Project :Electrostatic Problem
Main Problem:A parallel-plate capacitor of area A and spacing d is filled with five dielectrics as shown in Figure 1. Each occupies 1/5 of the volume.WHAT IS THE CAPACITANCE ?
For the five layers of dielectrics we have that: Material Glass 1 Air Glass2 Air Glass 3 K 10 1.000591 100 1.000591 1000
1 2 3 4 5
As for the input data, we are not interested in specifying the area A for the QuickField solution.We specify d= 2 cm an l=5cm. The shield that will surround the capacitor is a rectangle with a height of 12 cm and a width of 10 cm in it there is vacuum. The approach to solve the problem using QuickField: First of all, we crated a new empty electrostatics problem.
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,JL too.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 conditions for 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 nonhomogenous 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.
If we build the mesh in all the blocks, it looks like this:
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 as homogenous DIRICHLET boundary condition (U=0) and the labes Strip1 (the positively charged one) and Strip2( the negatively charged one) as conductor (Ui-=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 the problem and analyze the solution.
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
The results are:
COLOR MAP OF POTENTIAL:
INITIAL STATE VALUE OF POINT S1(-2,4)
Coordinates x y r q Voltage U Strength Strength Strength Field Gradient Displacement Displacement Displacement Energy Density -2 cm 4 cm 4.4721 cm 116.57 deg 9.842 e+10 V E 1.24e+12 Ex -3.48e+11 Ey -1.19e+12 D Dx Dy w 43.913 C/m2 -12.322 C/m2 -42.149 C/m2 2.722e+13
V/m V/m V/m
J/m3 /Permittivity
er
(the capacitance) = 0.102 e -10F INITIAL STATE VALUE OF POINT S2(-1,4)
Coordinates x y r q Voltage U Strength Strength Strength Field Gradient Displacement Displacement Displacement Energy Density -1 cm 4 cm 4.1231 cm 104.04 deg 1.021e+11 V E 1.164e+12 Ex -2.87e+11 Ey -1.13e+12 D Dx Dy w 10.317 C/m2 -2.5406 C/m2 -9.9989 C/m2 6.007e+12
V/m V/m V/m
J/m3
(the capacitance) =0.98e-11F INITIAL STATE VALUE OF POINT S2(0,6) Coordinates x 0 y 6 r 6 q 90 Voltage U Strength E Strength Ex Strength Ey Field Gradient Displacement D
cm cm cm deg 1.144e+11 1.127e+12 -1.07e+9 -1.13e+12 49.899 C/m2
V V/m V/m V/m
Displacement Dx Displacement Dy Energy Density Permittivity er
-0.047364 C/m2 -49.899 C/m2 w 2.812e+13 J/m3 5
(the capacitance) =0.876-11F
COLOR MAP OF POTENTIAL:
TEST1. Two Charges Placed outside the Capacitor
We place the following charges outside the capacitor (the vertices together with their respective charges will be given below):
POINT Q1 Q2 Z(cm) 4 -4 R(cm) 10 1 Spacing Automatic Automatic Electric charge (C) 22 22
The results are:
COLOR MAP OF POTENTIAL:
TEST1 STATE VALUE OF POINT S1(-2,4)
Local Values Coordinates x y r q U E Ex Ey D Dx Dy w -2 cm 4 cm 4.4721 cm 116.57 deg 1.617e+11 9.866e+11 -4.55e+11 -8.75e+11 34.942 C/m2 -16.131 C/m2 -30.996 C/m2 1.724e+13
Voltage Strength Strength Strength Field Gradient Displacement Displacement Displacement Energy Density
V V/m V/m V/m
J/m3
(the capacitance) =0,619e-11F TEST1 STATE VALUE OF POINT S2(-1,4)
Local Values Coordinates x y r q Voltage U Strength Strength Strength Field Gradient Displacement Displacement Displacement Energy Density -1 cm 4 cm 4.1231 cm 104.04 deg 1.664e+11 V E 8.983e+11 Ex -3.6e+11 V/m Ey -8.23e+11 D Dx Dy w 7.9588 C/m2 -3.1899 C/m2 -7.2916 C/m2 3.575e+12
V/m V/m
J/m3
(the capacitance) =0,601e-11 F TEST1 STATE VALUE OF POINT S3(0,6)
Coordinates x y r q Voltage U Strength Strength Strength Field Gradient Displacement Displacement 0 cm 6 cm 6 cm 90 deg 1.628e+11 V E 1.86e+12 Ex -5.62e+11 Ey 1.773e+12 D Dx 16.471 C/m2 -4.9805 C/m2
V/m V/m V/m
Displacement Dy Energy Density w Permittivity er
15.7 C/m2 1.532e+13 1
J/m3
(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)
The results are:
TEST2 STATE VALUE OF POINT S1(-2,4)
Coordinates x y r q Voltage U Strength Strength Strength Field Gradient Displacement Displacement Displacement Energy Density -2 cm 4 cm 4.4721 cm 116.57 deg 4.527e+10 V E 5.232e+11 Ex -1.61e+11 Ey 4.978e+11 D Dx Dy w 18.53 C/m2 -5.7015 C/m2 17.631 C/m2 4.847e+12
V/m V/m V/m
J/m3
(the capacitance) =0,22e-10F TEST2 STATE VALUE OF POINT S2(-1,4)
x y r q Voltage U Strength Strength Strength Field Gradient Displacement Displacement Displacement Energy Density Permittivity -1 cm 4 cm 4.1231 cm 104.04 deg 4.699e+10 V E 5.648e+11 Ex -1.3e+11 V/m Ey 5.497e+11 D Dx Dy w er 5.0042 C/m2 -1.1514 C/m2 4.8699 C/m2 1.413e+12 1.0006
V/m V/m
J/m3
(the capacitance) =0,212e-10F TEST2 STATE VALUE OF POINT S3(0,6)
X=0;y=6 cm Voltage U 4.738e+10 V Strength E 6.183e+11 Strength Ex 3.092e+10 Strength Ey 6.175e+11 Field Gradient Displacement D 27.372 C/m2 Displacement Dx 1.3689 C/m2 Displacement Dy 27.338 C/m2 Energy Density w 8.462e+12 Permittivity er 5
V/m V/m V/m
J/m3
(the capacitance) =0,211e-10F
