Electric field of a plane with a one-dimensional rectangular-periodic charge density
Markus Deserno
Max-Planck-Institut für Polymerforschung, Ackermannweg 10, 55128 Mainz, Germany
(Dated: January 12, 2005)
A charge density with one-dimensional periodicity can be expanded in a simple Fourier series.
Since for each mode the electric field can be calculated, the complete field and it’s square follow
readily. For the special case of “rectangular wave” (ut) the resulting polarization force and its
lateral average are calculated explicitly. To leading order (relevant for the far field) they decay
exponentially with a characteristic length equal to the rectangular wavelength divided by 4π.
Consider a periodic one-dimensional charge density in
the xy-plane, given by σk (x) = σ0 cos(kx), where σ0 is
the amplitude and k = 2π
λ the wave vector of the periodic
charge modulation (of wavelength λ). The electric field
E k at position (x, y, z)> above the plane is given by
1
x − x̄
dx̄ dȳ σ(x̄)
1
@ ȳ A
E k (x, y, z) =
4πε0 εr −∞ [(x − x̄)2 + ȳ 2 + z 2 ]3/2
z
0
1
Z ∞
−ξ
cos[k(ξ + x)] @
σ0
0 A
dξ
=
2πε0 εr −∞
ξ2 + z2
z
0
1
sin(kx)
σ0 −kz @
A .
0
e
=
(1)
2ε0 εr
cos(kx)
Z
0
∞
A charge density that is also periodic with wavelength
λ, but not a simple cosine function, can be expanded in
a Fourier cosine series in the following way:
σ(x) =
∞
X
σn cos(nkx) .
n=1
The corresponding electric field is then given by


∞
sin(nkx)
X
1
 .
0
E(x, y, z) =
σn e−nkz 
2ε0 εr n=1
cos(nkx)
(2)
Let us look at the example of a rectangular-periodic
charge density of amplitude σ0 . It’s Fourier expansion is
∞
4X
cos[(2n + 1)kx]
σut (x) = σ0
.
(−1)n
π n=0
2n + 1
(3)
Combining Eqns. (2,3), we obtain the electric field
E(x, y, z) =
0
1
∞
sin[(2n + 1)kx]
2σ0 X
e−(2n+1)kz @
A.
0
(−1)n
πε0 εr n=0
2n + 1
cos[(2n + 1)kx]
This expresses the result as a kind of “Laplace series”.
For large distances z the first term in the series is the
most important one, and up to the understandable prefactor 4/π it coincides exactly with the field of a single cosine mode of wavelength λ, see Eqn. (1). In other words:
The far field of a rectangular-periodic charge density is
4/π times the far field of a cosine-periodic charge density
of the same wavelength.
With the help of MathematicaTM , the Laplace series
can be written in a closed form:


sin(kx)
artanh cosh(kz)
4
σ0


0
E(x, y, z) =
×
×
 .
π 4ε0 εr
cos(kx)
arctan sinh(kz)
Assume that there’s a (point-sized) object of (scalar)
polarizability α at position (x, y, z)> above the plane. It
will develop a polarization P = αE and will thus have
an electrostatic energy
Z
1
E = − P · dE = − αE 2 .
2
The force in z direction on that object is thus given by
∂E(x, y, z)
F (x, y, z) = −
ez
∂z
„
«2
ˆ
˜−1
1
4
σ0
= − kα
× cos(2kx) + cosh(2kz)
×
×
2
π
2ε0 εr

cos(kx)
cos(kx) cosh(kz)
arctan
sinh(kz)
ff
sin(kx)
+ artanh
sin(kx) sinh(kz)
ez .
cosh(kz)
Unfortunately, it is hard to see what happens to this
expression after averaging over all x-positions. However,
we can first expand it for large kz, by which we obtain
«2 
4
4 σ0
× e−2kz − cos(2kx)e−4kz
π 2ε0 εr
3
ff
h1
i
6
−6kz
+ + cos(4kx) e
∓ · · · ez .
(4)
3
5
kzÀ1
F (x, y, z) = −kα
„
The leading order is independent of x and decays like
e−2kz , just as one would expect it from the leading order
electric field. The expansion (4) can now be averaged
over x quite easily. In fact, it turns out that the nonvanishing terms are very simple, hence the expansion can
be re-summed to obtain the exact result in closed form:
¶2 X
µ
∞
e−2(2n+1)kz
4 σ0
×
ez
hF (x, y, z)i = −kα
π 2ε0 εr
2n + 1
n=0
µ
¶2
4 σ0
−2kz
= −kα
× artanh
{ze } ez . (5)
|
π 2ε0 εr
= 21 log coth ekz