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6.013/ESD.013J Electromagnetics and Applications, Fall 2005
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6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn, Sept. 13, 2005
Lecture 2: Electromagnetic Field Boundary Conditions
I. Boundary Conditions
1. Gauss’ Continuity Condition
Zahn, Markus. Figs. 1.13-1.1.17, 1.19 (a) and (b), 1.23, 1.20, 2.19, 3.12 (a).
Electromagnetic Field Theory: A Problem Solving Approach. Robert E. Krieger
Publishing Company, Florida, 1987. Used with permission.
∫ ε E i da = ∫ σ dS ⇒ ε (E
0
s
S
S
0
2n
- E1n ) dS = σ sdS
ε0 (E2n - E1n ) = σ s ⇒ n i ⎡⎣ε0 (E2 - E1 ) ⎤⎦ = σ s
2. Continuity of Tangential E
Zahn, Markus. Figs. 1.13-1.1.17, 1.19 (a) and (b), 1.23, 1.20, 2.19, 3.12 (a).
Electromagnetic Field Theory: A Problem Solving Approach. Robert E. Krieger
Publishing Company, Florida, 1987. Used with permission.
∫ E i ds = (E
1t
- E2t ) dl = 0 ⇒ E1t - E2t = 0
C
(
)
n× E1 - E2 = 0
Equivalent to Φ1 = Φ2 along boundary
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 2
Page 1 of 8
3. Normal H
∫µ
0
H i da = 0
S
µ 0 (H an - Hbn ) A = 0
H an = Hbn
n i ⎡⎣H a - H b ⎤⎦ = 0
4. Tangential H
d
∫ H i ds = ∫ J i da + dt ∫ ε E i da
0
C
S
S
Hbt ds - H at ds = Kds
Hbt - H at = K
n × ⎡⎣H a - H b ⎤⎦ = K
5. Conservation of Charge Boundary Condition
d
∫ J i da + dt ∫ ρdV = 0
S
n i ⎡⎣ J a - J b ⎤⎦+
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
V
∂
σs = 0
∂t
Lecture 2
Page 2 of 8
II. Boundary Condition Problems
1. Electric Field from a Sheet of Surface Charge
a. Electric Field from a Line Charge
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 2
Page 3 of 8 dEr =
dq
(
4πε0 r + z
+∞
Er =
∫
z = −∞
2
2
)
λ r
dEr = 0
4πε0
λ0rdz
cos θ =
(
4πε0 r2 + z2
+∞
3
)
2
dz
∫
z = −∞
(r
2
+ z2
3
)
2
+∞
=
=
λ0r
4πε0
z
(
r2 z2 + r2
)
1
2
z = −∞
λ0
2πε0r
Another way: Gauss’ Law
∫ ε E i da = ε E 2πrL = λ L
0
0 r
0
S
Er =
λ0
2π ε0 r
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 2
Page 4 of 8
b. Electric Field from a Sheet Charge
dEy =
dλ
(
2πε0 x2 + y2
+∞
Ey =
∫
dEy =
x = −∞
=
)
σ0 y
2πε0
1
cos θ =
2
+∞
∫
x = −∞
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
(
2πε0 x2 + y2
)
dx
2
x + y2
σ0 y 1
x
tan−1
2πε0 y
y
⎧ σ0
⎪
⎪ 2ε0
=⎨
⎪− σ0
⎪ 2ε0
⎩
σ0 ydx
+∞
−∞
y>0
y<0
Lecture 2
Page 5 of 8
Checking Boundary condition at y=0
Ey ( y = 0+
) − Ey ( y = 0− ) =
⎛ σ
σ0
− ⎜⎜ − 0
2ε0 ⎝ 2ε0
σ0
ε0
⎞ σ0
⎟⎟ =
⎠ ε0
c. Two sheets of Surface Charge (Capacitor)
⎧ σ0 _
iy
⎪
⎪ 2ε0
E1 = ⎨
_
⎪− σ0 i y
⎪ 2ε
0
⎩
E = E1 + E2 =
⎧ σ0 _
iy
⎪−
⎪ 2ε0
, E2 = ⎨
_
⎪ σ0 i y
y < −a
⎪ 2ε
⎩ 0
y > −a
σ0
ε0
0
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
_
iy
y>a
y<a
y <a
y >a
Lecture 2
Page 6 of 8
2. Magnetic Field from a Sheet of Surface Current
From a line current I
Hφ =
I
2πr
_
_
_
i φ = − sin φ i x + cos φ i y
Thus from 2 symmetrically located line currents
dHx =
dI
(
2
2
2π x + y
)
1
2
( − sin φ )
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 2
Page 7 of 8
K dx
y
= − 0
2π x2 + y2
Hx = −
K0
y
2π
+∞
∫
x = −∞
dx
2
x + y2
+∞
=−
K0 y 1
x
tan−1
2π y
y
x = −∞
⎧ K0
⎪⎪− 2
= ⎨
⎪ + K0
⎪⎩ 2
y>0
y < 0
Check boundary condition at y=0:
Hx ( y = 0 + ) − Hx ( y = 0 − ) = −K 0
−
K 0 ⎛ K 0 ⎞
−⎜
⎟ = −K 0
2
⎝ 2 ⎠
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 2
Page 8 of 8