project electrostatics problem

8/3/2019 Project Electrostatics Problem

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The geometry class of the model is axis-symmetric.The precision is chosen to be

normal. Because higher precision leads to longer solution time and our current

problem is quite simple. Next, centimeters were chosen as units of length and

the Cartesian coordinates were preferred to the polar ones.

Now we can start editing the model and for this we select from the Edit menu

EditGeometryModel. We proceed with the geometry itself. At the beginning we

define the vertices specified in the drawing below,namely:A(-5,0), O(0,0),B(5,0),

C(5,12), D(-5,12), E(-3,5.2), F(-1.8,5.2),G(-0.6,5.2),H(0.6,5.2),I(1.8,5.2),J(3,5.2),

L(3,2.4),M(1.8,2.4),N(0.6,2.4),P(-0.6,2.4),Q(-1.8,2.4),R(-3,2.4)

We can now create the adges connecting the vertices. We create the edges of

the rectangle ABCD an then the edges EJ and RL. To be able to have different

layers of dielectrics, we are obliged to create the edges ER ,FQ,GP,HN,IM,JLtoo.Now that we are done with the models geometry, we can assign labels to

geometrical objects to describe material properties.

The model cantainsf ive blocks having different material properties: vacuum,

dry air, glass1 ,glass2,glass3. Although we take the vacuum and dry air as two

blocks, their properties are almost identical.We assign these labels to the

blocks.

Edge labes are used to define specific boundary conditions on inner and outward

boundaries of the region. In this case we need to specify boundary conditionsfor the shield (rectangle ABCD) and for the two conductive plates that form the

capacitor(EJ and RL)

We also need to assign vertex label toany vertex contacting the strips, to

specify that the strips be charged. No matter what vertices we choose, the

charge will br distributed through the entire conductor. We choose point

E,F,G,H,I,J asCharge+ and point R as Charge-, since the charges on the plates

of the capacitor have the same absolute value, but opposite signs.

Now we can proceed with building a mesh of finite elements.

TO define tha mesh density, we need to define spacing parameters in several

vertices of the model. We suppose that the electric field is most non-

homogenous near the ends of the conducting plates, so the mesh there must be

maximum dense. Therefore, the spacing valueo f 0.7 cm will br assigned to the

vertices E,J,L and R the value of 1.4 cm to the vertices A,B,C and D.


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If we build the mesh in all the blocks, it looks like this:


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The model is now ready, so we go on with defining data for material properties

and boundary conditions, using the pieces of information that were handed to us

at the beginning of the problem.

We continue with the edge label s data. We ll define the label Shield ashomogenous DIRICHLET boundary condition (U=0) and the labes Strip1 (the

positively charged one) and Strip2( the negatively charged one) as conductor (U-

i-=constant, i=1,2) conditions.

We also need to define the vertex labes Charge+ and Charge- to assign the

charge to the strips. While determing the capacitance, we will issue the value 1C

for Charge+ and -1C for Charge-.

All the data needed to solve the problem is now defined. At last we can solve theproblem and analyze the solution.


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Statistics of the model:

Blocks

6 blocks

6 meshed

5 labeled 225 nodes of mesh

Edges

20 edges

Vertices

16 vertices

7 labeled

0 isolated

Dimensions

Left:-5 cm Right: 5 cm

Top: 12 cm

Bottom:0 cm


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The results are:


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COLOR MAP OF POTENTIAL:


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INITIAL STATE VALUE OF POINT S1(-2,4)Coordinates

x -2 cm

y 4 cm

r 4.4721 cm

q 116.57 degVoltage U 9.842 e+10 V

Strength E 1.24e+12 V/m

Strength Ex -3.48e+11 V/m

Strength Ey -1.19e+12 V/m

Field Gradient

Displacement D 43.913 C/m2

Displacement Dx -12.322 C/m2

Displacement Dy -42.149 C/m2

Energy Density w 2.722e+13 J/m3 /Permittivity er 4

(the capacitance) = 0.102 e -10F

INITIAL STATE VALUE OF POINT S2(-1,4)

Coordinates

x -1 cm

y 4 cm

r 4.1231 cm

q 104.04 deg

Voltage U 1.021e+11 V




Field Gradient




Energy Density w 6.007e+12 J/m3

(the capacitance) =0.98e-11F

INITIAL STATE VALUE OF POINT S2(0,6)

Coordinates

x 0 cm

y 6 cm

r 6 cm

q 90 deg





Field GradientDisplacement D 49.899 C/m2


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Permittivity er 5

(the capacitance) =0.876-11F


TEST1. Two Charges Placed outside the Capacitor


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We place the following charges outside the capacitor (the vertices together

with their respective charges will be given below):

POINT Z(cm) R(cm) Spacing Electric charge (C)Q1 4 10 Automatic 22

Q2 -4 1 Automatic 22


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The results are:


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TEST1 STATE VALUE OF POINT S1(-2,4)

Local ValuesCoordinates

x -2 cm

y 4 cm

r 4.4721 cm

q 116.57 deg





Field Gradient


Displacement Dx -16.131 C/m2Displacement Dy -30.996 C/m2


(the capacitance) =0,619e-11F

TEST1 STATE VALUE OF POINT S2(-1,4)Local Values

Coordinates

x -1 cm

y 4 cm

r 4.1231 cm

q 104.04 degVoltage U 1.664e+11 V


Strength Ex -3.6e+11V/m


Field Gradient





(the capacitance) =0,601e-11 FTEST1 STATE VALUE OF POINT S3(0,6)Coordinates

x 0 cm

y 6 cm

r 6 cm

q 90 deg




Strength Ey 1.773e+12 V/m




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Displacement Dy 15.7 C/m2


Permittivity er 1

(the capacitance) =0,615e-11 F

Test 2 Two charged strips inside the capacitor

We place inside the capacitor two strip charges of length 2 cm the first one

(charge -20C) and the second one (charge 20C)


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The results are:


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TEST2 STATE VALUE OF POINT S1(-2,4)Coordinates

x -2 cm

y 4 cm

r 4.4721 cm

q 116.57 deg










TEST2 STATE VALUE OF POINT S2(-1,4)

x -1 cmy 4 cm

r 4.1231 cm

q 104.04 deg





Field Gradient




Energy Density w 1.413e+12 J/m3Permittivity er 1.0006


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TEST2 STATE VALUE OF POINT S3(0,6)

X=0;y=6cm



Strength Ex 3.092e+10 V/m


Field Gradient


Displacement Dx 1.3689 C/m2



Permittivity er 5


project electrostatics problem

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