Project Stuff Real cool

Solution of Laplace’s
Equation:
Finite Difference Method
Julia A. London
Applied Magnetism ECE 2317
Professor Jackson
Fall 2013

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I. Abstract
A problem was presented to prove the usefulness of the Laplace equation, using the
Finite Difference method, in equating the properties of a non-symmetrical surface. Several
equations were derived and given to the class to solve for several properties; the change in
potential, the electric field, the surface charge density and the capacitance per unit length.
Microsoft Excel was used to model and solve for these several properties by creating a gird
system. Different sets of node amounts were used to solve for the properties of the model, and as
the node amount increased so did the accuracy of the approximation of each property. After these
properties were all solved for, the model was compared with that of an ideal parallel-plate
capacitor. Throughout the project it was observed that the model was very similar to the top of an
ideal parallel-plate capacitor. A major difference that was found, during the calculations, was in
the flux plots, and the overall capacitance per unit length of the model. It was noted that the flux
lines for an ideal parallel-plate capacitor would go towards its second plate, whereas for the
magnetic strip’s flux lines went towards all the edges of the box. It was also noted that the model
had a higher capacitance than an ideal parallel-plate capacitor.
After comparing the model with an ideal parallel-plate capacitor, another model was
created, where the permittivity of the area under the magnetic strip was higher that then the rest
of the model. This part of the project was an extra credit assignment and the purpose of that
section will be explained in that section.

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II. Introduction
To find the capacitance between a parallel plate capacitor you divide the total charge
accumulated upon each surface of the capacitor by the voltage drop between the two plates.
Another method is by dividing the area of the plates by the distance between them and
multiplying this product by the permittivity between the two plates. These methods appear
simple but what if there was a much larger area between the capacitor plates, or what if the
capacitor didn’t have plates; instead it was some arbitrary shape. Sometimes in electrostatics
there will arise a problem that cannot be easily solved using symmetry, considering that not
many things in the world are perfect shapes. When more complicated boundaries come into play
more numerical methods are needed to analysis certain phenomenon. One of such methods is by
using the Laplace formula.
This project will be able to prove and display the usefulness of the Laplace equation in
solving the numerical electrostatic properties of non-symmetrical objects. To begin the original
Laplace equation,
∇2
𝛷 = 0, (1)
was used to derive several equations used to calculate the potential, the electric field, and the
capacitance per unit length. A model, an infinitely long magnetic strip with the potential of 1[V]
surrounded by a perfectly conducting box, was used to represent a nonsymmetrical figure.
Several different node amounts were used to calculate and test the accuracy of different
computations. With these nodes a grid was formed to calculate said unknowns using different
derivations of the Laplace formula which were input into an Excel spreadsheet.
The unknown properties that will be calculated throughout the project are the changes in
potential throughout the model, the charge density on the strip, and the capacitance per unit
length of the model. Each task will display different examples of different computations derived
from the Laplace formula. Each computation will generate a result depending on the amount of
nodes within the conducting box, in accordance to what was requested for that task.

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III. Results:
Task1
In Task 1 the problem stated that the potential of each node throughout the model needed
to be found. The model can be shown in Figure 1
with N equaling the number of segments on the magnetic strip. N was solved for using equation
𝑁 = 5𝑚, (2)
with m being a given.
For the model used in this task m equaled 3, thus resulting in the first column of Tables 1
and 2,
L
w
H
ԑ0
y
x
Figure 1 – The model for an example of an non-symmetrical object. £
equals 6h and w equals 5h.
h
££
£

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Unknowns
Number of Segments
Within
m 3 5 7
h=N∆ 3 5 7
£=6h 18 30 42
w=5h=N 15 25 35
L 51 85 119
H 21 35 49
Upon knowing this, the potential throughout the model was solved using equation
𝛷( 𝑥, 𝑦) =
𝛷( 𝑥+∆,𝑦)+𝛷( 𝑥,𝑦+∆)+𝛷( 𝑥−∆,𝑦)+𝛷( 𝑥,𝑦−∆)
4
[V], (3)
which was derived from the original Laplace equation. ∆ was equal to the shift in Excel cells,
with ∆ = 1 cell. Knowing that the magnetic strip has a potential of 1 [V] and that the inner wall
of the box had zero potential, this equation was input into an Excel spreadsheet in order to find
the potential of each node within the model. Table 3 represents the potential of each node
aligned with the magnetic strip
Potential [V] (nodes 1-13)
0.000 0.005 0.010 0.016 0.022 0.029 0.037 0.046 0.058 0.071 0.089 0.111 0.140
0.179 0.232 0.308 0.422 0.614 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000
1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 0.614 0.422 0.308 0.232 0.179
Unknowns
Number of Nodes (Excel
Cells) Within
m 3 5 7
h 4 6 8
£ 19 31 43
w 16 26 36
L 52 86 120
H 22 36 50
Table 1 – The Number of Segments used for
each m.
Table 2 – The Number of Nodes (Excel
spreadsheet Cells) used for each m.
Table 3 – The potential at each node in the row with the magnetic strip.

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0.140 0.111 0.089 0.071 0.058 0.046 0.037 0.029 0.022 0.016 0.010 0.005 0.000
After finding the potentials it can be noted that the closer the nodes are to the magnetic
strip the higher their potential. This is also true for an ideal plate capacitor. It was also seen that
the change in potential across each node became more drastic the closer the wall of the
conducting box is to the magnetic strip. This change is due to the idea that with each node a
charge moving through the area, it is getting closer to the potential of zero volts; thus the
potential of each node must be decreasing the further it is from the magnetic strip.
Task 2
After being able to find the potential of each node throughout the model, it is now
possible to create a flux plot depicting the electric field. Since the equation for the electric field is
𝐸𝑓𝑖𝑒𝑙𝑑 =
𝑉
𝑑
[
𝑉
𝑚
] (4)
with V equaling the potential at a certain point a d distance away from its origin.
It was requested that in this task that m equal to 7, thus the third columns from
Tables 1 and 2 were used to generate a table of potential values similar to that of Task 1. With
these potential values a contour plot, in which a flux plot could be made to modeling the change
in electric field, was constructed. Figure 2 displays the equipotential and flux lines needed to
represent a flux plot of the model.

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This flux plot represents the change in the electric field of the non-symmetrical object
model. It can be noted that the electric field around the edges, and below the strip, is much
stronger than that at the top of strip. This outcome is to be expected because eth distance of the
top of the conducting box is much farther from the magnetic strip than the bottom of the box.
With this knowledge and upon using Equation 4 proves the electric field must be weaker above
the magnetic strip. Said result is not much different from that of an ideal parallel-plate capacitor.
In fact this flux plot appears similar to that of the top half of an ideal parallel capacitor. The most
notable difference between this model and an ideal parallel-plate capacitor is the direction of flux
lines.
For an ideal parallel-plate capacitor, flux lines all went from the highest potential plate to
the lower potential plate. For this model all of the flux lines went towards the lower potential
around the edges of the box. This means that the electric field goes only as far as the box will
allow it but is strongest directly below the plate and at the edges. Due to its lack of symmetry the
electric field is also not symmetrical. Whereas for the ideal parallel-plate capacitor the electric
field is strongest in between the two plates, and this electric field would symmetrical via the
center of a plate.
Flux Plot
0.8-1
0.6-0.8
0.4-0.6
0.2-0.4
0-0.2
0.80[V]-1.00[V]
0.60[V]-0.80[V]
0.40[V]-0.60[V]
0.20[V]-0.40[V]
0.00[V]-0.20[V]
`Figure 2 – displays the change in potential and the flux lines for the model
depicting the strength of the electric field throughout the model.

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Task 3
For Task 3 the charge density at the top and bottom of the strip were to be found. These
charge densities can be found by deriving another formula from the original Laplace formula and
using the equation
𝑝𝑠 = 𝐷 ∙ ṉ̂. (5)
Thus the equations
𝑝𝑠
𝑡𝑜𝑝
= −𝜀0 (
𝛷( 𝑥,𝑦+∆)−𝛷( 𝑥,𝑦)
∆
)[
𝐶
𝑚2 ], (6)
and
𝑝𝑠
𝑏𝑜𝑡𝑡𝑜𝑚
= 𝜀0(
𝛷( 𝑥,𝑦+∆)−𝛷( 𝑥,𝑦)
∆
) [
𝐶
𝑚2 ]. (7)
where derived.
Now w has a width of one meter and, as instructed, m equals 7 thus the third column of
Tables 1 and 2 are used to generate a table of the potential of each node throughout the model.
The equation
∆= 𝑤/𝑁, (8)
with N equaling the amount of nodes across the magnetic strip. Knowing this, the charge
densities at the top and bottom of the magnetic strip we’re able to be calculated, as seen in Table
4.
Surface Charge Density
Top [pC/m2
] 48.42 32.84 25.31 21.07 18.39 16.55 15.23 14.23 13.46
Bottom
[pC/m2
] 72.95 58.53 52.06 48.79 47.00 45.96 45.33 44.94 44.70
Total
[pC/m2
] 121.4 91.37 77.37 69.86 65.39 62.51 60.55 59.17 58.15
Position on
the Strip [m] -0.500 -0.471 -0.443 -0.414 -0.386 -0.357 -0.329 -0.300 -0.271
Table 4 – displays the total surface charge density and the densities at the top and
bottom of the magnetic strip corresponding to its location.

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Top [pC/m2
] 12.85 12.37 11.99 11.69 11.45 11.27 11.14 11.06 11.02
Bottom
[pC/m2
] 44.54 44.44 44.38 44.34 44.32 44.30 44.29 44.29 44.28
Total
[pC/m2
] 57.40 56.82 56.37 56.03 55.77 55.57 55.43 55.34 55.30
Position on
the Strip [m]
-
0.243 -0.214 -0.186 -0.157 -0.129 -0.100 -0.071 -0.043 -0.014
Top [pC/m2
] 11.02 11.06 11.14 11.27 11.45 11.69 11.99 12.37 12.85
Bottom
[pC/m2
] 44.28 44.29 44.29 44.30 44.32 44.34 44.38 44.44 44.54
Total [pC/m2
] 55.30 55.34 55.43 55.57 55.77 56.03 56.37 56.82 57.40
Position on
the Strip [m] 0.014 0.043 0.071 0.100 0.129 0.157 0.186 0.214 0.243
Top [pC/m2
] 13.46 14.23 15.23 16.55 18.39 21.07 25.31 32.84 48.42
Bottom
[pC/m2
] 44.70 44.94 45.33 45.96 47.00 48.79 52.06 58.53 72.95
Total [pC/m2
] 58.15 59.17 60.55 62.51 65.39 69.86 77.37 91.37 121.4
Position on
the Strip [m] 0.271 0.300 0.329 0.357 0.386 0.414 0.443 0.471 0.500
As expected from the flux plot there is a higher charge density at the bottom of the strip. And as
expected the total charge density at the tips of the strip were far greater than that anywhere else. This
change in charge density can also be seen in Figure 3

London 9
Knowing that the charge density is much greater at the edges, means the electrical field
should be greatest on the edges, which was already noted in Task 2.
These outcomes were also as predicted due to previous knowledge of the subject matter.
The ‘knife edge’ effect is what causes the charge density to accumulate around the edges of the
strip, thus causing an increase in the electric field. The ‘knife edge’ effect is the idea that charges
like to be evenly spaced on a surface. So even on a very small surface, such as a point or an edge
in this case, charges will try to group up evenly, thus; the smaller the surface the higher the
density.
Task 4
As requested, in the final task the capacitance per unit length will be calculated using another
derivation of the Laplace equation for finding the amount charge at each node across the magnetic strip
𝑄 ≈ 𝜀0 ∑(𝛷( 𝑥 𝑛, 𝑦𝑠) −
𝑁
𝑛=1
𝛷( 𝑥 𝑛, 𝑦𝑠 + ∆)) + 𝜀0 ∑(𝛷( 𝑥 𝑛,𝑦𝑠) −
𝑁
𝑛=1
𝛷( 𝑥 𝑛,𝑦𝑠 − ∆))
+𝜀0[𝛷( 𝑥1, 𝑦𝑠) − 𝛷( 𝑥0,𝑦𝑠)] + 𝜀0[𝛷( 𝑥 𝑁, 𝑦𝑠) − 𝛷( 𝑥 𝑁+1,𝑦𝑠)][𝐶]. (8)
0.00
20.00
40.00
60.00
80.00
100.00
120.00
140.00
-0.500 -0.357 -0.214 -0.071 0.071 0.214 0.357 0.500
SurfaceChargeDensity[pC/m^2]
Node Position [m]
Top
Bottom
Total
Figure 3 – displays the total surface charge density and the densities at the top and
bottom of the magnetic strip corresponding to its location.
asd.

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This equation will be used in
𝐶 =
𝑄
𝑉
[𝐹], (8)
in order to find the capacitance per unit length, with the potential across the strip equal to 1[V]. The
capacitance will be calculated for each value of m on Table 1. With this, it will be possible to make a
prediction of the magnetic strip’s capacitance per unit length.
The capacitance for each value of m can be seen in Table 5
m ∆=1/N Value [m] Capacitance [pF/m]
3 0.0667 72.93
5 0.04 71.88
7 0.0286 71.44
Two graphs were to be made to estimate the capacitance per unit length of the magnetic strip vs.
its delta. The capacitance of the strip as N approaches infinity can now be estimated by creating a
quadratic line that fits all three of the data points, as seen in Figure 4
5
Table 5 – displays the capacitance per unit length for each
value of m.
asd.

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and a trend line that fits the last two data points, as seen in Figure 5
As seen from Figures 4 and 5, it can be estimated that as N increases the capacitance will hit
either y intercept of each of the graphs. For Figure 4, as N increases,it’s estimated that the capacitance
Figure 4 – displays the capacitance per unit length for all three values of ∆ with a
quadratic trend line to help predict the capacitance per unit length at higher N amounts.
The equation of the trend line is also displayed.
asd.
Figure 5 – displays the capacitance for the m values 5 and 7 in correspondence with
their ∆s with a linear trend line to help predict the capacitance at different N amounts.
The equation of the trend line is also displayed.
asd.
y = 24.336x2 + 36.937x + 70.364
71.2
71.4
71.6
71.8
72
72.2
72.4
72.6
72.8
73
73.2
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07
Capacitance[pF]
Delta [m]
Capacitance per unit Length
Capacitance [pF/m]
Poly. (Capacitance [pF/m])
y = 38.65x + 70.335
71.4
71.45
71.5
71.55
71.6
71.65
71.7
71.75
71.8
71.85
71.9
71.95
0 0.01 0.02 0.03 0.04 0.05
Capacitance[pF]
Delta [m]
Capacitance per unit Length
Capacitance [pF/m]
Linear (Capacitance
[pF/m])

London 12
per unit length will reach 70.364 [pF/m]. For Figure 5 the capacitance per unit length is estimated to reach
70.335[pF/m]. The percent difference between these two values can be found using
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = |
𝑉𝑎𝑙𝑢𝑒1−𝑉𝑎𝑙𝑢𝑒2
𝑉𝑎𝑙𝑢𝑒1+𝑉𝑎𝑙𝑢𝑒2
2⁄
| ∙ 100 [%] (9)
In using this equation the percent difference between these two values is .0412%. From prior knowledge
it appears as though these methods of extrapolation seem valid. Although out of the two graphs Figure 5
appears to be the most accurate,because it has more data in which to create a justifiable trend line. And
with more data it becomes easier to accurately create an estimate.
For an ideal parallel-plate capacitor
𝐶 = 𝜀0
𝐿
ℎ
[𝐹/𝑚] (10)
can be used to find the capacitance per unit length. L would be the width of the strip, 1[m]. The h is equal
to .2[m] as defined by Table 1 if w equaled 1[m]. There are two values of h because there’s a model of an
ideal parallel-plate capacitor above and below the strip. And with the dimensions given the total
capacitance per unit length of an ideal parallel-plate capacitor would be 44.27[pF/m] which is about two
thirds the average of the extrapolated values of the model.
IV. Conclusions
Throughout the project, if the derivations were done correctly, it appears as though the Laplace
formula, paired with an accurate enough grid system, is very useful in finding the properties of non-
symmetrical objects. By using a grid the changes throughout and around the object can be approximated
by the amount of nodes in the system. This is similar to how integration works, adding up pieces of an
area to get an approximate area. And with the Laplace formulas, the nodes in the grid can be used to find
important components of the model such as the potential and the charge density or other components that
gradually change. Upon knowing how to solve for such properties, other properties that depend on
gradually changing properties can be solved for. And as the node amount increases as should the accuracy
of the approximation.

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V. Extra Credit
As an extra credit assignment, a new model was made differing from the original model only by
increasing the permittivity under the strip, a new ԑr. While ԑr=2.1 the potential was recalculated and
the new capacitance was also solved for. A new table of potential was made along with two new
extrapolation graphs. Once the calculations were finished this new model was compared with the
old model to observe what differences the change in permittivity created. In the end the changes
between the two models will be compared with ideal parallel-plate capacitors with similar
charges. Because the permittivity was the only change Tables 1 and 2 both can still be used to
find the dimensions and values of the unknowns of the new model. To have a distinction
between the two models the original model was named Model A and the new model was named
Model B.
VI. Results
Task 1
For this task it was requested to recalculate the potential of the nodes aligned with the magnetic
strip. This recalculated potential can be seen in Table 6.
0.000 0.005 0.011 0.017 0.024 0.031 0.040 0.050 0.062 0.076 0.095 0.118 0.149
0.188 0.242 0.318 0.431 0.618 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000
1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 0.618 0.430 0.318 0.242 0.188
0.148 0.118 0.094 0.076 0.061 0.049 0.039 0.031 0.023 0.017 0.011 0.005 0.000
Table 6 – displays the potential for each node in the same row with the magnetic
strip. In this calculation the ԑr under the strip was 2.1.
asd.

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In comparison with Model A it appears as though the change in potential for each of the nodes
was higher for the Model B. This means that since the permittivity below the strip increased thus the
charge density closer to the higher permittivity should also increase. And with an increase in charge
density there is an increase in the electric field and, since the electric field and potential have a direct
relationship, the potential across each node increased as well. These relationships can be proven using
𝜌𝑠0 = 𝜀0 𝜀𝑟 𝐸𝑓𝑖𝑒𝑙𝑑[
𝐶
𝑚2
], (11)
and
𝑉𝐴𝐵 = ∫ 𝐸 ∙ 𝑑𝑟
𝐵
𝐴 [𝑉]. (12)
Thus because there’s a portion in Model B that has a higher permittivity than Model A, Model B
will have both a higher charge density below and to the edges of the strip. Model B will also have a
stronger electric field than Model A in these locations.
Task 2
For this task the capacitance per unit length for the three different values of m, as shown by Table
1, were found for Model B. With these capacitances two extrapolation charts were created,to predict the
actual capacitance per unit length. The capacitance per unit length for each delta can be seen in Table 7
An extrapolating quadratic tend line was created to estimate the capacitance per unit length as N went
towards infinity, as seen in Figure 6
m ∆=1/N Value [m] Capacitance [pF/m]
3 0.066667 73.551
5 0.04 72.889
7 0.028571 72.403
Table 7 – displays the capacitance per unit length for each value of m. In this
calculation the area that lies beneath the magnetic strip has an ԑr equal to 2.1.
asd.

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and a extrapolating trend line was created to do the same, as seen in Figure 7
From observing Figure 6 and 7 it can be seen that the new capacitances per unit lengths
are 70.661[pF/m] and 71.19[pF/m]. Upon using Equation (9) the percent difference between
these two amounts is .7459%. Comparing these values with that from Figures 4 and 5 it can be
Figure 6 – displays the capacitance per unit length for all three values of ∆ with a quadratic
trend line to help predict the capacitance per unit length for Model B. The equation of the
trend line is also displayed.
asd.
Figure 7 – displays the capacitance per unit length for the m values 5 and 7 in correspondence
with their ∆s with a linear trend line to help predict the capacitance per unit length for Model
B. The equation of the trend line is also displayed.
asd.
y = -463.1x2 + 74.219x + 70.661
72.2
72.4
72.6
72.8
73
73.2
73.4
73.6
73.8
0 0.02 0.04 0.06 0.08
Capacitance[pF/m]
Delta [m]
Capacitance per unit Length (ԑr=2.1)
Capacitance [pF/m]
Poly. (Capacitance
[pF/m])
y = 42.464x + 71.19
72.3
72.4
72.5
72.6
72.7
72.8
72.9
73
0 0.01 0.02 0.03 0.04 0.05
Capacitance[pF/m]
Delta [m]
Capacitance per unit Length (ԑr=2.1)
Capacitance [pF/m]
Linear (Capacitance
[pF/m])

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seen that the capacitance per unit length has increased. With this observation it can be noted that
with a higher permeability there will be a higher capacitance.
For an ideal parallel-plate capacitor Equation 10 is used to give the capacitance per unit
length, which would be equal to 92.969[pF/m]. This capacitance is about 131.08% the average
capacitance of Model B. This change could be caused by the dispersion of charge throughout
Model B. For Model B charges can be dispersed throughout the model whereas for an ideal
parallel-plate capacitor the charges are concentrated around the top, bottom and in between
plates. Due to this dispersion the charge density below and above the strip will not be as large as
the charge densities on the top and bottom plates of a parallel-plate capacitor.
VII. Conclusions
It appears as though with an increase in permittivity almost every property of nodes
increases with it. Even though the change in permittivity only appeared below the strip it affected
the properties of the nodes above it as well. And due to this change the capacitance per unit
length, the charge density, and the electric field all increased throughout Model B. This change
can also be true for an ideal parallel-plate capacitor. If the permittivity changed so that the
different permittivities are in parallel with the strip, or for an ideal parallel-plate capacitor, its
plates, then this different permittivity splits the capacitor in two. And when two capacitors are in
parallel with each other they add together thus increasing the total capacitance.
In turn, these test shows that the model, if split into two different permittivity areas will
not have a greater capacitance per unit length than an ideal parallel-plate capacitor with the same
divisions. Although without the division if the permittivity throughout the model and the parallel
plate capacitor is constant then the model will have a greater capacitance per unit length than an
ideal parallel-plate capacitor.

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VI. References
1. Professor Jackson. 2013. Capacitance Notes 23. Available
http://www0.egr.uh.edu/courses/ECE/ECE2317/SectionJackson/Class%20Notes/notes%2
023%202317%20Capacitance.pdf
2. Professor Jackson.2013.Flux Notes 9. Available
http://www0.egr.uh.edu/courses/ECE/ECE2317/SectionJackson/Class%20Notes/notes%2
09%202317%20Flux.pdf
3. Pierce, Rod. 2013.Percent Difference. Available http://www.mathsisfun.com/percentage-
difference.html

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VI. Academic Honesty Statement
This project represents completely my own work, and I have abided by the University of Houston
Academic Honesty Policy in doing this work and writing this report.

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