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EE204B
Electromagnetics
Electromagnetics
Lesson 11 Electrostatic BoundaryValue Problems
Poisson’s and Laplace’s Equation
Instructor: Prof. Yong H. Won Lesson 11 Topics
Electrostatic BoundaryValue
Problems (Chap. 6) • Poisson’s and Laplace’s Equations
Eq
• Uniqueness Theorem
• Solution of Poisson’s and Laplace’s
Equations EE204B Electromagnetics Prof. Yong H. Won Spring 2010 2 Lesson 11 Poisson’s and Laplace’s Equations
So far we learned Electrostatics in free space and in material space.
Given a certain charge distribution, we know how to find E by using
either Coulomb’s law or Gauss’s law. However, in reality, often we don’t
know charge distribution.
Consider the following problem.
z 1V z=d z=0 The upper plate at z = d is attached to 1 V.
The lower plate at z = 0 is attached to ground
What is E between the two plates?
EE204B Electromagnetics Prof. Yong H. Won Spring 2010 3 Lesson 11 Poisson’s and Laplace’s Equations EE204B Electromagnetics Prof. Yong H. Won Spring 2010 4 Lesson 11 Poisson’s and Laplace’s Equations Poisson’s Equation in
inhomogeneous medium EE204B Electromagnetics Prof. Yong H. Won Spring 2010 5 Lesson 11 Poisson’s and Laplace’s Equations
In Cartesian Coordinate System ∂V r ∂V r ∂V r
ax +
ay +
az
∂x
∂y
∂z
∂ 2V ∂ 2V ∂ 2V
∇ 2V = ∇ ⋅ (∇V ) = 2 + 2 + 2
∂x
∂y
∂z
∇V = In Cylindrical Coordinate System 1∂
∂V
1 ∂ 2V ∂ 2V
∇V=
(ρ
)+ 2
+2
2
∂z
ρ ∂ρ ∂ρ ρ ∂φ
2 In Spherical Coordinate System 1 ∂ 2 ∂V
1
∂
∂V
1
∂ 2V
∇ 2V = 2
(r
)+ 2
(sin θ
)+ 2 2
r ∂r
∂r
r sin θ ∂θ
∂θ
r sin θ ∂φ 2
EE204B Electromagnetics Prof. Yong H. Won Spring 2010 6 Lesson 11 Uniqueness Theorem
Consider the problem introduced at the beginning.
th
th
z It can be described by the following
z=d ∇ V =0 0≤ z≤d V =0 z = 0 ⎫ Boundary
⎬ Conditions
z = d ⎭ (A & B are 2 V = 1V 1V z=0 r
E =? boundaries
boundaries
of 0 ≤ z ≤ d) What is in the box is so called as a boundary value problem. The
solution is unique!
solution is unique!
The solution to Laplace’s equation or Poisson’s equation)
subject to certain boundary conditions is unique!
subject to certain boundary conditions is unique!
EE204B Electromagnetics Prof. Yong H. Won Spring 2010 7 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
Steps for solving either Poisson’s or Laplace’s equation analytically: 1. Choose the most appropriate representation of
Laplacian based on any symmetry.
2. Perform the integration of the differential equation to
obtain the most general solution for the potential.
3. Let this general solution to satisfy the boundary
conditions to find constants of integration.
diti
fi
Addi
Additionally, numerical methods, such as Finite Difference Method, Finite
Fi
Diff
Fi
Element Method, Method of Moments, are developed to solve Boundary
Value problems in situations where analytical solution is hard to find. EE204B Electromagnetics Prof. Yong H. Won Spring 2010 8 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
Example 111
z
z=d z=0 1V
Boundary value problem
∇2 V = 0
V=0
V=1V 0≤ z ≤d
z=0
z=d Solution
If ignoring the fringing field (valid when the plate width is much
larger than the distance d), V can be regarded as a function of z only. EE204B Electromagnetics Prof. Yong H. Won Spring 2010 9 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
∂ 2V
∇ V = 2 =0
∂z
2 ⇒ V = Az + B To determine A and B, use boundary conditions
determine and use boundary conditions z = 0, V = 0 ⇒ o=B − − − (1) z = d, V =1 ⇒ 1 = Ad + B − − − (2) From (1) and (2) 1
⎧
⎪A =
⇒⎨
d
⎪
⎩B=0
1
∴V = z
d
EE204B Electromagnetics r
∂V r
1r
E = −∇V = −
az = − az
∂z
d
Prof. Yong H. Won Spring 2010 E 10 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
Question
What is the surface charge density ρs on the two plates?
On the upper plate,
th
Assuming region is inside the plate and r
an On the lower plate
the lower plate r
an EE204B Electromagnetics region is in air
an is normal to the interface pointing to .
r
rr
From D2n D1n = ρs
an ⋅ ( D2 − D1 ) = ρ s
r
rr
r
1r
Q D1 = 0 ⇒ ρ s = an ⋅ D2 = −az ⋅ (− azε 0 )
d
ε0
∴ ρs =
d
r
r
r
From an ⋅ ( D2 − D1 ) = ρ s (Remember an is from 1 to 2)
r
rr
r
1r
Q D1 = 0 ⇒ ρ s = an ⋅ D2 = az ⋅ (− azε 0 )
d
ε0
∴ ρs = −
d
Prof. Yong H. Won Spring 2010 11 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
Example 112
An infinitely long coaxial cylindrical structure having an inner
conductor of radius a, and an outer conductor of radius b. The inner
Th
conductor is attached to V0; the outer one is grounded. a
V0
b What is E in between a and b?
EE204B Electromagnetics Prof. Yong H. Won Spring 2010 12 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
Solution
∇ 2V = 0 a≤ ρ ≤b 1∂
∂V
1 ∂ 2V ∂ 2V
∇V=
(ρ
)+ 2
+ 2 =0
2
ρ ∂ρ ∂ρ ρ ∂φ
∂z
2 The structure has a cylindrical symmetry
∴ ∂V
= 0,
∂φ i.e. V does not change with φ The structure has no variation along
The structure has no variation along z
∂V
=0
∂z
1∂
∂V
∂V
(ρ
)=0 ⇒ ρ
∴
= C (constant)
ρ ∂ρ ∂ρ
∂ρ
C
∴V = ∫ d ρ = C ln ρ + D
∴ ρ EE204B Electromagnetics Prof. Yong H. Won Spring 2010 13 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
To determine C and D, use boundary conditions, ρ = a,
ρ = b, V = V0 V =0 Substitute them into the general solution
th
th C ln a + D = V0 ⎫
⎬
C ln b + D = 0 ⎭
V= V0
ln(a / b)
V0
D=−
ln b
ln(a / b) C= V0
V0
ln ρ −
ln b
ln(a / b)
ln(a / b) r
∴ E = −∇V = − EE204B Electromagnetics a≤ ρ ≤b V0
1r
aρ
ln(a / b) ρ
Prof. Yong H. Won Higher potential
to lower potential,
Perpendicular to
the conducting
surface
Spring 2010 14 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
Example 113
Calculate the potential variation
between two infinite parallel
metal
metal plates in a vacuum.
x=0 x = x0 We assume no resistance,
therefore, no variation in y and z
directions. Since no charges exist between plates, we need to solve the Laplace’s equation: d 2V
=0
2
dx
Boundary conditions: V = V0 at x = 0; V = 0 at x = x0
EE204B Electromagnetics Prof. Yong H. Won Spring 2010 15 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
The solution of the Laplace’s equation will be in the form: V = C1 x + C2
Here C1 and C2 are integration constant that can be found from boundary cond. ⎧V0 = C1 ⋅ 0 + C2
⎧C2 = V0
⇒⎨
⎨
⎩0 = C1 ⋅ x0 + C2
⎩C1 = − V0 x0
Finally, the solution: The electric field is: EE204B Electromagnetics ⎛
x⎞
V = V0 ⎜1 − ⎟
⎝ x0 ⎠
V0
dV
E=−
ux = ux
dx
x0
Prof. Yong H. Won Spring 2010 16 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
Let’s assume that there is a charge with ρv uniformly distributed between plates
if
di ρ
d 2V
Poisson’s equation:
=− v
dx 2
ε0
ρv x 2
+ C3 x + C4
The solution will be in form: V = −
ε0 2
from the boundary
conditions: ⎧
ρv 02
⎧C4 = V0
⎪V0 = − ε 2 + C3 ⋅ 0 + C4
⎪
⎪
0
⇒⎨
ρv x0 2 ⎞
1⎛
⎨
2
⎪0 = − ρv x0 + C x + C
⎪C3 = − x ⎜ V0 − ε 2 ⎟
30
4
0⎝
0
⎠
⎩
⎪
ε0 2
⎩ ⎛
⎞⎛
ρv
x⎞
V = V0 ⎜1 +
x0 x ⎟⎜1 − ⎟
⎝ 2ε 0V0
⎠⎝ x0 ⎠ EE204B Electromagnetics V0 ⎛ ρ v x0 2 ⎛
x ⎞⎞
Ε = ⎜1 −
⎜ 2ε V ⎜1 − 2 x ⎟ ⎟ u x
⎟
x0 ⎝
0 0⎝
0 ⎠⎠ Prof. Yong H. Won Spring 2010 17 Lesson 11 Solving Poisson’s or Laplace ’s Eq . EE204B Electromagnetics Prof. Yong H. Won Spring 2010 18 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
Example 114
11
z
d
a ε2 ++++++++++++++++++++
 ε1 ρs 0 Both plates are grounded. In between there are two media
characterized by ε1 and ε2, respectively. There exists surface charges
on the dielectric interface, the density of which is
on the dielectric interface, the density of which is ρs (C/m2).
What is E between two plates?
(Note: this example is slightly different from Text Example 6.2. ε1 and ε2
this example is slightly different from Text Example
are exchanged. We used zaxis instead of xaxis.)
EE204B Electromagnetics Prof. Yong H. Won Spring 2010 19 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
Solution
The problem is inhomogeneous by itself, i.e., two different media are
involved. However, it can be treated as two homogeneous regions:
0 ≤ z ≤ a and a ≤ z ≤ d
In each region, V satisfies ∇ 2V = 0
Si
Since ρv in each region is 0.
So we will solve Lapalace’s equation ∇ 2V = 0 in each region, then
combine the solutions.
For 0 ≤ z ≤ a
∇ 2V = 0
∂V ∂V ∂V
+ 2 + 2 =0
2
∂x
∂y
∂z
2 2 2 EE204B Electromagnetics The structure is infinitely long along x, y
V has no x, ydependence
(i.e., V does not change with x and y.)
Prof. Yong H. Won Spring 2010 20 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
∂ 2V
∂V
∴ 2 =0 ⇒
= A (constant)
∂z
∂z
∴V = ∫ Adz = Az + B ( B is another constant) This is V for the region 0 ≤ z ≤ a, Let us denote it by V1. ∴V1 = Az + B
For a ≤ z ≤ d
∇ V =0
2 ⇒ ∂ 2V
=0
2
∂z ⇒ V = Cz + D
Denoting it by V2
V2 = Cz + D
EE204B Electromagnetics Prof. Yong H. Won Spring 2010 21 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
To determine A,B,C and D, use boundary conditions,
z = a, V1 = V2
Aa + B = Ca + D r
r
r
D2 = ε 2 E2 = ε 2 (−∇V2 ) = ε 2 (−Caz )
r
r
r
D1 = ε1 E1 = ε1 (−∇V1 ) = ε1 (− Aaz )
r
rr
r
r
r
From an ⋅ ( D2 − D1 ) = ρ s ⇒ az ⋅ (−Cε 2 az + Aε1a ) = ρ s
⇒ − Cε 2 + Aε1 = ρ s
Solve A, B, C , D from the above 4 box equations
a
d −a
C = − ρs
A = ρs
(ε 2 − ε1 )a + ε1d
(ε 2 − ε1 )a + ε1d
B=0
EE204B Electromagnetics D = ( A − C )a = ρ s
Prof. Yong H. Won d
(ε 2 − ε1 )a + ε1d
Spring 2010 22 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
d −a
V1 = ρ s
z
(ε 2 − ε1 )a + ε1d 0≤ z≤a a
d
V2 = − ρ s
z + ρs
(ε 2 − ε1 )a + ε1d
(ε 2 − ε1 )a + ε1d
r
d −a
⎧r
⎪ E = −∇V = − ρ s (ε − ε )a + ε d az
⎪
2
1
1
∴⎨
r
r
a
⎪ E = −∇V = − ρ
az
s
⎪
(ε 2 − ε1 )a + ε1d
⎩ a≤z≤d 0≤ z≤a
a≤z≤d
d
a
0 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 23 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
Example 115
V0 a
b Two concentric spherical shells of radius a and b.
The inner shell is attached to V0.
The outer shell is grounded.
What is E in region a ≤ r ≤ b ? EE204B Electromagnetics Prof. Yong H. Won Spring 2010 24 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
Solution
∇ 2V = 0 a≤r ≤b 1 ∂ 2 ∂V
1
∂
∂V
1
∂ 2V
=0
(r
)+ 2
(sin θ
)+ 2 2
2
2
∂r
∂θ
r ∂r
r sin θ ∂θ
r sin θ ∂φ
Because of spherical symmetry
∂V
∂V
= 0,
=0
∂θ
∂φ
1 ∂ 2 ∂V
2 ∂V
∴2
(r
)=0 ⇒ r
= C (constant)
r ∂r
∂r
∂r
C
C
∴V = ∫ 2 dr = − + D
r
r
EE204B Electromagnetics Prof. Yong H. Won Spring 2010 25 Lesson 11 Solving Poisson’s or Laplace ’s Eq .
To determine C and D, use boundary conditions, r = a, V = V0 r = b, V =0 ab
⎧
⎪ C = V0 a − b
⎪
⇒⎨
⎪D = V a
0
⎪
a −b
⎩
ab 1
a
∴V = −V0
+ V0
a≤r ≤b
a −b r
a −b
r
ab 1 r
∂V r
∴ E = −∇V = −
ar = −V0
a
2r
∂r
a −b r
EE204B Electromagnetics Prof. Yong H. Won r
E Spring 2010 V0 26 ...
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This note was uploaded on 04/02/2010 for the course EE 204 taught by Professor Won during the Spring '10 term at 카이스트, 한국과학기술원.
 Spring '10
 Won
 Electromagnet

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