The restoring force on a dielectric
in a parallel plate capacitor
L. P. Staunton
Department of Physics & Astronomy
Drake University, Des Moines, Iowa 50311
July 11, 2011 Revised July 16, 2013
We wish to investigate the restoring force on a dielectric slab which is being pulled out from
a parallel plate capacitor connected to a battery. The treatment in most textbooks for undergraduates is either missing or completely wrong. It assumes a uniform E field with no
fringing effects and predicts a force which fails to vanish for zero displacement. In order to
obtain the correct answer it is necessary to treat the fringing effects correctly.1
To this end we use a conformal mapping solution first suggested by J. C. Maxwell.2 For this
introduction we largely follow the published lecture of Bradley.3
Let Z= x + i y, and define the analytic mapping f(Z ) = u(x,y) + i Φ(x,y) º W. The mapping
suggested by Maxwell is Z =1 + W + ãW . Then
x = 1 + u + ãucos(Φ)
and
y = Φ + ãusin(Φ)
(1)
We take partial derivatives
¶x
¶x
¶x
¶y
=1 =
¶y
¶x
¶y
¶y
=0 =
=0 =
=1 =
¶u
¶x
¶u
¶y
+
¶Φ
¶x
¶Φ
¶y
+
+
+
¶u u
ã cos(Φ) ¶x
¶u u
ã cos(Φ) ¶y
¶Φ
¶x
¶Φ
¶y
ãu sin(Φ)
(2)
ãu sin(Φ)
(3)
¶u
¶x
¶u
¶y
¶Φ
¶x
¶Φ
¶y
ãu cos(Φ)
(4)
ãu cos(Φ)
(5)
¶Φ
¶y
) ãu sin(Φ) = 0
(6)
) ãu sin(Φ) = 0
(7)
ãu sin(Φ) +
ãu sin(Φ) +
Add (3) to (4) to obtain
¶Φ
)
¶x
¶u
( ¶y
+
(1 + ãu cos(Φ)) + ( ¶u
¶x
and subtract (5) from (2) to obtain
(
¶u
¶x
-
¶Φ
¶y
Multiply (6) by (
(
¶u
¶x
-
¶Φ
¶y
¶u
) (1 + ãu cos(Φ)) - ( ¶y
+
¶u
¶x
-
¶Φ
¶y
¶u
) ( ¶y
+
¶Φ
¶x
) to obtain
¶Φ
)
¶x
and multiply (7) by ( ¶u +
¶Φ
(1 + ãu cos(Φ)) + JJ ¶u
¶x
) to obtain
¶Φ
¶y
NN ãu sin(Φ) = 0
2
2
¶u
¶x
ConformalCapacitor.nb
Multiply (6) by (
(
¶u
¶x
-
¶Φ
¶y
¶Φ
¶y
-
¶u
) ( ¶y
+
) to obtain
¶Φ
)
¶x
¶Φ
)
¶x
¶u
and multiply (7) by ( ¶y
+
¶u
( ¶y
+
¶Φ
)
¶x
(
¶u
¶x
-
(1 + ãu cos(Φ)) + JJ ¶u
¶x
¶Φ
¶y
to obtain
¶u
) (1 + ãu cos(Φ)) -J J ¶y
+
Then we subtract the 2 equations to obtain
[JJ ¶u
¶x
¶Φ
¶y
¶Φ
¶y
¶u
NN +J J ¶y
+
2
¶Φ
¶x
NN ãu sin(Φ) = 0
2
NN ãu sin(Φ) = 0
2
NN ]ãu sin(Φ) = 0
2
¶Φ
¶x
Since this must hold for all values of u and Φ, we must have
¶u
¶x
=
¶Φ
¶y
¶u
¶y
and
=-
¶Φ
¶x
(8)
These equations are known as the Cauchy-Reimann conditions. Although we have derived
them for the particular mapping above, they hold for any mapping f(Z) which is analytic,
i.e., has continuous derivatives in the complex plane. Functions u and Φ which satisfy (8)
are called harmonic functions.
Then
Ñ2 u =
¶2 u
+
2
¶x
¶2 u
2
¶y
=
¶2 Φ
¶x ¶y
¶2 Φ
- ¶y ¶x = 0,
since cross partials are equal if they
both exist and are continuous,
and also
Ñ2 Φ =
¶2 Φ
¶x2
+
¶2 Φ
¶y2
2
u
= - ¶x¶ ¶y
+
¶2 u
¶y ¶x
= 0.
Since both u and Φ satisfy Laplace's equation, each of them represents a possible choice for
an electric potential function. The difference lies in the boundary conditions satisfied. For
the parallel plate capacitor problem, the proper choice is a function proportional to Φ.
Note that x, y, u, and Φ are all dimensionless quantities. Physical quantities may be introduced via X=x ( 2dΠ ), and Y = y ( 2dΠ ), where d is in meters, and F = ( 2VΠ0 ) Φ , where V0 is in Volts. Then
E = -õF
(9)
where the gradient is to be taken with respect to X and Y. It follows that
EX = - (
V0 ¶Φ
)
d
¶x
and EY = - (
V0 ¶Φ
)
d
¶y
(10)
It is sufficient to consider values of Φ between -Π and Π. Since y = Φ + ãusin(Φ), clearly, y =
0 for Φ = 0 for any value of u. For Φ = Π, y = Π for any value of u, and for Φ = -Π, y = -Π
for any value of u also. Define
ConformalCapacitor.nb
In[1]:=
x@u_, Φ_D := 1 + u + ãu Cos@ΦD;
y@u_, Φ_D := Φ + ãu Sin@ΦD;
Plot@x@u, ΠD, 8u, - 3, 3<, AxesLabel ® 8u, x<, PlotStyle ® 8RGBColor@0, 0, 0D<D
x
u
-3
-2
1
-1
2
3
-2
-4
-6
-8
Plot@x@u, - ΠD, 8u, - 3, 3<, AxesLabel ® 8u, x<, PlotStyle ® 8RGBColor@0, 0, 0D<D
x
u
-3
-2
1
-1
2
3
-2
-4
-6
-8
At Φ = y = ±Π, x varies from -¥ to zero for all values of u, and is zero only at u = 0.
3
4
ConformalCapacitor.nb
ParametricPlot@88x@u, - ΠD, y@u, - ΠD<, 8x@u, ΠD, y@u, ΠD<<, 8u, - 8, 0<,
PlotRange ® 8- 7, 7<, AxesLabel ® 8x, y<, PlotStyle ® 8RGBColor@0, 0, 0D<D
y
6
4
2
x
-6
-4
2
-2
4
6
-2
-4
-6
It is clear then that F = ( 2VΠ0 ) Φ is the electrical potential function for a semi - infinite parallel
plate capacitor whose plates are separated by a distance d, and are held at ± ( V20 ) . The uniqueness theorem guaranties that F is the only solution. A sampling of lines of constant Φ is
displayed via
ConformalCapacitor.nb
In[7]:=
g1 = ParametricPlot@
88x@u, - ΠD, y@u, - ΠD<, 8x@u, - 0.8 ΠD, y@u, - 0.8 ΠD<, 8x@u, - 0.6 ΠD, y@u, - 0.6 ΠD<,
8x@u, - 0.4 ΠD, y@u, - 0.4 ΠD<, 8x@u, - 0.2 ΠD, y@u, - 0.2 ΠD<, 8x@u, 0D, y@u, 0D<,
8x@u, 0.2 ΠD, y@u, 0.2 ΠD<, 8x@u, 0.4 ΠD, y@u, 0.4 ΠD<, 8x@u, 0.6 ΠD, y@u, 0.6 ΠD<,
8x@u, 0.8 ΠD, y@u, 0.8 ΠD<, 8x@u, ΠD, y@u, ΠD<<, 8u, - 5.5, 1<,
PlotRange ® 8- 7, 7<, AxesLabel ® 8x, y<, PlotStyle ® 8RGBColor@0, 0, 0D<D
y
6
4
2
Out[7]=
x
-6
-4
2
-2
4
6
-2
-4
-6
Now, from (8),
so that
¶u
¶x
¶Φ ¶Φ
¶y ¶x
õu =
õu ·õΦ =
¶u
j
¶y
¶Φ ¶Φ
¶x ¶y
¶Φ
¶y
i+
=
-
º0
i-
¶Φ
¶x
j,
and
õΦ =
¶Φ
¶x
i+
¶Φ
j
¶y
which means that lines of constant u are everywhere perpendicular to lines of constant Φ
wherever they cross. It follows that since lines of constant F are equipotentials, lines of
constant u are proportional to electric field lines of force. They are exhibited via
5
6
ConformalCapacitor.nb
In[8]:=
g2 = ParametricPlot@88x@1, ΦD, y@1, ΦD<, 8x@0.5, ΦD, y@0.5, ΦD<, 8x@0, ΦD, y@0, ΦD<,
8x@- 0.5, ΦD, y@- 0.5, ΦD<, 8x@- 1, ΦD, y@- 1, ΦD<, 8x@- 1.5, ΦD, y@- 1.5, ΦD<,
8x@- 2, ΦD, y@- 2, ΦD<, 8x@- 2.5, ΦD, y@- 2.5, ΦD<, 8x@- 3, ΦD, y@- 3, ΦD<,
8x@- 3.5, ΦD, y@- 3.5, ΦD<, 8x@- 4, ΦD, y@- 4, ΦD<, 8x@- 4.5, ΦD, y@- 4.5, ΦD<,
8x@- 5, ΦD, y@- 5, ΦD<, 8x@- 5.5, ΦD, y@- 5.5, ΦD< 8x@- 6, ΦD, y@- 6, ΦD<<, 8Φ, - Π, Π<,
PlotRange ® 8- 7, 7<, AxesLabel ® 8x, y<, PlotStyle ® 8RGBColor@0, 0, 0D<D
y
6
4
2
Out[8]=
x
-6
-4
2
-2
-2
-4
-6
4
6
ConformalCapacitor.nb
In[9]:=
7
Show@8g1, g2<D
y
6
4
2
Out[9]=
x
-6
-4
2
-2
4
6
-2
-4
-6
This graph represents part of the exact solution to the (semi - infinite) parallel plate capacitor
and reproduces Maxwell’s result.4 Note carefully that the electric field lines are essentially
vertical for x b -4. Therefore we may approximate a finite capacitor by adjoining a mirror
image graph to the left. The gap shown below represents the intermediate part which may
be as wide as desired so as to minimize the error.
8
ConformalCapacitor.nb
In[10]:=
Out[13]=
shiftX = - 10;
g3 = ParametricPlot@88- x@u, - ΠD + shiftX, y@u, - ΠD<,
8- x@u, - 0.8 ΠD + shiftX, y@u, - 0.8 ΠD<, 8- x@u, - 0.6 ΠD + shiftX, y@u, - 0.6 ΠD<,
8- x@u, - 0.4 ΠD + shiftX, y@u, - 0.4 ΠD<, 8- x@u, - 0.2 ΠD + shiftX, y@u, - 0.2 ΠD<,
8- x@u, 0D + shiftX, y@u, 0D<, 8- x@u, 0.2 ΠD + shiftX, y@u, 0.2 ΠD<,
8- x@u, 0.4 ΠD + shiftX, y@u, 0.4 ΠD<, 8- x@u, 0.6 ΠD + shiftX, y@u, 0.6 ΠD<,
8- x@u, 0.8 ΠD + shiftX, y@u, 0.8 ΠD<, 8- x@u, ΠD + shiftX, y@u, ΠD<<, 8u, - 5.5, 1<,
PlotRange ® 88- 15, - 5<, 8- 7, 7<<, Axes ® None, PlotStyle ® 8RGBColor@0, 0, 0D<D;
g4 = ParametricPlot@88- x@1, ΦD + shiftX, y@1, ΦD<,
8- x@0.5, ΦD + shiftX, y@0.5, ΦD<, 8- x@0, ΦD + shiftX, y@0, ΦD<,
8- x@- 0.5, ΦD + shiftX, y@- 0.5, ΦD<, 8- x@- 1, ΦD + shiftX, y@- 1, ΦD<,
8- x@- 1.5, ΦD + shiftX, y@- 1.5, ΦD<, 8- x@- 2, ΦD + shiftX, y@- 2, ΦD<,
8- x@- 2.5, ΦD + shiftX, y@- 2.5, ΦD<, 8- x@- 3, ΦD + shiftX, y@- 3, ΦD<,
8- x@- 3.5, ΦD + shiftX, y@- 3.5, ΦD<, 8- x@- 4, ΦD + shiftX, y@- 4, ΦD<,
8- x@- 4.5, ΦD + shiftX, y@- 4.5, ΦD<, 8- x@- 5, ΦD + shiftX, y@- 5, ΦD<,
8- x@- 5.5, ΦD + shiftX, y@- 5.5, ΦD< 8- x@- 6, ΦD + shiftX, y@- 6, ΦD<<, 8Φ, - Π, Π<,
PlotRange ® 88- 15, - 5<, 8- 7, 7<<, Axes ® None, PlotStyle ® 8RGBColor@0, 0, 0D<D;
Show@8g1, g2, g3, g4<, PlotRange ® 88- 15, + 5<, 8- 7, 7<<, Axes ® NoneD
ConformalCapacitor.nb
9
We position a dielectric slab of dielectric constant Κ and very small depth L = { ( 2dþ ) in the
direction perpendicular to the figure. The dielectric is inserted between the plates half
filling the gap, i.e., for y Ε [-Π/2, Π/2] .
10
ConformalCapacitor.nb
In[14]:=
poly0 = Polygon@88- 10, - Π  2<, 80, - Π  2<, 80, Π  2<, 8- 10, Π  2<<D;
g5 = Graphics@8GrayLevel@0.8D, poly0<D;
Show@8g1, g2, g3, g4, g5<, PlotRange ® 88- 15, + 5<, 8- 7, 7<<, Axes ® NoneD
Out[16]=
If the voltage is kept constant as the dielectric is inserted, then the E field everywhere
remains exactly the same. Within the dielectric, however, the D field increases by the factor
Κ. Then the stored energy inside the dielectric, U =
1
D × E â Vol
2 Ù
, increased by the factor Κ
also. Now if the dielectric is moved a small distance Dx to the right, we have
ConformalCapacitor.nb
In[17]:=
verticalx@xvalue_, t_D := xvalue
verticaly@t_D := t
horizontalx@t_D := t
horizontaly@yvalue_, t_D := yvalue
g6 = ParametricPlot@
88verticalx@- 10, tD, verticaly@tD<, 8verticalx@- 9.5, tD, verticaly@tD<<,
8t, - Π  2, Π  2<, Axes ® None, PlotStyle ® 8RGBColor@0, 0, 0D<D;
g7 = ParametricPlot@88horizontalx@tD, horizontaly@- Π  2, tD<,
8horizontalx@tD, horizontaly@Π  2, tD<<,
8t, - 10, - 9.5<, Axes ® None, PlotStyle ® 8RGBColor@0, 0, 0D<D;
poly1 = Polygon@88- 9.5, - Π  2<, 80, - Π  2<, 80, Π  2<, 8- 9.5, Π  2<<D;
g8 = Graphics@8GrayLevel@0.8D, poly1<D;
poly2 = Polygon@880, - Π  2<, 80.5, - Π  2<, 80.5, Π  2<, 80, Π  2<<D;
g9 = Graphics@8GrayLevel@0.6D, poly2<D;
Show@8g1, g2, g3, g4, g6, g7, g8, g9<, PlotRange ® 88- 15, + 5<, 8- 7, 7<<, Axes ® NoneD
Out[27]=
11
12
ConformalCapacitor.nb
In the shaded area on the right, of width Dx, the stored energy increases by Κ. Since there
was no dielectric present before the displacement, the change in stored energy, ( final initial) is (Κ-1) times the energy originally present without the dielectric. In the rectangular
space shown boxed to the left, also of width Dx, the stored energy decreases. The change in
the energy stored there,( final - initial) is (1-Κ) times the energy present after the dielectric is
removed. There is no change anywhere else. Note that in both cases, the energy change
needed for the calculation is that which is present without the dielectric.
Now the energy originally present in any region of depth L absent a dielectric is given by
U=
1
2
Ε0 L Ù IE2X + EY2 M â X â Y
where the integration is over the area of the region in question. Since we need to calculate
the change in the stored energy, U, as the dielectric is removed, there will be no change in
the perpendicular direction and we can ignore the contribution of the perpendicular component of the electric field. We have also assumed that the plates are of a sufficiently large
depth that the field in the small depth L near the center is not a function of the perpendicular
variable.
We define the dimensionless quantities
Ε
x
=
¶Φ
¶x
Ε
and
y
=
¶Φ
¶y
(11)
Then we have
EX = - (
V0
)
d
Ε
x
and EY = - (
V0
)
d
Ε
y
(12)
so that
U=
1
2
Ε0 { V02 (
d
8 Π3
) ÙI
Ε +Ε Mâxâ y
2
x
2
y
(13)
Ε
x
=
Ε
and
¶x
y
=
(11)
¶y
ConformalCapacitor.nb
Then we have
EX = - (
V0
)
d
Ε
x
and EY = - (
V0
)
d
Ε
13
(12)
y
so that
U=
1
2
Ε0 { V02 (
d
8 Π3
) ÙI
Ε +Ε Mâxâ y
2
x
2
y
(13)
Now x and y are implicit functions of u and Φ. Then U =
1
2
Ε0 { V02 (
d
8 Π3
) ÙI
Ε + Ε M J âu âΦ
2
x
2
y
where J is the Jacobian of the transformation, and the integration is now over the corresponding region in the ( u, Φ ) plane which maps into the original region in the ( x, y ) plane.
The Jacobian of the transformation is
J=
¶x ¶y
¶u ¶Φ
-
¶x ¶y
¶Φ ¶u
= (1 + ãucos(Φ))(1 + ãucos(Φ)) + ã2 u sin2 (Φ)
= 1 + 2 ãucos(Φ) + ã2 u cos2 (Φ) + ã2 u sin2 (Φ)
J
=1 + 2 ãucos(Φ) + ã2 u
(14)
which is non-zero for all values of the parameters for which y Ε A- Π2 ,
Π
]
2
.
Now we need to calculate the integrand IΕx2 + Ε y2 )J. From (2) and (3) and (8) and (10)
1 = Εx ãu sin(Φ) - Ε y (1 + ãu cos(Φ))
0 = Ε y ãu sin(Φ) + Εx (1 + ãu cos(Φ))
(15)
(16)
Then
and
ãu sin(Φ) = Εx ã2 u sin2 (Φ) - Ε y Hãu sin(Φ) + ã2 u sin(Φ) cos(Φ))
0 = Ε y (1 + ãu cos(Φ)) ãu sin(Φ) + Εx (1 + ãu cosHΦLL2
or
and
ãu sin(Φ) = Εx ã2 u sin2 (Φ) - Ε y ãu sin(Φ) - Ε y ã2 u sin(Φ) cos(Φ)
0 = Ε y ãu sin(Φ) + Ε y ã2 u sin(Φ) cos(Φ) + Εx + 2 Εx ãu cos(Φ) + Εx ã2 u cos2 (Φ)
Addingãu sin(Φ) = Εx ã2 u sin2 (Φ) + Εx + 2 Εx ãu cos(Φ) + Εx ã2 u cos2 (Φ)
ãu sin(Φ) = Εx (1 + 2 ãu cos(Φ) + ã2 u)
or
ãu sin(Φ) = Εx J
so that
Εx 2 J 2 = ã
2u
sin2 (Φ)
(17)
Similarly, from (15) and (16)
- (1 + ãu cos(Φ)) = - Εx (1 + ãu cos(Φ)) ãu sin(Φ) + Ε y (1 + ãu cosHΦLL2
and
0 = Ε y ã2 u sin2 (Φ) + Εx ãu sin(Φ) (1 + ãu cos(Φ))
or
and
- (1 + ãu cos(Φ)) = - Εx ãu sin(Φ) -Εx ã2 u sin(Φ) cos(Φ) + Ε y (1 + ãu cosHΦLL2
0 = Ε y ã2 u sin2 (Φ) + Εx ãu sin(Φ) + Εx ã2 u sin(Φ) cos(Φ)
Adding- (1 + ãu cos(Φ)) = Ε y ã2 u sin2 (Φ) + Ε y +2 Ε y ãu cos(Φ) +
- (1 + ãu cos(Φ)) = Ε y (1 + 2 ãu cos(Φ) + ã2 u )
or
- (1 + ãu cos(Φ)) = Ε y J
2
2
u
2
Ε
y
ã2 u cos2 (Φ)
or
and
14
- (1 + ã cos(Φ)) = - Εx ã sin(Φ) -Εx ã sin(Φ) cos(Φ) + Ε y (1 + ã cosHΦLL
0 = Ε y ã2 u sin2 (Φ) + Εx ãu sin(Φ) + Εx ã2 u sin(Φ) cos(Φ)
ConformalCapacitor.nb
Adding- (1 + ãu cos(Φ)) = Ε y ã2 u sin2 (Φ) + Ε y +2 Ε y ãu cos(Φ) +
- (1 + ãu cos(Φ)) = Ε y (1 + 2 ãu cos(Φ) + ã2 u )
or
- (1 + ãu cos(Φ)) = Ε y J
so that
Ε y2 J 2 = (1 + ã
u
Ε
y
ã2 u cos2 (Φ)
cosHΦLL2
(18)
It follows from (14) and (17) and (18) that
Εx2 J 2 + Ε y2 J 2 = ã sin (Φ) + 1 + 2 ã cos(Φ) + ã
IΕx2 + Ε y2 ) J 2 = (1 + 2 ã cos(Φ) + ã ) = J
2u
2
u
u
2u
cos2 (Φ)
2u
so that at all points where J ¹ 0,
IΕx2 + Ε y2 ) J = 1
(19)
Equation (19) is the simplifying result which facilitates a simple numerical solution to the
problem.5 It follows that the electrical energy stored in any dielectric free region in the (x,y)
plane is given by
1
2
Ε0 { V02 (
d
8 Π3
)
sponding region in the (u, Φ) plane.
times the simple area integral Ù 1 â u â Φ over the corre-
Consider the first shaded region of width Dx into which the dielectric is moved. We will
perform the area integration considering u to be the independent variable and Φ to be the
dependent variable. Since the area of the lower half is equal to that of the upper half, we
will just calculate for the upper half, i.e., for y Ε [0, Π2 ] and drop the factor
1
2
in the energy
expression. We first determine the lower and upper limits of the variable u along the vertical
line at x = 0, which we shall call the mid line.
In[28]:=
Out[28]=
In[30]:=
Out[30]=
FindRoot@80 Š x@u, ΦD, y@u, ΦD Š 0<, 88u, - 1<, 8Φ, 0<<D
miduL = u . %;
8u ® - 1.27846, Φ ® 0.<
Π
midrule = FindRootB:0 Š x@u, ΦD, y@u, ΦD Š
miduH = u . midrule;
8u ® - 1.10298, Φ ® 1.2553<
2
>, :8u, - 1<, :Φ,
Π
2
>>F
Next, we generate a table of rules for 1000 points (u, Φ) along the line
In[32]:=
mid =
Π
TableBFindRootB:0 Š x@u, ΦD, y@u, ΦD Š Ξ
2
>, :8u, - 1<, :Φ, Ξ
Π
2
>>F, 8Ξ, 0, 1, 0.001<F;
Finally we define an interpolating function for Φ versus u along the line and plot it.
ConformalCapacitor.nb
In[33]:=
15
midtable = Table@8u, Φ< . midD;
midf = Interpolation@midtableD;
midplot =
Plot@midf@uD, 8u, miduL, miduH<, AxesLabel ® 8u, Φ<, PlotStyle ® 8RGBColor@0, 0, 0D<D
Φ
1.2
1.0
0.8
Out[35]=
0.6
0.4
0.2
-1.20
-1.15
u
-1.10
Now we will do the same things for the right hand vertical boundary line at x = Dx, a line we
will call right1:
In[44]:=
deltaX = 0.01;
16
ConformalCapacitor.nb
In[45]:=
FindRoot@8deltaX Š x@u, ΦD, y@u, ΦD Š 0<, 88u, - 1<, 8Φ, 0<<D
right1uL = u . %;
Π
Π
r1rule = FindRootB:deltaX Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F
2
2
right1uH = u . r1rule;
right1 = TableBFindRootB
:deltaX Š x@u, ΦD, y@u, ΦD Š Ξ
Out[45]=
Out[47]=
Π
>, :8u, - 1<, :Φ, Ξ
Π
2
2
right1table = Table@8u, Φ< . right1D;
right1f = Interpolation@right1tableD;
right1plot = Plot@right1f@uD, 8u, right1uL, right1uH<,
AxesLabel ® 8u, Φ<, PlotStyle ® 8RGBColor@0, 0, 0D<D
>>F, 8Ξ, 0, 1, 0.001<F;
8u ® - 1.27065, Φ ® 0.<
8u ® - 1.09461, Φ ® 1.2529<
Φ
1.2
1.0
0.8
Out[52]=
0.6
0.4
0.2
u
-1.20
-1.15
Finally, we do the same thing for the line at the top where y =
-1.10
Π
2
:
ConformalCapacitor.nb
In[53]:=
17
topright1 = TableB
>>F, 8Ξ, 0, deltaX, 0.001<F;
2
2
topright1table = Table@8u, Φ< . topright1D;
topright1f = Interpolation@topright1tableD;
topright1plot = Plot@topright1f@uD, 8u, miduH, right1uH<,
AxesLabel ® 8u, Φ<, PlotStyle ® 8RGBColor@0, 0, 0D<D
Π
FindRootB:Ξ Š x@u, ΦD, y@u, ΦD Š
>, :8u, - 1<, :Φ,
Π
Φ
1.2550
1.2545
Out[56]=
1.2540
1.2535
1.2530
u
-1.100
-1.098
-1.096
The area of interest is then
Show@8midplot, topright1plot, right1plot<, PlotRange ® AllD
Φ
1.2
1.0
0.8
0.6
0.4
0.2
u
-1.20
-1.15
-1.10
Since the bottom of the region is along the u axis, i.e., the line Φ = 0, we can just integrate
the midline function, add to it the integral of the topline function and subtract from that sum
the integral of the right1line function:
18
ConformalCapacitor.nb
In[57]:=
Out[60]=
midarea = NIntegrate@midf@uD, 8u, miduL, miduH<D;
topr1area = NIntegrate@topright1f@uD, 8u, miduH, right1uH<D;
right1area = NIntegrate@right1f@uD, 8u, right1uL, right1uH<D;
firstrightarea = midarea + topr1area - right1area
0.0102862
Now we turn our attention to the area on the left from which the dielectric is removed.
Since the left half of the diagram is just a reflection of the right half, for the area calculation
we may use the area of width Dx just to the left of the x=0 midline. We repeat the above
calculations for the vertical line we call left1:
In[61]:=
FindRoot@8- deltaX Š x@u, ΦD, y@u, ΦD Š 0<, 88u, - 1<, 8Φ, 0<<D;
left1uL = u . %;
Π
Π
left1rule = FindRootB:- deltaX Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F;
2
2
left1uH = u . left1rule;
left1 = TableBFindRootB
:- deltaX Š x@u, ΦD, y@u, ΦD Š Ξ
Π
2
>, :8u, - 1<, :Φ, Ξ
Π
2
>>F, 8Ξ, 0, 1, 0.001<F;
left1table = Table@8u, Φ< . left1D;
left1f = Interpolation@left1tableD;
left1plot = Plot@left1f@uD, 8u, left1uL, left1uH<,
AxesLabel ® 8u, Φ<, PlotStyle ® 8RGBColor@0, 0, 0D<D;
Π
Π
topleft1 = TableBFindRootB:Ξ Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F,
2
2
8Ξ, - deltaX, 0, 0.001<F;
topleft1table = Table@8u, Φ< . topleft1D;
topleft1f = Interpolation@topleft1tableD;
topleft1plot = Plot@topleft1f@uD, 8u, left1uH, miduH<,
AxesLabel ® 8u, Φ<, PlotStyle ® 8RGBColor@0, 0, 0D<D;
Show@8left1plot, topleft1plot, midplot<, PlotRange ® AllD
Φ
1.2
1.0
0.8
Out[73]=
0.6
0.4
0.2
-1.20
-1.15
u
-1.10
ConformalCapacitor.nb
In[74]:=
Out[76]=
19
left1area = NIntegrate@left1f@uD, 8u, left1uL, left1uH<D;
topleft1area = NIntegrate@topleft1f@uD, 8u, left1uH, miduH<D;
firstleftarea = left1area + topleft1area - midarea
0.0103228
The net change in stored energy is DU = Ε0 { V02 ( 8dΠ3 ) (Κ-1) rightarea + Ε0 { V02 ( 8dΠ3 )
(1-Κ) leftarea, which with
In[77]:=
Out[77]=
netfirstarea = Hfirstrightarea - firstleftareaL
- 0.0000366405
evaluates to
In[78]:=
Out[78]=
DU1 = Ε0 { V02
d
8 Π3
HΚ - 1L HnetfirstareaL
- 0.0000366405 { H- 1 + ΚL Ε0 V20 B
d
8 Π3
F
It is, of course, negative since the capacitance of the device decreases as the dielectric is
removed. The force responsible is the negative gradient of the potential energy, and is
therefore directed to the right. The restoring force which we seek is in the opposite direction. We obtain then the numerical approximation (Fx)1
(Fx)1
=
DU
DX
=
(2Π)
d
DU
Dx
(20)
In[79]:=
Out[79]=
=
Ε0 {J 4 Π0 2 N HΚ - 1LH fxL1
V2
where the dimensionless force H fxL1 is given by
firstforce = HnetfirstareaL  deltaX
- 0.00366405
Note that for Κ >1 the force is in the negative x direction, as it should be.
Now in order to investigate the x dependence of the force, we will repeat the calculations
above for 2 successive small displacements Dx.
20
ConformalCapacitor.nb
In[80]:=
FindRoot@82 deltaX Š x@u, ΦD, y@u, ΦD Š 0<, 88u, - 1<, 8Φ, 0<<D;
right2uL = u . %;
Π
Π
right2rule = FindRootB:2 deltaX Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F;
2
2
right2uH = u . right2rule;
right2 = TableBFindRootB
:2 deltaX Š x@u, ΦD, y@u, ΦD Š Ξ
>>F, 8Ξ, 0, 1, 0.001<F;
2
2
right2table = Table@8u, Φ< . right2D;
right2f = Interpolation@right2tableD;
right2plot = Plot@right2f@uD, 8u, right2uL, right2uH<,
AxesLabel ® 8u, Φ<, PlotStyle ® 8RGBColor@0, 0, 0D<D;
Π
Π
topright2 = TableBFindRootB:Ξ Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F,
2
2
Π
>, :8u, - 1<, :Φ, Ξ
Π
8Ξ, deltaX, 2 deltaX, 0.001<F;
topright2table = Table@8u, Φ< . topright2D;
topright2f = Interpolation@topright2tableD;
topright2plot = Plot@topright2f@uD, 8u, right1uH, right2uH<,
AxesLabel ® 8u, Φ<, PlotStyle ® 8RGBColor@0, 0, 0D<D;
Show@8right1plot, topright2plot, right2plot<, PlotRange ® AllD
Φ
1.2
1.0
0.8
Out[92]=
0.6
0.4
0.2
u
-1.20
In[93]:=
Out[95]=
-1.15
right2area = NIntegrate@right2f@uD, 8u, right2uL, right2uH<D;
topright2area = NIntegrate@topright2f@uD, 8u, right1uH, right2uH<D;
secondrightarea = right1area + topright2area - right2area
0.0102495
-1.10
ConformalCapacitor.nb
In[96]:=
21
FindRoot@8- 2 deltaX Š x@u, ΦD, y@u, ΦD Š 0<, 88u, - 1<, 8Φ, 0<<D;
left2uL = u . %;
Π
Π
left2rule = FindRootB:- 2 deltaX Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F;
2
2
left2uH = u . left2rule;
left2 = TableBFindRootB
:- 2 deltaX Š x@u, ΦD, y@u, ΦD Š Ξ
Π
2
>, :8u, - 1<, :Φ, Ξ
Π
2
>>F, 8Ξ, 0, 1, 0.001<F;
left2table = Table@8u, Φ< . left2D;
left2f = Interpolation@left2tableD;
left2plot = Plot@left2f@uD, 8u, left2uL, left2uH<,
AxesLabel ® 8u, Φ<, PlotStyle ® 8RGBColor@0, 0, 0D<D;
Π
Π
topleft2 = TableBFindRootB:Ξ Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F,
2
2
8Ξ, - 2 deltaX, - deltaX, 0.001<F;
topleft2table = Table@8u, Φ< . topleft2D;
topleft2f = Interpolation@topleft2tableD;
topleft2plot = Plot@topleft2f@uD, 8u, left2uH, left1uH<,
AxesLabel ® 8u, Φ<, PlotStyle ® 8RGBColor@0, 0, 0D<D;
Show@8left1plot, topleft2plot, left2plot<, PlotRange ® AllD
Φ
1.2
1.0
0.8
Out[108]=
0.6
0.4
0.2
u
-1.20
In[109]:=
Out[111]=
In[112]:=
Out[112]=
In[113]:=
Out[113]=
-1.15
left2area = NIntegrate@left2f@uD, 8u, left2uL, left2uH<D;
topleft2area = NIntegrate@topleft2f@uD, 8u, left2uH, left1uH<D;
secondleftarea = left2area + topleft2area - left1area
0.0103594
netsecondarea = secondrightarea - secondleftarea
- 0.000109919
secondforce = netsecondarea  deltaX
- 0.0109919
22
ConformalCapacitor.nb
For the final small displacement Dx, we have:
In[114]:=
FindRoot@83 deltaX Š x@u, ΦD, y@u, ΦD Š 0<, 88u, - 1<, 8Φ, 0<<D;
right3uL = u . %;
Π
Π
right3rule = FindRootB:3 deltaX Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F;
2
2
right3uH = u . right3rule;
right3 = TableBFindRootB
:3 deltaX Š x@u, ΦD, y@u, ΦD Š Ξ
>>F, 8Ξ, 0, 1, 0.001<F;
2
2
right3table = Table@8u, Φ< . right3D;
right3f = Interpolation@right3tableD;
right3plot = Plot@right3f@uD, 8u, right3uL, right3uH<,
AxesLabel ® 8u, Φ<, PlotStyle ® 8RGBColor@0, 0, 0D<D;
Π
Π
topright3 = TableBFindRootB:Ξ Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F,
2
2
Π
>, :8u, - 1<, :Φ, Ξ
Π
8Ξ, 2 deltaX, 3 deltaX, 0.001<F;
topright3table = Table@8u, Φ< . topright3D;
topright3f = Interpolation@topright3tableD;
topright3plot = Plot@topright3f@uD, 8u, right2uH, right3uH<,
AxesLabel ® 8u, Φ<, PlotStyle ® 8RGBColor@0, 0, 0D<D;
Show@8right2plot, topright3plot, right3plot<, PlotRange ® AllD
Φ
1.2
1.0
0.8
Out[126]=
0.6
0.4
0.2
u
-1.20
In[127]:=
Out[129]=
-1.15
-1.10
right3area = NIntegrate@right3f@uD, 8u, right3uL, right3uH<D;
topright3area = NIntegrate@topright3f@uD, 8u, right2uH, right3uH<D;
thirdrightarea = right2area + topright3area - right3area
0.0102126
ConformalCapacitor.nb
In[130]:=
FindRoot@8- 3 deltaX Š x@u, ΦD, y@u, ΦD Š 0<, 88u, - 1<, 8Φ, 0<<D;
left3uL = u . %;
Π
Π
left3rule = FindRootB:- 3 deltaX Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F;
2
2
left3uH = u . left3rule;
left3 = TableBFindRootB
:- 3 deltaX Š x@u, ΦD, y@u, ΦD Š Ξ
Π
2
>, :8u, - 1<, :Φ, Ξ
Π
2
>>F, 8Ξ, 0, 1, 0.001<F;
left3table = Table@8u, Φ< . left3D;
left3f = Interpolation@left3tableD;
left3plot = Plot@left3f@uD, 8u, left3uL, left3uH<,
AxesLabel ® 8u, Φ<, PlotStyle ® 8RGBColor@0, 0, 0D<D;
Π
Π
topleft3 = TableBFindRootB:Ξ Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F,
2
2
8Ξ, - 3 deltaX, - 2 deltaX, 0.001<F;
topleft3table = Table@8u, Φ< . topleft3D;
topleft3f = Interpolation@topleft3tableD;
topleft3plot = Plot@topleft3f@uD, 8u, left3uH, left2uH<,
AxesLabel ® 8u, Φ<, PlotStyle ® 8RGBColor@0, 0, 0D<D;
Show@8left3plot, topleft3plot, left2plot<, PlotRange ® AllD
Φ
1.2
1.0
0.8
Out[142]=
0.6
0.4
0.2
u
-1.25
In[143]:=
Out[145]=
In[146]:=
Out[146]=
In[147]:=
Out[147]=
-1.20
-1.15
left3area = NIntegrate@left3f@uD, 8u, left3uL, left3uH<D;
topleft3area = NIntegrate@topleft3f@uD, 8u, left3uH, left2uH<D;
thirdleftarea = left3area + topleft3area - left2area
0.0103958
netthirdarea = thirdrightarea - thirdleftarea
- 0.000183191
thirdforce = netthirdarea  deltaX
- 0.0183191
Finally, we examine a graph of the dimensionless restoring force versus displacement.
23
24
ConformalCapacitor.nb
In[148]:=
In[151]:=
In[152]:=
list1 = ListPlot@880, 0<, 8deltaX, firstforce<,
82 deltaX, secondforce<, 83 deltaX, thirdforce<<, PlotStyle ®
8PointSize@0.02D, RGBColor@0, 0, 0D<, Axes ® False, AxesLabel ® 8None<D;
list2 = ListPlot@880, 0<, 8deltaX, firstforce<,
82 deltaX, secondforce<, 83 deltaX, thirdforce<<, PlotJoined ® True,
PlotStyle ® 8Dashing@80.02, 0.02<D, RGBColor@0, 0, 0D<, AxesLabel ® 8x, fx <D;
Show@8list1, list2<, Axes ® True, AxesLabel ® 8x, fx <D
fx
x
0.005
0.010
0.015
0.020
0.025
0.030
-0.005
Out[152]=
-0.010
-0.015
Now the 3 force values lie along a straight line, which however, fails to line up with the
(0,0) to firstforce line. However, we must keep in mind that our values for the forces are for
finite regions of width Dx. The mean value theorem assures us that the value computed is
the exact value somewhere in the interval. Since the points are linear, if we make the simple
guess that the forces should be associated with the midpoints of the intervals, we obtain
In[153]:=
In[154]:=
list3 = ListPlot@880, 0<, 8deltaX  2, firstforce<,
83 deltaX  2, secondforce<, 85 deltaX  2, thirdforce<<, PlotStyle ®
8PointSize@0.02D, RGBColor@0, 0, 0D<, Axes ® False, AxesLabel ® 8None<D;
list4 = ListPlot@880, 0<, 8deltaX  2, firstforce<,
83 deltaX  2, secondforce<, 85 deltaX  2, thirdforce<<, PlotJoined ® True,
PlotStyle ® 8Dashing@80.02, 0.02<D, RGBColor@0, 0, 0D<, AxesLabel ® 8x, fx <D;
ConformalCapacitor.nb
In[155]:=
25
Show@8list3, list4<, Axes ® True, AxesLabel ® 8x, fx <D
fx
0.005
0.010
0.015
0.020
0.025
-0.005
Out[155]=
-0.010
-0.015
This is the proper form of a Hooke's Law restoring force, which is what is expected from
general principles, for small displacements. We wish to emphasize that there is no error
here due to the mirror image "join" because we have calculated areas using only the right
half of the diagram using the exact conformal mapping result.
Now let's automate the process. We will suppose that the dielectric is 30 units wide, i.e.,
the mirror image "join" is at x = -15. Then when the right end is at x = +15, the left end is at
x = -15, and the force should begin to level off and decrease in strength. The module below
will suffice to calculate the force at various values of x, and we will investigate a representative sample.
In[156]:=
xJoin = 15;
In[157]:=
forcefunction@xlow_, xhigh_D :=
ModuleB8deltaX, u, Φ, rightminusLrule, rightminusuL, rightminusHrule,
rightminusuH, rightminus, Ξ, rightminustable, rightminusf, rightLrule,
rightuL, rightHrule, rightuH, right, righttable, rightf, topright,
toprighttable, toprightf, rightminusarea, toprightarea, rightarea,
netrightarea, leftvalue, leftLrule, leftuL, leftHrule, leftuH, left,
lefttable, leftf, leftplusLrule, leftplusuL, leftplusHrule, leftplusuH,
leftplus, leftplustable, leftplusf, topleft, toplefttable, topleftf,
leftarea, topleftarea, leftplusarea, netleftarea, netarea, force, midpoint<,
Clear@deltaX, u, Φ, rightminusLrule, rightminusuL, rightminusHrule,
rightminusuH, rightminus, Ξ, rightminustable, rightminusf, rightLrule,
rightuL, rightHrule, rightuH, right, righttable, rightf, topright,
toprighttable, toprightf, rightminusarea, toprightarea, rightarea,
netrightarea, leftvalue, leftLrule, leftuL, leftHrule, leftuH, left,
lefttable, leftf, leftplusLrule, leftplusuL, leftplusHrule, leftplusuH,
leftplus, leftplustable, leftplusf, topleft, toplefttable, topleftf,
leftarea, topleftarea, leftplusarea, netleftarea, netarea, force, midpointD;
;
;
26
ConformalCapacitor.nb
deltaX = xhigh - xlow;
rightminusLrule = FindRoot@8xlow Š x@u, ΦD, y@u, ΦD Š 0<, 88u, - 1<, 8Φ, 0<<D;
rightminusuL = u . rightminusLrule;
Π
Π
rightminusHrule = FindRootB:xlow Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F;
2
2
rightminusuH = u . rightminusHrule;
rightminus = TableB
>>F, 8Ξ, 0, 1, 0.001<F;
2
2
rightminustable = Table@8u, Φ< . rightminusD;
rightminusf = Interpolation@rightminustableD;
rightLrule = FindRoot@8xhigh Š x@u, ΦD, y@u, ΦD Š 0<, 88u, - 1<, 8Φ, 0<<D;
rightuL = u . rightLrule;
Π
Π
rightHrule = FindRootB:xhigh Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F;
2
2
rightuH = u . rightHrule;
Π
FindRootB:xlow Š x@u, ΦD, y@u, ΦD Š Ξ
>, :8u, - 1<, :Φ, Ξ
Π
right = TableBFindRootB
:xhigh == x@u, ΦD, y@u, ΦD Š Ξ
Π
2
>, :8u, - 1<, :Φ, Ξ
righttable = Table@8u, Φ< . rightD;
rightf = Interpolation@righttableD;
Π
topright = TableBFindRootB:Ξ Š x@u, ΦD, y@u, ΦD Š
2
8Ξ, xlow, xhigh, 0.001<F;
Π
2
>>F, 8Ξ, 0, 1, 0.001<F;
>, :8u, - 1<, :Φ,
Π
2
>>F,
toprighttable = Table@8u, Φ< . toprightD;
toprightf = Interpolation@toprighttableD;
rightminusarea = NIntegrate@rightminusf@uD, 8u, rightminusuL, rightminusuH<D;
toprightarea = NIntegrate@toprightf@uD, 8u, rightminusuH, rightuH<D;
rightarea = NIntegrate@rightf@uD, 8u, rightuL, rightuH<D;
netrightarea = rightminusarea + toprightarea - rightarea;
If@xhigh > xJoin, leftvalue = xhigh - 2 xJoin, leftvalue = - xhighD;
If@xhigh > 2 xJoin, leftvalue = xhigh - 2 xJoin - deltaXD;
leftLrule = FindRoot@8leftvalue Š x@u, ΦD, y@u, ΦD Š 0<, 88u, - 1<, 8Φ, 0<<D;
leftuL = u . leftLrule;
Π
Π
leftHrule = FindRootB:leftvalue Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F;
2
2
leftuH = u . leftHrule;
left = TableBFindRootB
:leftvalue Š x@u, ΦD, y@u, ΦD Š Ξ
Π
2
>, :8u, - 1<, :Φ, Ξ
Π
2
>>F, 8Ξ, 0, 1, 0.001<F;
lefttable = Table@8u, Φ< . leftD;
leftf = Interpolation@lefttableD;
leftplusLrule =
FindRoot@8leftvalue + deltaX Š x@u, ΦD, y@u, ΦD Š 0<, 88u, - 1<, 8Φ, 0<<D;
leftplusuL = u . leftplusLrule;
leftplusHrule =
Π
Π
FindRootB:leftvalue + deltaX Š x@u, ΦD, y@u, ΦD Š >, :8u, - 1<, :Φ, >>F;
2
2
leftplusuH = u . leftplusHrule;
Π
leftplus = TableBFindRootB:leftvalue + deltaX == x@u, ΦD, y@u, ΦD Š Ξ >,
2
Π
:8u, - 1<, :Φ, Ξ >>F, 8Ξ, 0, 1, 0.001<F;
2
;
;
ConformalCapacitor.nb
leftplustable = Table@8u, Φ< . leftplusD;
leftplusf = Interpolation@leftplustableD;
Π
topleft = TableBFindRootB:Ξ Š x@u, ΦD, y@u, ΦD Š
8Ξ, leftvalue , leftvalue + deltaX, 0.001<F;
2
>, :8u, - 1<, :Φ,
Π
2
27
>>F,
toplefttable = Table@8u, Φ< . topleftD;
topleftf = Interpolation@toplefttableD;
leftarea = NIntegrate@leftf@uD, 8u, leftuL, leftuH<D;
topleftarea = NIntegrate@topleftf@uD, 8u, leftuH, leftplusuH<D;
leftplusarea = NIntegrate@leftplusf@uD, 8u, leftplusuL, leftplusuH<D;
netleftarea = leftarea + topleftarea - leftplusarea;
netarea = netrightarea - netleftarea;
force = netarea  deltaX;
midpoint = Hxlow + xhighL  2;
ans = 88midpoint, force<<;
In[158]:=
In[159]:=
In[160]:=
Out[160]=
In[161]:=
In[162]:=
In[163]:=
F
ResultTable = 880, 0<<;
Do@forcefunction@i deltaX, Hi + 1L deltaXD;
ResultTable = Join@ResultTable, ansD, 8i, 0, 9<D
ResultTable
880, 0<, 80.005, - 0.00366405<, 80.015, - 0.0109919<, 80.025, - 0.0183191<,
80.035, - 0.0256451<, 80.045, - 0.0329694<, 80.055, - 0.0402917<,
80.065, - 0.0476113<, 80.075, - 0.0549279<, 80.085, - 0.062241<, 80.095, - 0.0695502<<
list5 = ListPlot@ResultTable, PlotStyle ® 8PointSize@0.02D, RGBColor@0, 0, 0D<,
Axes ® False, AxesLabel ® 8None<D;
list6 = ListPlot@ResultTable, PlotJoined ® True,
PlotStyle ® 8Dashing@80.02, 0.02<D, RGBColor@0, 0, 0D<, AxesLabel ® 8x, fx <D;
Show@8list5, list6<, Axes ® True, AxesLabel ® 8x, fx <D
fx
x
0.02
0.04
0.06
0.08
-0.01
-0.02
-0.03
Out[163]=
-0.04
-0.05
-0.06
-0.07
Evidentally, the Hooke's Law behavior continues for 10 displacements of Dx, out to x = 0.1
We will track it further but since the calculation takes a rather long time, with a much different mesh. We will compile a table of force values for strips of width Dx ending at each
value of one tenth out to x = 45.
28
ConformalCapacitor.nb
In[164]:=
In[165]:=
In[166]:=
In[167]:=
Out[167]=
In[168]:=
Evidentally, the Hooke's Law behavior continues for 10 displacements of Dx, out to x = 0.1
We will track it further but since the calculation takes a rather long time, with a much different mesh. We will compile a table of force values for strips of width Dx ending at each
value of one tenth out to x = 45.
deltaX = 0.1;
ResultTable = 880, 0<<;
Do@forcefunction@i - deltaX, i D;
ResultTable = Join@ResultTable, ansD, 8i, deltaX, 1, deltaX<D
ResultTable
880, 0<, 80.05, - 0.0366212<, 80.15, - 0.10963<, 80.25, - 0.181947<,
80.35, - 0.253138<, 80.45, - 0.322805<, 80.55, - 0.390598<,
80.65, - 0.456222<, 80.75, - 0.519439<, 80.85, - 0.580068<, 80.95, - 0.637983<<
ListPlot@ResultTable, PlotJoined ® True,
PlotStyle ® 8RGBColor@0, 0, 0D<, AxesLabel ® 8x, fx <D
fx
0.2
0.4
0.6
x
0.8
-0.1
-0.2
Out[168]= -0.3
-0.4
-0.5
-0.6
In[169]:=
In[170]:=
In[171]:=
Do@forcefunction@i - deltaX, i D;
ResultTable = Join@ResultTable, ansD, 8i, 1 + deltaX, xJoin, deltaX<D
list7 = ListPlot@ResultTable, PlotJoined ® True, PlotRange ® 880, xJoin<, 80, - 1.6<<,
PlotStyle ® 8RGBColor@0, 0, 0D<, AxesLabel ® 8x, fx <D;
list8 = ListPlot@Table@8i, - Π  2<, 8i, 0, 15, 0.1<D,
PlotJoined ® True, PlotRange ® 880, xJoin<, 80, - 1.6<<, PlotStyle ®
8Dashing@80.02, 0.02<D, RGBColor@0, 0, 0D<, Axes ® False, AxesLabel ® 8None<D;
ConformalCapacitor.nb
In[172]:=
Show@8list7, list8<, Axes ® True, AxesLabel ® 8x, fx <D
29
fx
0.0
2
4
6
8
x
10
12
14
-0.5
Out[172]=
-1.0
-1.5
This behavior is physically what is expected. At x = +15, the dielectric is half removed.
The dimensionless force is converging to - Π2 , the dashed line, which is the expected result.
6
30
ConformalCapacitor.nb
In[173]:=
Out[173]=
ResultTable
880, 0<, 80.05, - 0.0366212<, 80.15, - 0.10963<, 80.25, - 0.181947<,
80.35, - 0.253138<, 80.45, - 0.322805<, 80.55, - 0.390598<,
80.65, - 0.456222<, 80.75, - 0.519439<, 80.85, - 0.580068<, 80.95, - 0.637983<,
81.05, - 0.693109<, 81.15, - 0.745412<, 81.25, - 0.7949<, 81.35, - 0.841608<,
81.45, - 0.8856<, 81.55, - 0.926957<, 81.65, - 0.965775<, 81.75, - 1.00216<,
81.85, - 1.03623<, 81.95, - 1.06811<, 82.05, - 1.0979<, 82.15, - 1.12574<,
82.25, - 1.15173<, 82.35, - 1.17599<, 82.45, - 1.19864<, 82.55, - 1.21977<,
82.65, - 1.23949<, 82.75, - 1.25789<, 82.85, - 1.27507<, 82.95, - 1.29111<,
83.05, - 1.30609<, 83.15, - 1.32008<, 83.25, - 1.33316<, 83.35, - 1.34538<,
83.45, - 1.35682<, 83.55, - 1.36752<, 83.65, - 1.37754<, 83.75, - 1.38693<,
83.85, - 1.39573<, 83.95, - 1.40398<, 84.05, - 1.41173<, 84.15, - 1.41901<,
84.25, - 1.42584<, 84.35, - 1.43227<, 84.45, - 1.43832<, 84.55, - 1.44402<,
84.65, - 1.44939<, 84.75, - 1.45446<, 84.85, - 1.45923<, 84.95, - 1.46375<,
85.05, - 1.46801<, 85.15, - 1.47204<, 85.25, - 1.47586<, 85.35, - 1.47947<,
85.45, - 1.4829<, 85.55, - 1.48614<, 85.65, - 1.48922<, 85.75, - 1.49215<,
85.85, - 1.49493<, 85.95, - 1.49757<, 86.05, - 1.50009<, 86.15, - 1.50248<,
86.25, - 1.50476<, 86.35, - 1.50693<, 86.45, - 1.50901<, 86.55, - 1.51099<,
86.65, - 1.51288<, 86.75, - 1.51468<, 86.85, - 1.51641<, 86.95, - 1.51806<,
87.05, - 1.51964<, 87.15, - 1.52115<, 87.25, - 1.5226<, 87.35, - 1.52399<,
87.45, - 1.52533<, 87.55, - 1.52661<, 87.65, - 1.52783<, 87.75, - 1.52901<,
87.85, - 1.53015<, 87.95, - 1.53124<, 88.05, - 1.53229<, 88.15, - 1.5333<,
88.25, - 1.53427<, 88.35, - 1.5352<, 88.45, - 1.53611<, 88.55, - 1.53697<,
88.65, - 1.53781<, 88.75, - 1.53862<, 88.85, - 1.5394<, 88.95, - 1.54016<,
89.05, - 1.54088<, 89.15, - 1.54159<, 89.25, - 1.54227<, 89.35, - 1.54292<,
89.45, - 1.54356<, 89.55, - 1.54417<, 89.65, - 1.54477<, 89.75, - 1.54534<,
89.85, - 1.5459<, 89.95, - 1.54644<, 810.05, - 1.54697<, 810.15, - 1.54747<,
810.25, - 1.54797<, 810.35, - 1.54844<, 810.45, - 1.54891<, 810.55, - 1.54936<,
810.65, - 1.54979<, 810.75, - 1.55022<, 810.85, - 1.55063<, 810.95, - 1.55103<,
811.05, - 1.55141<, 811.15, - 1.55179<, 811.25, - 1.55216<, 811.35, - 1.55252<,
811.45, - 1.55286<, 811.55, - 1.5532<, 811.65, - 1.55353<, 811.75, - 1.55385<,
811.85, - 1.55416<, 811.95, - 1.55446<, 812.05, - 1.55476<, 812.15, - 1.55504<,
812.25, - 1.55532<, 812.35, - 1.5556<, 812.45, - 1.55586<, 812.55, - 1.55612<,
812.65, - 1.55637<, 812.75, - 1.55662<, 812.85, - 1.55686<, 812.95, - 1.5571<,
813.05, - 1.55732<, 813.15, - 1.55755<, 813.25, - 1.55777<, 813.35, - 1.55798<,
813.45, - 1.55819<, 813.55, - 1.55839<, 813.65, - 1.55859<, 813.75, - 1.55878<,
813.85, - 1.55897<, 813.95, - 1.55916<, 814.05, - 1.55934<, 814.15, - 1.55951<,
814.25, - 1.55969<, 814.35, - 1.55986<, 814.45, - 1.56002<, 814.55, - 1.56018<,
814.65, - 1.56034<, 814.75, - 1.5605<, 814.85, - 1.56065<, 814.95, - 1.56079<<
There will be some small error as we cross the "join".
In[174]:=
Do@forcefunction@i - deltaX, i D;
ResultTable = Join@ResultTable, ansD, 8i, xJoin + deltaX, 2 xJoin, deltaX<D
ConformalCapacitor.nb
In[175]:=
31
ListPlot@ResultTable, PlotJoined ® True, AxesLabel ® 8x, fx <,
PlotRange ® All, PlotStyle ® 8RGBColor@0, 0, 0D<D
fx
x
5
10
15
20
25
30
-0.5
Out[175]=
-1.0
-1.5
The force does indeed begin to decrease in magnitude, but due to the small "join" error, not
exactly at the half-way mark.
In[176]:=
In[177]:=
Do@forcefunction@i - deltaX, i D;
ResultTable = Join@ResultTable, ansD, 8i, 2 xJoin + 0.1, 3 xJoin, 0.1<D
ListPlot@ResultTable, PlotJoined ® True, AxesLabel ® 8x, fx <,
PlotRange ® All, PlotStyle ® 8RGBColor@0, 0, 0D<D
fx
x
10
20
30
40
-0.5
Out[177]=
-1.0
-1.5
This is qualitatively the expected result. Since the fringes continue to x ® ¥, the force will
approach but never reach zero. The full result is given below.
In[178]:=
Out[178]=
ResultTable
880, 0<, 80.05, - 0.0366212<, 80.15, - 0.10963<, 80.25, - 0.181947<, 80.35, - 0.253138<,
80.45, - 0.322805<, 80.55, - 0.390598<, 80.65, - 0.456222<, 80.75, - 0.519439<,
80.85, - 0.580068<, 80.95, - 0.637983<, 81.05, - 0.693109<, 81.15, - 0.745412<,
,
,
,
,
Out[178]=
32
ConformalCapacitor.nb
81.25, - 0.7949<, 81.35, - 0.841608<, 81.45, - 0.8856<, 81.55, - 0.926957<,
81.65, - 0.965775<, 81.75, - 1.00216<, 81.85, - 1.03623<, 81.95, - 1.06811<,
82.05, - 1.0979<, 82.15, - 1.12574<, 82.25, - 1.15173<, 82.35, - 1.17599<,
82.45, - 1.19864<, 82.55, - 1.21977<, 82.65, - 1.23949<, 82.75, - 1.25789<,
82.85, - 1.27507<, 82.95, - 1.29111<, 83.05, - 1.30609<, 83.15, - 1.32008<,
83.25, - 1.33316<, 83.35, - 1.34538<, 83.45, - 1.35682<, 83.55, - 1.36752<,
83.65, - 1.37754<, 83.75, - 1.38693<, 83.85, - 1.39573<, 83.95, - 1.40398<,
84.05, - 1.41173<, 84.15, - 1.41901<, 84.25, - 1.42584<, 84.35, - 1.43227<,
84.45, - 1.43832<, 84.55, - 1.44402<, 84.65, - 1.44939<, 84.75, - 1.45446<,
84.85, - 1.45923<, 84.95, - 1.46375<, 85.05, - 1.46801<, 85.15, - 1.47204<,
85.25, - 1.47586<, 85.35, - 1.47947<, 85.45, - 1.4829<, 85.55, - 1.48614<,
85.65, - 1.48922<, 85.75, - 1.49215<, 85.85, - 1.49493<, 85.95, - 1.49757<,
86.05, - 1.50009<, 86.15, - 1.50248<, 86.25, - 1.50476<, 86.35, - 1.50693<,
86.45, - 1.50901<, 86.55, - 1.51099<, 86.65, - 1.51288<, 86.75, - 1.51468<,
86.85, - 1.51641<, 86.95, - 1.51806<, 87.05, - 1.51964<, 87.15, - 1.52115<,
87.25, - 1.5226<, 87.35, - 1.52399<, 87.45, - 1.52533<, 87.55, - 1.52661<,
87.65, - 1.52783<, 87.75, - 1.52901<, 87.85, - 1.53015<, 87.95, - 1.53124<,
88.05, - 1.53229<, 88.15, - 1.5333<, 88.25, - 1.53427<, 88.35, - 1.5352<,
88.45, - 1.53611<, 88.55, - 1.53697<, 88.65, - 1.53781<, 88.75, - 1.53862<,
88.85, - 1.5394<, 88.95, - 1.54016<, 89.05, - 1.54088<, 89.15, - 1.54159<,
89.25, - 1.54227<, 89.35, - 1.54292<, 89.45, - 1.54356<, 89.55, - 1.54417<,
89.65, - 1.54477<, 89.75, - 1.54534<, 89.85, - 1.5459<, 89.95, - 1.54644<,
810.05, - 1.54697<, 810.15, - 1.54747<, 810.25, - 1.54797<, 810.35, - 1.54844<,
810.45, - 1.54891<, 810.55, - 1.54936<, 810.65, - 1.54979<, 810.75, - 1.55022<,
810.85, - 1.55063<, 810.95, - 1.55103<, 811.05, - 1.55141<, 811.15, - 1.55179<,
811.25, - 1.55216<, 811.35, - 1.55252<, 811.45, - 1.55286<, 811.55, - 1.5532<,
811.65, - 1.55353<, 811.75, - 1.55385<, 811.85, - 1.55416<, 811.95, - 1.55446<,
812.05, - 1.55476<, 812.15, - 1.55504<, 812.25, - 1.55532<, 812.35, - 1.5556<,
812.45, - 1.55586<, 812.55, - 1.55612<, 812.65, - 1.55637<, 812.75, - 1.55662<,
812.85, - 1.55686<, 812.95, - 1.5571<, 813.05, - 1.55732<, 813.15, - 1.55755<,
813.25, - 1.55777<, 813.35, - 1.55798<, 813.45, - 1.55819<, 813.55, - 1.55839<,
813.65, - 1.55859<, 813.75, - 1.55878<, 813.85, - 1.55897<, 813.95, - 1.55916<,
814.05, - 1.55934<, 814.15, - 1.55951<, 814.25, - 1.55969<, 814.35, - 1.55986<,
814.45, - 1.56002<, 814.55, - 1.56018<, 814.65, - 1.56034<, 814.75, - 1.5605<,
814.85, - 1.56065<, 814.95, - 1.56079<, 815.05, - 1.56094<, 815.15, - 1.56108<,
815.25, - 1.56122<, 815.35, - 1.56136<, 815.45, - 1.56149<, 815.55, - 1.56162<,
815.65, - 1.56175<, 815.75, - 1.56187<, 815.85, - 1.562<, 815.95, - 1.56212<,
816.05, - 1.56224<, 816.15, - 1.56235<, 816.25, - 1.56246<, 816.35, - 1.56258<,
816.45, - 1.56268<, 816.55, - 1.56279<, 816.65, - 1.5629<, 816.75, - 1.563<,
816.85, - 1.5631<, 816.95, - 1.5632<, 817.05, - 1.5633<, 817.15, - 1.56339<,
817.25, - 1.56348<, 817.35, - 1.56358<, 817.45, - 1.56367<, 817.55, - 1.56375<,
817.65, - 1.56384<, 817.75, - 1.56393<, 817.85, - 1.56401<, 817.95, - 1.56409<,
818.05, - 1.56417<, 818.15, - 1.56425<, 818.25, - 1.56433<, 818.35, - 1.56441<,
818.45, - 1.56448<, 818.55, - 1.56455<, 818.65, - 1.56463<, 818.75, - 1.5647<,
818.85, - 1.56477<, 818.95, - 1.56483<, 819.05, - 1.5649<, 819.15, - 1.56497<,
819.25, - 1.56503<, 819.35, - 1.56509<, 819.45, - 1.56515<, 819.55, - 1.56521<,
819.65, - 1.56527<, 819.75, - 1.56533<, 819.85, - 1.56539<, 819.95, - 1.56544<,
820.05, - 1.5655<, 820.15, - 1.56555<, 820.25, - 1.5656<, 820.35, - 1.56565<,
820.45, - 1.5657<, 820.55, - 1.56575<, 820.65, - 1.56579<, 820.75, - 1.56584<,
820.85, - 1.56588<, 820.95, - 1.56592<, 821.05, - 1.56596<, 821.15, - 1.566<,
821.25, - 1.56603<, 821.35, - 1.56607<, 821.45, - 1.5661<, 821.55, - 1.56613<,
821.65, - 1.56616<, 821.75, - 1.56618<, 821.85, - 1.5662<, 821.95, - 1.56622<,
822.05, - 1.56623<, 822.15, - 1.56625<, 822.25, - 1.56625<, 822.35, - 1.56626<,
822.45, - 1.56626<, 822.55, - 1.56625<, 822.65, - 1.56624<, 822.75, - 1.56623<,
822.85, - 1.5662<, 822.95, - 1.56617<, 823.05, - 1.56614<, 823.15, - 1.56609<,
823.25, - 1.56604<, 823.35, - 1.56597<, 823.45, - 1.56589<, 823.55, - 1.56581<,
823.65, - 1.56571<, 823.75, - 1.56559<, 823.85, - 1.56546<, 823.95, - 1.56531<,
824.05, - 1.56514<, 824.15, - 1.56495<, 824.25, - 1.56473<, 824.35, - 1.56449<,
824.45, - 1.56422<, 824.55, - 1.56392<, 824.65, - 1.56359<, 824.75, - 1.56321<,
,
,
,
,
ConformalCapacitor.nb
824.85,
825.25,
825.65,
826.05,
826.45,
826.85,
827.25,
827.65,
828.05,
828.45,
828.85,
829.25,
829.65,
830.05,
830.45,
830.85,
831.25,
831.65,
832.05,
832.45,
832.85,
833.25,
833.65,
834.05,
834.45,
834.85,
835.25,
835.65,
836.05,
836.45,
836.85,
837.25,
837.65,
838.05,
838.45,
838.85,
839.25,
839.65,
840.05,
840.45,
840.85,
841.25,
841.65,
842.05,
842.45,
842.85,
843.25,
843.65,
844.05,
844.35,
844.65,
33
- 1.56279<, 824.95, - 1.56233<, 825.05, - 1.56181<, 825.15, - 1.56123<,
- 1.56059<, 825.35, - 1.55989<, 825.45, - 1.5591<, 825.55, - 1.55823<,
- 1.55726<, 825.75, - 1.55619<, 825.85, - 1.55501<, 825.95, - 1.5537<,
- 1.55225<, 826.15, - 1.55065<, 826.25, - 1.54888<, 826.35, - 1.54693<,
- 1.54476<, 826.55, - 1.54238<, 826.65, - 1.53974<, 826.75, - 1.53683<,
- 1.53362<, 826.95, - 1.53008<, 827.05, - 1.52618<, 827.15, - 1.52188<,
- 1.51714<, 827.35, - 1.51192<, 827.45, - 1.50618<, 827.55, - 1.49986<,
- 1.49292<, 827.75, - 1.48529<, 827.85, - 1.47691<, 827.95, - 1.46772<,
- 1.45765<, 828.15, - 1.44662<, 828.25, - 1.43456<, 828.35, - 1.42138<,
- 1.40701<, 828.55, - 1.39135<, 828.65, - 1.37433<, 828.75, - 1.35587<,
- 1.33588<, 828.95, - 1.31429<, 829.05, - 1.29107<, 829.15, - 1.26615<,
- 1.23952<, 829.35, - 1.21119<, 829.45, - 1.18119<, 829.55, - 1.14959<,
- 1.11649<, 829.75, - 1.08204<, 829.85, - 1.0464<, 829.95, - 1.0098<,
- 1.00981<, 830.15, - 0.97247<, 830.25, - 0.934639<, 830.35, - 0.896579<,
- 0.858546<, 830.55, - 0.820785<, 830.65, - 0.783523<, 830.75, - 0.746964<,
- 0.711284<, 830.95, - 0.676629<, 831.05, - 0.643116<, 831.15, - 0.610835<,
- 0.579847<, 831.35, - 0.550192<, 831.45, - 0.521888<, 831.55, - 0.494939<,
- 0.469329<, 831.75, - 0.445036<, 831.85, - 0.422026<, 831.95, - 0.400258<,
- 0.379686<, 832.15, - 0.360261<, 832.25, - 0.341932<, 832.35, - 0.324646<,
- 0.308351<, 832.55, - 0.292994<, 832.65, - 0.278524<, 832.75, - 0.264891<,
- 0.252046<, 832.95, - 0.239943<, 833.05, - 0.228539<, 833.15, - 0.21779<,
- 0.207657<, 833.35, - 0.198101<, 833.45, - 0.189088<, 833.55, - 0.180583<,
- 0.172554<, 833.75, - 0.164972<, 833.85, - 0.157809<, 833.95, - 0.151039<,
- 0.144637<, 834.15, - 0.13858<, 834.25, - 0.132848<, 834.35, - 0.127419<,
- 0.122276<, 834.55, - 0.117401<, 834.65, - 0.112777<, 834.75, - 0.108389<,
- 0.104224<, 834.95, - 0.100267<, 835.05, - 0.0965073<, 835.15, - 0.092932<,
- 0.0895307<, 835.35, - 0.0862933<, 835.45, - 0.0832103<, 835.55, - 0.080273<,
- 0.0774729<, 835.75, - 0.0748025<, 835.85, - 0.0722544<, 835.95, - 0.0698218<,
- 0.0674985<, 836.15, - 0.0652783<, 836.25, - 0.0631559<, 836.35, - 0.0611258<,
- 0.0591833<, 836.55, - 0.0573237<, 836.65, - 0.0555426<, 836.75, - 0.053836<,
- 0.0522<, 836.95, - 0.0506312<, 837.05, - 0.049126<, 837.15, - 0.0476813<,
- 0.0462941<, 837.35, - 0.0449615<, 837.45, - 0.043681<, 837.55, - 0.0424499<,
- 0.0412659<, 837.75, - 0.0401269<, 837.85, - 0.0390306<, 837.95, - 0.037975<,
- 0.0369584<, 838.15, - 0.0359788<, 838.25, - 0.0350346<, 838.35, - 0.0341243<,
- 0.0332463<, 838.55, - 0.0323991<, 838.65, - 0.0315814<, 838.75, - 0.030792<,
- 0.0300296<, 838.95, - 0.029293<, 839.05, - 0.0285812<, 839.15, - 0.0278931<,
- 0.0272277<, 839.35, - 0.026584<, 839.45, - 0.0259613<, 839.55, - 0.0253586<,
- 0.0247751<, 839.75, - 0.02421<, 839.85, - 0.0236626<, 839.95, - 0.0231323<,
- 0.0226183<, 840.15, - 0.02212<, 840.25, - 0.0216368<, 840.35, - 0.0211682<,
- 0.0207135<, 840.55, - 0.0202723<, 840.65, - 0.019844<, 840.75, - 0.0194281<,
- 0.0190243<, 840.95, - 0.0186321<, 841.05, - 0.018251<, 841.15, - 0.0178806<,
- 0.0175206<, 841.35, - 0.0171706<, 841.45, - 0.0168302<, 841.55, - 0.0164992<,
- 0.0161771<, 841.75, - 0.0158637<, 841.85, - 0.0155587<, 841.95, - 0.0152618<,
- 0.0149727<, 842.15, - 0.0146912<, 842.25, - 0.0144169<, 842.35, - 0.0141498<,
- 0.0138895<, 842.55, - 0.0136358<, 842.65, - 0.0133884<, 842.75, - 0.0131473<,
- 0.0129122<, 842.95, - 0.0126829<, 843.05, - 0.0124592<, 843.15, - 0.012241<,
- 0.012028<, 843.35, - 0.0118202<, 843.45, - 0.0116173<, 843.55, - 0.0114193<,
- 0.0112259<, 843.75, - 0.011037<, 843.85, - 0.0108525<, 843.95, - 0.0106723<,
- 0.0104962<, 844.15, - 0.0103241<, 844.25, - 0.0101559<,
- 0.00999154<, 844.45, - 0.00983082<, 844.55, - 0.00967368<,
- 0.00952<, 844.75, - 0.0093697<, 844.85, - 0.00922268<, 844.95, - 0.00907884<<
1. S Marguiles, Am. J. Phys. 52 (6), 515-518, (1984). This parer presents a physical analysis showing that the edge effects are responsible for the restoring force. A detailed calculation of the effect is not included.
2. J.C. Maxwell, A Treatise on Electricity and Magnetism, 3rd Edition, Vol. 1, reprinted in
the USA by Oxford University Press in the Oxford Classic Series, (1998), Chapter XII, pp.
309-310. Rogowski extended Maxwell’s sketch of the method in W. Rogowski, “Archiv
fur Electrotechnik,” Vol. 12, 1-8, (1923). See also E. Albayrak, Turk J. Phy., 25, 181-193,
(2001).
tion of the effect is not included.
34
ConformalCapacitor.nb
2. J.C. Maxwell, A Treatise on Electricity and Magnetism, 3rd Edition, Vol. 1, reprinted in
the USA by Oxford University Press in the Oxford Classic Series, (1998), Chapter XII, pp.
309-310. Rogowski extended Maxwell’s sketch of the method in W. Rogowski, “Archiv
fur Electrotechnik,” Vol. 12, 1-8, (1923). See also E. Albayrak, Turk J. Phy., 25, 181-193,
(2001).
3. Michael P. Bradley. Go to http://physics.usask.ca/~bradley/ then follow the link entitled “Lecture on Solution of Parallel Plate Capacitor Edge Effects via Conformal Mapping”
4. J.C. Maxwell, ibid, appendix PLATES, Fig. XII.
5. This particular result appears either not to have been noticed previously, or not to have
been exploited, possibly due to a lack of computational resourses.
6. E. R. Dietz, Am. J. Phys. 72 (12), 700-701, (2004).