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6.641 Electromagnetic Fields, Forces, and Motion, Spring 2005
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6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 4: The Scalar Electric Potential and the Coulomb Superposition
Integral
I. Quasistatics
Electroquasistatics (EQS)
0
∇ × E = −
∂
∂ t
(
µ
0
H
)
≈ 0
∇ i
(
0
E
)
= ρ
∇ × H = J +
∂
∂ t
(
ε
0
E
)
∇ i J +
∂ρ
∂ t
= 0
Magnetquasistatics (MQS)
∇ × E = −
∂
∂ t
∇ × H = J +
∇ i =
∇ i
(
0
E
)
= ρ
(
µ
0
H
)
0
∂
∂ t
(
0
E
)
∇ i
(
0
H
)
= 0
II. Irrotational EQS Electric Field
1. Conservative Electric Field
0
C
∫ E ds d dt
∫
S
µ
0
H da 0
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 4
Page 1 of 6

C
∫ E ds a
I b
∫ E ds ∫ a i = 0 ⇒ b
∫ i b
II a
I
Electromotive Force
(EMF)
= ∫ b
E ds a
II
EMF between 2 points (a, b) independent of path
E field is conservative
Φ
( )
− Φ
( r
)
= r
∫ ref r
E i ds
Scalar electric potential b
∫ a i r ref a
∫ i r
∫ b ref
E ds ( ) − Φ
( ) ( )
− Φ b = Φ ( )
2. The Electric Scalar Potential r =
_ x i x
+
_ y i y
+
_ z i z
∆ r = ∆
_ x i x
+ ∆
_ y i y
+ ∆
_ z i z
∆ = ∆ r cos θ
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 4
Page 2 of 6

∆Φ = Φ ( x + ∆ x, y + ∆ y, z + ∆ z ) − Φ ( x, y, z )
= Φ
( x, y, z
)
+
∂Φ
∂ x
∆ x +
∂Φ
∂ y
∆ y +
∂Φ
∂ z
∆ z − Φ
( x, y, z
)
=
∂Φ
∂ x
∆ x +
∂Φ
∂ y
∆ y +
∂Φ
∂ z
∆ z
⎡∂Φ
⎣ ∂ x
_ i x
+
∂Φ
∂ y
_ i y
+
∂Φ
∂ z
_ i z
⎤
⎦ i ∆ r grad Φ = ∇Φ
_
∇ = i x
∂
∂ x
_
+ i y
∂
∂ y
_
+ i z
∂
∂ z grad
_
Φ = ∇Φ = i x
∂Φ
∂ x
_
+ i y
∂Φ
∂ y
+
_ i z
∂Φ
∂ z r +∆ r
∫ E ds r
( )
− Φ
( r + ∆ r
)
= −∆Φ = −∇Φ i ∆ = E i ∆ r
E = −∇Φ
∇Φ =
∆Φ
∆ n n =
∂Φ
∂ n n
The gradient is in the direction perpendicular to the equipotential surfaces.
III.
Vector Identity
∇ × E 0
∆Φ =
∆Φ
∆ n
∆ r cos θ =
∆Φ
∆ n n i ∆ r = ∇Φ i ∆ r
E = −∇Φ
∇ ×
(
∇Φ
)
= 0
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 4
Page 3 of 6

IV.
Sample Problem
Φ
( x, y
)
= 0 (Equipotential lines hyperbolas: xy=constant)
E = −∇Φ = −
⎡ ∂Φ
⎣ x
_ i x
+
∂Φ
∂ y
_ i y
⎤
⎥
=
− a
V
0
2
⎛
⎜
⎝
_ y i +
_ x i
⎞
⎟
⎠
Electric Field Lines [lines tangent to electric field] dy dx
=
E
E y x
= x y
⇒ ydy = xdx y 2
2
=
2
+ C y 2 − x 2 = y 2
0
− x 2
0
[lines pass through point
(hyperbolas orthogonal to xy)
( x , y
0
) ]
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 4
Page 4 of 6

V. Poisson’s Equation
∇ i E = ∇ i
(
−∇Φ
)
= ρ ε
0
⇒ ∇ Φ = − ρ ε
0
∇ Φ = ∇ i
(
∇Φ
)
=
⎡
⎢
⎣ i
_ x
∂
∂ x
+ i
_ y
∂
∂ y
+
_ i z ∂
∂ z
⎤
⎦ i
⎡
⎣
∂Φ
∂ x
_ i x
+
∂Φ
∂ y
_ i y
+
∂Φ
∂ z
_ i z
⎦
⎤
⎥
=
∂ Φ
∂ x 2
+
∂ Φ
∂ y 2
+
∂ Φ
∂ z 2
VI. Coulomb Superposition Integral
1. Point Charge
= −
∂Φ
∂ r
= q
π ε 2
⇒ Φ = q
π ε
+ C
Take reference Φ
( r → ∞
)
= 0 ⇒ C 0
Φ =
π q
ε
2. Superposition of Charges
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 4
Page 5 of 6

d Φ
T
( )
=
1
4 π ε
0
⎢
⎢
⎡
⎢
⎢
⎣ q
−
1
1
+ q
−
2
2
+ ... dq
1
−
'
1
+ dq
2
'
2
+ ...
⎤
⎥
⎥
⎥
⎥
⎦
Φ
T
( )
=
4 π
1
ε
0
⎢
⎢
⎡
⎢
⎢
N
∑ q n
⎢ n 1 r r n
+ ∫ all line, surface, and volume ch arg es dq
− ⎥
⎥
⎤
⎥
⎥
⎥
=
1
4 π ε
0
⎢
⎢
⎡
⎢
⎢
N
∑ q n n
+
L
∫
λ
⎛ ⎞
⎝ ⎠
'
+ r − r'
∫
S
σ s
⎛ ⎞
⎝ ⎠
'
+
V
∫
ρ '
⎝ ⎠
'
⎤
⎥
⎥
⎥
⎦
Φ
( )
=
V
∫
ρ ' '
4 π ε
0 r − r'
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 4
Page 6 of 6