Dennis C. Prieve
June 24, 2006
DC Polarization of Planar Electrodes
Separated by Low-Dielectric Fluid: Revised
In this Note, we review the derivation of the equations used by Jun Kim et al. Langmuir 21,
8620 (2005).
Capacitance with Neutral Fluid
Consider a parallel plate capacitor consisting of two
parallel plates, located at x=0 and at x=. The fluid in between
the plates has permittivity  while the plates themselves have
permittivity 0. Suppose we apply a voltage drop V between the
two plates. What charge will accumulate on the plates at steady
state after any electrical current has subsided?
V
0
0


Assuming that E = Ex(x)ex and no free charges in any of
the media [i.e. (x) = 0], then Gauss’s equation [see (12) in
Electrostatics of Continuous Media.doc] * becomes
0H
0
x
0F
+
–
dEx
0
dx
Ex(x) = const = –V/
or
(1)
0
V
The value of the integration constant is set by the boundary conditions on the electrostatic
potential. Assuming that V>0, this means that Ex is pointing in the –x direction. From
(anti-)symmetry (or overall electroneutrality), we expect that equal but opposite surface charge
densities arise on the two surfaces:
– = –+
– > 0 and + < 0 if V>0. The electric field outside the two surfaces vanishes according to
1H
EofCM-(8) and (9):
for x<0 or x>:
Ex(x) = 0
The magnitude of the charge density is related to the electric field strength by EofCM-(7):
2H
     

Ex  

 
2
2

Eliminating Ex between (1) and (2):
4H
5H
 
V

* Files referred to in this document can be found online at http://www.andrew.cmu.edu/user/dcprieve/Notes/.
(2)
2
The capacitance of the device is defined as the charge separated per unit area per unit voltage
applied:


c  
V

Capacitance with Thin Double Layers
Jun uses the differential capacitance calculated from the Gouy-Chapman
model. The potential profile in the Gouy-Chapman model is sketched in the
figure at right. The usual definition of zeta potential is the difference in
electrostatic potential of the inner edge (adjacent to the solid surface) of the
diffuse layer and the electrically neutral solution outside the diffuse layer.
So, as drawn at right, the zeta potential is negative: the surface is more
negative than the solution. The surface charge density in the GouyChapman model is given by [see (16) from Electrohydrodynamics.doc]:
–1
–
3H

2 kT
 ze 
sinh 

ze
 2kT 
(3)
Note: the charge appearing in this equation is the charge borne by the solid (possibly from
electrons a metal). The diffuse cloud has an equal but oppose charge density. The differential
capacitance is defined as
cdl 
d
ze 2kT
 ze

cosh 
d  2kT ze
 2kT

 ze 
   cosh 

 c0
 2kT 
(4)
dl
In particular, Jun used the differential capacitance corresponding to   0:
0
cdl
 
(5)
Comparing (5) with (3), we see that the differential capacitance from the Gouy-Chapman theory
corresponds to an apparent plate spacing equal to the Debye length:  = –1.
6H
7H
The rationale for using the differential capacitance to interpret Jun’s experiments is not
obvious, although it seems like a good guess. Below we develop a model for slow changes in
current during polarization of the double layers.
3
Slow Polarization of Thin Double Layers
In the absence of current, each electrode
might have a different charge and a different zeta
potential. The blue curve in the figure at right
represents the profile of electrostatic potential for
open-circuit case (i.e. the electrodes are not
connected to each other). Notice then that any
difference in zeta potential between the two
electrodes leads to a nonzero voltage difference
between the electrodes themselves.

–1
–1
+––
––
–+
––
–+
Instead of an open circuit, if we short-circuit
the two electrodes to each other, we force their
potentials to be equal. We then alter the
potential profile to what is seen at left. This
causes a nonzero potential drop across the
electrically neutral solution outside either double
layer which, in turn, will generate a current
through the solution. From Ohm’s law, the
current density (current per unit area) is
x
  
ix  KEx  K 

where we have assumed the diffuse clouds are very thin compared to the electrode spacing  or
>>1. Reality check: in the two drawings above, we assumed that both zetas are negative and
+ is less negative than –. That leads to the negative slope in the potential profile and a positive
current in the x-direction. Using these same inequalities in the equation above, we obtain ix > 0.
Now suppose we externally apply a voltage
difference between the two electrodes: say we
raise the voltage of the right electrode by V
relative that that of the left electrode. Of course
this causes the voltage drop across the solution to
change by V and leads to current given by
    V
ix  K 

+–––V
–+
––
(6)
In the absence of any Faradaic reaction to
–
convert ionic charge to electronic charge, a
positive electric current causes positive charges
to build up next to the right electrode at a rate given by
d 
 ix
dt
V
x
+
(7)
4
and negative charges to build up next to the left electrode at a rate given by
d 
 ix
dt
(8)
d     
0
dt
Adding (7) and (8) we obtain
12H
so the system as a whole remains electrically neutral although charges are separated. Current
through the solution causes the charges and potential drop across the diffuse layer next to either
electrode to change with time. Integrating (7) and equating the result with (3) after reversing the
sign because now  denotes the charge density of the diffuse cloud (not the solid surface):
12H
9H
t
  t     0    ix  t   dt   
0
2kT
 ze

 0     t  
sinh 
ze
 2kT

Differentiating with respect to t:
ix  
2kT
 ze 
cosh 
ze
 2kT
 ze d  
 ze   d  
 2kT dt   cosh  2kT  dt



(9)
Similarly for (8):
t
  t     0    ix  t   dt   
0
and
ix  
2kT
 ze

 0     t  
sinh 
ze
 2kT

2kT
 ze   ze d  
 ze 
cosh 
  cosh 

ze
 2kT  2kT dt
 2kT
 d 
 dt

(10)
Eliminating ix between (9) and (10):
 ze   d  
 ze   d  
 cosh 
 cosh 


 2kT  dt
 2kT  dt
(11)
Recalling that cosh>0 for all real arguments, we see that the direction of changes in the two zeta
potentials are opposite. Furthermore, the changes in the two zetas are not necessarily equal but
opposite in sign. Substituting (9) and (10) into (6):
    V
 ze   d  
 ze   d  
 cosh 
  cosh 
K 



 2kT  dt
 2kT  dt
Dividing by 2K/:
5
 ze   d  
 ze   d         V
 cosh 
  cosh 



2
 2kT  dt
 2kT  dt

where

2K
Multiply by ze/2kT:
 cosh y
dy
dy
y  y  2v
  cosh y   
dt
dt
2
(12)
which represent two coupled ODE’s in the two zeta potentials, where
y 
ze 
2kT
and
v
zeV
4kT
Special Case: +0 = –0 = 0
If the initial zeta potentials are equal to zero, then (11) predicts the initial rates of change in
zetas will be equal in absolute magnitude but opposite in sign. They will grow at equal rates but
with opposite signs. Subsequent values of the two cosh’s in (11) will be equal and cancel out,
leaving
d 
d
 
dt
dt
at all times. Integrating while taking + = –– = 0 at t = 0:
+ = –– for all t
So
+ – – = 2+ = –2–
and the two ODE’s of (12) can be decoupled into a single ODE:
dy
 y  v
dt 
(13)
dy
  y  v
dt 
(14)
 cosh y
or
where
Let
Then either (13) or (14) becomes
cosh y
t 
t

u = y– + v = –y+ + v
(15)
6
cosh  u  v 
du
u  0
dt 
(16)
Integrating this separable FOODE using u(0) = v as the initial condition yields
v
cosh  u  v 
u  t 
u

tm
im  j
At right is the solution u(t') to the above
integral for various values of v = u(0). The
initial rate can be deduced directly from (16)
since at this instant cosh(u-v) = 1:
du  t 
 ts
v = 100
100
um
N im  j
im  j
 u

 j
 j
ufn ( t  j)  SplineFit tm  um  t
N im  j
Current vs time for various v
10
du
 u  0   v
dt  t   0
or
or
(17)
1
d ln u
 1
dt  t   0
0.1
d log u
 2.303
dt  t   0
v = 0.1
0.01
0
2
4
6
8
10
t/tau
Notice that for v<<1, the curves are linear on
Figure 1
these semilog coordinates and the slope is
-2.303 as expected. For v>>1, this value of
the initial slope is not apparent, but the initial value predicted by (17) must be true: in particular
for v = 100, the curve appears to be completely horizontal, suggesting a slope of zero. The
reason for this is that the slope for v = 100 decays very rapidly from its initial value as u becomes
nonzero and cosh grows much larger than unity.
Initial Rate of Decay
We can estimate the time for the initial slope to differ significantly from (17). To be
specific, suppose we wish to know how long it takes for the slope to change by 10%. Dividing
(16) by itself evaluated at t=0:
cosh  u  v 
du
dt  t 
du
1 
dt t   0

u  t  
u  0 
or
du
u t 
dt  t 

 0.9
du
u  0  cosh u  t    v 
dt  t   0
There are two limiting case. For v<<1, the cosh does not differ significantly from unity,
regardless of the t. For this same case, we have

7
u  t   u  0 et 
for v<<1:
and
0
du
du

e t 
dt  t  dt  t   0
Then the time for a 10% drop in rate is simply –ln(0.9) = 0.105.
t
Initial Decay
Initial Rate of Decay
0.1
0.08
v  0.1
J
0.085
du/dt
u(t)
0.095
0.09
0.085
0.08
 vJ 0.9
0.09
0.095
0
0.05
0.1
0.15
0.1
0.2
0
0.05
t/tau
0.1
0.15
0.2
t/tau
Numerical
Exponential Approximation
Numerical
Above we see how u and du/dt actually decay at short times. Note that the initial rate is –u(0) =
–v (as expected) and that the rate decays to 90% of its initial value in about t' = 0.1 as expected.
In the opposite limit of v>>1, we see (from the top curve in the previous figure) that u itself
does not decrease noticeably before the initial rate almost vanishes. In this case, it is the growth
of the cosh that is reducing the rate. Let z = u – v, du = dz and approximate u = z + v  v; (16)
becomes
cosh z
dz
v 0
dt 
 cosh z  dz  vdt 
or
Integrating both sizes, taking z = 0 at t' = 0:
sinh z = –v t'
or
or

z   ln  vt  

 vt 2  1   vt   16  vt  3  403  vt  5  O  vt  7 


u  z  v  v  ln  vt  

 vt  2  1 

0
1
2
3
0
2·10 -3
4·10 -3
6·10 -3
4
5
6
7
8
9
8·10 -3
0.01
0.012
0.014
0.016
0.018
10
11
12
13
14
0.02
0.022
0.024
0.026
0.028
15
0.03
8
u
1 
 1  ln  vt  
v
v 
or
 vt  2  1 

Since we approximated z + v  v, this is only good when z << v and v >> 1.
Returning to our estimate of how long it takes for the slope to decay to 90% of its initial
value: For small values of its argument, cosh can be approximated as
cosh z  1 
 
z2
 O z4
2
1
 0.9
cosh z
Then
implies
z  u  v  u  0 
But
 
1
z2
 1
 O z4
cosh z
2
and
z  0.2  0.447
du
t   u  0   u  0  t 
dt  t   0
u  0 
v
u  t 
z  u  0 t   0.447
Solving
t 
for v>>1:
for t' yields
0.447
u  0
For u(0) = 100, this yields t' = 0.0045, which is considerably shorter than t' = 0.1 at the other
extreme.
Initial Decay
Initial Rate of Decay
70
v  100
J
99.8
80
u(t)
du/dt
99.6
 vJ 0.9
90
99.4
100
99.2
99
0
0.002
0.004
0.006
0.008
110
0
0.002
t/tau
Numerical
Exponential Approximation
0.004
0.006
t/tau
Numerical
0.008
9
Above we see how u and du/dt decay at short times for large v (v = 100). Note that the initial
rate is –u(0) = –v (as expected) and that the rate decays to 90% of its initial value in about t' =
0.0045, also as expected.
Uncertainty in Initial Current
Owing to the finite rate of sampling of current, the first sample might not correspond to t' =
0. A more useful “initial” condition is to prescribe u(t'1) (which must be smaller in magnitude
than v but of the same sign), where t'1 corresponds to the first sample. which leads to
u  t1 

cosh  u  v 
u  t 
u
du  t   t1
(18)
Notice that we did not replace the v in the integrand, because this represents the true initial value
of u which must be used even if it is not measurable. Once u(t) is known, we are most interested
in predicting the evolution of current. Nondimensionalizing (6) taking + = –:
ix  t   K
2   V 4kTK
4kTK

u t 
  y  v   

ze
ze
Normalizing by the current measured at the first sampling point:
ix  t 
known:
ix  t1 

u t 
 s  t  
u  t1 
(19)
This quantity and t' (defined as t' – t'1) are known experimentally.
v
 s0
u  t1 
Unknown is
Then
u  v  su  t1   s0u  t1    s  s0  u  t1    s  s0 
u
v
Normalizing all of the u's by u(t'1), (18) becomes
1

s  t  
 s
 
cosh   1 v 
 
 s0
ds  t 
s
Generally we know the value of v, but not s0 (which must be 1).

v  s
   1 v
s0  s0

10
if s(t' ) = s0 then
Current vs time for various s0
2
0
0.278
1.746
11.35
1
0.5
t'1 = –t'
20
10
0
Thus we determined the
following relationship summar–
ized by the table at left. Keep in
mind that this relationship only applies for v = 9.735.
To confirm that we haven’t made any errors,
we test the above results by realizing that all
of the above curves should map into a single
universal curve corresponding to that curve in
Figure 1 on page 6 having v = 9.735. The
mapping is as follows: the ordinate of the
universal plot is the current normalized by the
true initial current. So instead of plotting s
defined by (19) we renormalize to
u  t 
u  t   u  t1  s  t  


u  0
v u  t1 
s0  t  
To convert the abscissa (t' ) of the above plot
into that of the universal plot (t' ), we use the
definition of t'; thus we obtain
10
20
30
40
50
dtprime = (t-t1)/tau
Renormalized Current vs t for various s0
1
0
0.8
0.6
s/s0
t'1
 1 
1.2 
s0  
 1.5 
 2 
 
1
0
s0
1
1.2
1.5
2
0
1.5
s(dtprime) = I(t)/I(t1)
The plot at right shows the effect of
uncertainty in the initial current for the case of
v = 9.735 (V = 1 volt). The lowest curve
corresponds to measuring the true initial
current (i.e. s0 = 1). The higher curves
correspond to larger values of s0 and more
offset in the time (i.e. greater values of t'1).
We have extended the calculations to t' < 0
to see when s(t' ) = s0. The corresponding
value of t' tells us how large t'1 is:
0.4
0.2
0
0
10
20
30
40
50
t/tau
s0 = 1
s0 = 2
t' = t' + t'1
Through this mapping, the figure above is transformed into the figure at right. Although only the
renormalized results for s0 = 2 are shown (as the blue points), all four curves map into one
universal curve (see Polarization Model3.mcd).
Test “Point Charge” Assumption
Jun et al. point out that the nonlinear analysis is not valid because very early in the
polarization process, the volume fraction of ions in the diffuse layer becomes comparable to
unity, whereas the Gouy-Chapman theory assumes “point charges” — in other words, zero
x
11
volume fraction. A major conclusion from Jun’s paper is that the charge carriers for OLOA in
dodecane are unicharged (z=1) micelles whose diameter is about 20 nm, independent of OLOA
concentration. A closed packed array of these spheres corresponds to a concentration of
Cmax 
1
 20 nm 3
 0.21 mM
In 0.5 wt% OLOA in dodecane, Jun determined the bulk concentration of charge carriers to be
210–5 mM which is 104 times smaller than Cmax, so there’s no problem in the bulk. But, as the
electrodes begin to polarize, much higher concentrations of carriers can arise next to the
polarized electrodes. In the Gouy-Chapman model, the local concentration of counterions is
related to the local electrostatic potential by Boltzmann’s equation:
 ze  
Cwall  Cbulk exp 

 kT 
Taking Cbulk to equal 210–5 mM and Cwall to equal Cmax, we can calculate the zeta potential
leading to closed-packed conditions:
e
C
 ln wall  9.2
kT
Cbulk
or
 
kT Cwall
ln
 0.2 volt
e
Cbulk
Unless the applied voltage is very small compared to twice this value, we really can’t expect the
Gouy-Chapman model to hold throughout the polarization process. More generally, we must
restrict our analysis to conditions in which the dimensionless zeta potential remains much
smaller than this. According to the nondimensionalization of (15), we need
for 0.5 wt% OLOA:
u  v  y  y 
e
2kT
4.6
At higher concentrations of OLOA, this constraint will be tighter because Cbulk is higher.
Special Case: +0 = –0
If the initial zeta potentials are equal, then (11) predicts the initial rates of change in zetas
will be equal in absolute magnitude but opposite in sign. Subsequent changes in the two zetas
will not be equal in magnitude nor sign: one increases and one decreases. Thus we really do
have to solve two coupled ODE’s — not one. So this particular special case affords no
simplification. Solving (12) for each of the two derivatives:
dy
y  y  2v
 
dt 
2cosh y
(20)
Current
Current
15
12
15
dy y  y  2v

dt 
2cosh y
(21)
20
20
8
6
2
0
1.5
1
0.5
0
Below is the
solution
for three
different
initial2zeta potentials (assumed
equal4and denoted
by
y0). 10
Note that when both initial
deduced above.
t/tau
t/tauzeta potentials are zero, y+ = –y– for all time as
When the initial zeta potentials are equal but nonzero, the initial rates are equal but of opposite
0 potentials tend to ±5,
y0 =zeta
y0 = 0evolution is not symmetric. At long times, both
sign but subsequent
y0 = 2
= 2 value.
y0initial
regardless of the
y0 = 3
y0 = 3
5
0
0
5
D'less Zetas
D'less Zetas
5
0
1
0.5
1.5
2
0
0
5
0
8
10
yplus for y0=0
yminus for y0=0
yplus for y0=2
yminus for y0=2
yplus for y0=3
yminus for y0=3
yplus for y0=0
yminus for y0=0
yplus for y0=2
yminus for y0=2
yplus for y0=3
yminus for y0=3
Nondimensionalizing:
6
4
t/tau
t/tau
The current is given by (6):
2
    V
ix  K 

zeix
 y  y  2v
2kTK
(22)
ze dix dy dy


2kTK dt 
dt 
dt 
(23)
Differentiating with respect to t':
Assuming that the surfaces are charged at t=0 with
at t' = 0:
y+ = y– = y0
(24)
13
then (20) and (21) yield
dy
dy
v
 

dt  t   0
dt  t   0 cosh y0
ze dix
2v

2kTK dt  t   0 cosh y0
and (23) yields
(25)
With the initial condition prescribed by (24), the initial current calculated from (22) is:
ze
i x  0   2 v
2kTK
or in terms of symbols having units:*
ix  0   
(26)
KV

1 dix
1

ix  0  dt  t   0 cosh y0
Dividing (25) by (26):
or in terms of symbols having units:
1 dix
1
2 K 
2 K 



ix  0  dt t  0  cosh y0  cosh y0
cdl
Thus any nonzero initial charge on the electrodes will decrease (perhaps dramatically) the initial
rate of change of current (because cosh in the denominator becomes larger than unity). Recall
from (4) that  coshy0 is the differential capacitance cdl of the double layer. A nonzero charge
on the surfaces, increases the differential capacitance and decreases the initial rate of decay of
current. Interpreting the initial rate of decay of current assuming y0 is zero when it is not will
cause  to be overestimated which, in turn, leads to an overestimate in the ionic strength.
* Recall that i is current density (i.e. current per unit area).
x