integral
is 2(σq/e)ln(c 0 /c∗). To simplify the second term, use the Poisson–Boltzmann equation to write
c+=− 4 π^1 Bd
(^2) V
dx^2 .Next integrate by parts, obtaining
F/(kBT·area) = 2
σq
e
[
ln
c 0
c∗−
1
2
V(D)
]
+
1
8 π BdV
dxV
∣∣
∣∣
D−D−
1
8 π B∫D
−Ddx(
dV
dx) 2
.
Weevaluate the boundary terms using Equation 7.24 atx=−Dand its analog on the other surface;
they equal−σeqV(D).
Todo remaining integral, recall Equation 7.41: it’s−
∫D
−Ddx(c+−c^0 ), or 2(Dc^0 −(σq/e)).
Combining these results gives
F/(kBT·area) = 2Dc 0 +2
σq
e(
ln
c 0
c∗−V(D)− 1
)
=const + 2Dc 0 +2
σq
e
ln
c+(D)
c∗.
The concentration at the wall can again be found from Equations 7.41 and 7.24: c+(D)=c 0 +
1
8 πB(dV/dx)
(^2) =c 0 +2π B(σq/e) (^2).
Afew abbreviations will make for shorter formulas. Letγ=2π Bσq/e,u=βD,whereβ=
√
2 π Bc 0 as before. Thenuandβdepend on the gap spacing, whereasγdoes not. With these
abbreviations
F/(kBT·area) = 2Dc 0 +
γ
π B
ln
c 0 +γ^2 /(2π B)
c∗
.
Wewantto compute the derivative of this expression with respect to the gap spacing, holdingσq
(and henceγ)fixed. We find
p
kBT=−
1
kBTd(
F/(kBT·area))
d(2D)
=−c 0 −(
D+
γ
2 π Bc 0 +γ^2)
dc 0
dD.
In the last term we need
dc 0
dD
= d
dD(
u^2
D^22 π B)
= u
π BD^3(
Ddu
dD
−u)
.
Tofind du/dDwewrite the boundary condition (Equation 7.29) asγD=utanuand differen-
tiate to find
du
dD=
γ
tanu+usec^2 u=γu
Dγ+u^2 +(Dγ)^2.(^10) Notice that adding any constant toVleaves this formula unchanged, because the integral∫c+dx=σq/eis a
constant, by charge neutrality. To understand the reason for the factor 1/2 in the first and last terms, think about
twopoint chargesq 1 andq 2. Their potential energy at separationrisq 1 q 2 /(4πεr)(plus a constant). This isone
halfof the sumq 1 V 2 (r 1 )+q 2 V 1 (r 2 ). (The same factor of 1/2 also appeared in the Example on page 229.)