View
6
Download
0
Category
Preview:
Citation preview
10/5/2019
1
Electromagnetics:
Electromagnetic Field Theory
Electrostatic Devices
Outline
• Laplace’s Equation• Derivation•Meaning• Solving Laplace’s equation
•Resistors•Capacitors
Slide 2
1
2
10/5/2019
2
Derivation of Laplace’s Equation
Slide 3
Derivation of Poisson’s Equation (1 of 2)
Slide 4
In electrostatics, the field around charges is described by Gauss’ law
vD
In LI media, the constitutive relation is 𝐷 𝜀𝐸 so Gauss’ law can be written in terms of 𝐸.
vE
In electrostatics, the electric field is related to electric potential through 𝐸 ∇𝑉. This definition can be used to put the above equation solely in terms of the electric potential.
vV
3
4
10/5/2019
3
Derivation of Poisson’s Equation (2 of 2)
Slide 5
The previous slides leads to Poisson’s equation for inhomogeneous media
vV vV
If the medium is homogeneous, is a constant and can be brought to the righthand side of the equation.
vV
2 vV
Poisson’s equation for inhomogeneous media
Poisson’s equation for homogeneous media
Derivation of Laplace’s Equation
Slide 6
In the absence of charge, v = 0 and Poisson’s equation reduces to Laplace’s equation.
0V vV
2 vV
2 0V
Laplace’s equation for inhomogeneous media
Laplace’s equation for homogeneous media
2 is called “the Laplacian”
5
6
10/5/2019
4
No Charge in Electrostatics?
Slide 7
+ + + + + + + + + + + + + + + + + +
‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐
No charge in the region between the plates.
2 0V
Notes
• Poisson’s and Laplace’s equations describe how electric potential varies throughout a volume.
• These are scalar differential equations and usually easier to solve than vector differential equations.
• Use Poisson’s equation when there is charge and Laplace’s equation when there is not.
• Laplace’s equation is particularly important in electrostatics because it can be used to calculate electric potential around conductors maintained at different voltages.
• Uniqueness theorem states that there exists only one solution.
Slide 8
7
8
10/5/2019
5
Meaning of Laplace’s Equation
Slide 9
Meaning of Laplace’s Equation
10
2 0u
Laplace’s equation is
2 is a 3D second‐order derivative.
A second‐order derivative quantifies curvature.
But, we set the second‐order derivative to zero.
Functions satisfying Laplace’s equation vary linearly.
9
10
10/5/2019
6
Problem Setup
11
Suppose we know the value of V(x,y) at some points in space.
What does the function look like at every other point?
Figure it out by solving Laplace’s equation.
2 , 0V x y
Solution of Laplace’s Equation
12
Laplace’s equation is sort of a “number filler inner.”
Laplace’s equation fills in the numbers so they vary linearly between known regions.
11
12
10/5/2019
7
Another Example
13
Solving Laplace’s Equation
Slide 14
13
14
10/5/2019
8
Recipe for Solving Laplace’s Equation
Slide 15
Laplace’s equation is solved as a boundary value problem (i.e. partial differential equation plus boundary conditions).
2. Solve Laplace’s equation 2V = 0 in each homogeneous region.
a. When V is a function of only one variable, use direct integration.
b. Otherwise, use separation of variables.
3. Apply the boundary conditions at the edges of the homogeneous regions.
4. Calculate 𝐸 from V using 𝐸 ∇𝑉.
5. Calculate 𝐷 from 𝐸 using 𝐷 𝜀𝐸.
1. Choose a coordinate system that will simplify the math.
Example #1 – Voltage Between the Plates of a Capacitor
Slide 16
Suppose there exists a medium with permittivity and thickness d.
15
16
10/5/2019
9
Example #1 – Voltage Between the Plates of a Capacitor
Slide 17
Then apply a voltage V0 across that medium.
Suppose there exists a medium with permittivity and thickness d.
Example #1 – Voltage Between the Plates of a Capacitor
Slide 18
Then apply a voltage V0 across that medium, which puts charge on the plates.
Suppose there exists a medium with permittivity and thickness d.
17
18
10/5/2019
10
Example #1 – Voltage Between the Plates of a Capacitor
Slide 19
Calculate the electric potential and electric field between the plates.
Then apply a voltage V0 across that medium, which puts charge on the plates.
Suppose there exists a medium with permittivity and thickness d.
Example #1 – Voltage Between the Plates of a Capacitor
Slide 20
Step 1 – Choose a coordinate system.
Cartesian
19
20
10/5/2019
11
Example #1 – Voltage Between the Plates of a Capacitor
Slide 21
Step 2 – Solve Laplace’s equation
2
0
0
0 0
V
V
V d V
If we assume the device is uniform in the x and ydirections, Laplace’s equation reduces to
2 2 2
2 2 20
V V V
x y z
2
2 0d V
dz
Example #1 – Voltage Between the Plates of a Capacitor
Slide 22
Step 2 – Solve Laplace’s equation
Integrate to get
2
20
d VV z az b
dz
21
22
10/5/2019
12
Example #1 – Voltage Between the Plates of a Capacitor
Slide 23
Step 3 – Apply boundary conditions.
First boundary condition…
0 0 0 0 0V V a b b
Example #1 – Voltage Between the Plates of a Capacitor
Slide 24
Step 3 – Apply boundary conditions.
Second boundary condition…
00 0 0
VV d V V d a d V a
d
23
24
10/5/2019
13
Example #1 – Voltage Between the Plates of a Capacitor
Slide 25
Step 3 – Apply boundary conditions.
Altogether, the solution is
0 0V
V z z z dd
Example #1 – Voltage Between the Plates of a Capacitor
Slide 26
Step 4 – Calculate 𝐸 from V.
The electric field intensity is
0 0 0 ˆ z z z
V V Vd dE V E V E z E a
dz dz d d d
Observe that 𝐸 does not
depend on .
25
26
10/5/2019
14
Example #1 – Voltage Between the Plates of a Capacitor
Slide 27
Step 5 – Calculate 𝐷 from 𝐸.
Applying the constitutive relation, we get the electric flux density
0 0ˆ ˆ z z
V VD E D a D a
d d
Resistors
Slide 28
27
28
10/5/2019
15
What is a Resistor?
Slide 29
A resistor is a passive two‐terminal electrical component that limits the conductivity so as to limit current flow.
Analysis Setup
Slide 30
S
J
+‐ V
I
?V
RI
29
30
10/5/2019
16
Derivation of Resistance for Uniform Conductivity
Slide 31
Electric Current Density
IJ
S
Ohm’s Law
J E
E
Electric Field Intensity
VE
V
I V
S
V
I S
R RS S
Derivation of Resistance for Nonuniform Conductivity
Slide 32
Voltage across conductor
V E d
Now we must use electromagnetic analysis to derive V and I.
Current through conductor
S S
I J ds E ds
S
E dV
RI E ds
31
32
10/5/2019
17
Recipe for Analyzing Resistors
1. Choose a convenient coordinate system.2. Assume V0 as the potential difference across the
terminals of the conductor.3. Calculate electric potential V by solving Laplace’s
equation 2V = 0.4. Calculate 𝐸 using 𝐸 ∇𝑉.5. Calculate I from .
6. Calculate R using R = V0/I.
Slide 33
S
I E ds
Note: The final equation for R should not contain V0 or I. Use this as a self‐check.
The Parallel Plate Resistor
Slide 34
dR
S
surface areaS
33
34
10/5/2019
18
Capacitors
Slide 35
What is a Capacitor?
Slide 36
A capacitor is a passive two‐terminal electrical component that can store and release electric energy. It supplies current so as to keep the voltage across its terminals constant.
35
36
10/5/2019
19
Capacitance, C
Slide 37
Capacitance is defined as the magnitude of the charge on one of the plates to the potential difference between the two plates.
Q
Q
0
QC
V
We do not care about the signs to calculate capacitance.
Recipe for Analyzing Capacitors
1. Choose a convenient coordinate system.
2. Let the plates carry charges +Q and -Q.
3. Calculate 𝐷 using Gauss’ law.
4. Calculate 𝐸 using 𝐸 𝐷/𝜀.
5. Calculate V0 using .
6. Calculate C using .
Slide 38
0
L
V E d
0C Q VNote: The final equation for C should not contain Q or V0. Use this as a self‐check.
37
38
10/5/2019
20
Some Simple Capacitors
Slide 39
SC
d
2
ln
LC
ba
39
Recommended