Prof. David R. Jackson ECE Dept. Fall 2014 Notes 11 ECE 2317 Applied Electricity and Magnetism 1.
22
Prof. David R. Jackson ECE Dept. Fall 2014 Notes 11 ECE 2317 Applied Electricity and Magnetism 1
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Transcript of Prof. David R. Jackson ECE Dept. Fall 2014 Notes 11 ECE 2317 Applied Electricity and Magnetism 1.
Prof. David R. JacksonECE Dept.
Fall 2014
Notes 11
ECE 2317 Applied Electricity and Magnetism
1
Example
Assume ˆD D
Infinite uniform line charge
ˆ encl
S
D n dS Q
Find the electric field vector
2
x
y
z
S
l = l0 [C/m]
h
r
Example (cont.)
3
ˆ ˆˆLHS
ˆ ˆ
ˆ ˆ
c
t
b
S S
S
S
D n dS D dS
D z dS
D z dS
h
St
Sb
Sc
r
z
ˆ encl
S
D n dS Q
Side view
ˆ
ˆˆ
ˆ
z
n
z
top
side
bottom
Example (cont.)
ˆ ˆLHS 2c c cS S S
D dS D dS D dS D h
Hence,
We then have
0
0
V/mˆ2
lE
4
0
0
2
2
l
l
D h h
D
h
St
Sb
Sc
r
z
0RHS encl lQ h
5
Note About Cylindrical Coordinates
Note: In cylindrical coordinates, the LHS never changes.
ˆˆ 2S S
LHS D n dS D dS D h
ˆ ˆ( )zD D in the direction, and a function of only, not and
, , ( )v zz f a function of only, not and
Hence, the left-hand side of Gauss’s law will always be the same.
Assume:
Example
x
y
z
l0 -h
h
When Gauss’s Law is not useful
ˆ
ˆ
encl
S
D ndS Q
D D
E has more than one component !
E is not a function of only !
6
Finite uniform line charge
Although Gauss’s law is still valid, it is not useful in helping us to solve the problem.
We must use Coulomb’s law.
Example
v = 3 2 [C/m3] , < a
Infinite cylinder of non-uniform volume charge density
ˆ encl
S
D n dS Q
x
y
z
S
h
a
r
Find the electric field vector everywhere
7
Note: This problem would be very difficult
to solve using Coulomb’s law!
2 enclD h Q
Example (cont.)
(a) < a
/2 2
/2 0 0
0
2
0
4
0
4
RHS
2
2 3
32
4
3RHS
2
encl v
V
h
v
h
v
encl
Q dV
d d dz
h d
h d
h
Q h
8
S
h
r
z
Side view
so
Example (cont.)
Hence
so
LHS 2D h
3
0
V/m3
ˆ ,4
E a
9
4
3
32
23
4
D h h
D
43RHS
2h
x
y
z
a
r
a
Example (cont.)
(b) > a
43
2enclQ h a
432
2D h ha
S
h
r
z
10
4
0
3RHS 2
4
a
enclQ h
LHS RHS
so
LHS 2D h
Example (cont.)
11
4
0
V/m3
ˆ ,4
aE a
x
y
z
a
r
a
4
4
32
234
D h ha
aD
Hence, we have
Example
y
z
x
s = s0 [C/m2]Assume
ˆ zD z D z
ˆ encl
S
D n dS Q S
Ar
Find the electric field vector everywhere
Consider first z > 0
Infinite sheet of uniform surface charge density
12
ˆˆz encl
S
D z n dS Q
Example (cont.)
Assume
S
A
r
D
D
Then we have
13
so
2 z enclAD Q
z z enclD A D A Q
z zD D
ˆ ˆLHS
ˆ ˆ
top
bottom
z
S
z
S
D z z dS
D z z dS
02 z sAD A
Example (cont.)
Hence, we have
0
0
ˆ [V/m];2
0, 0
sE z
z z
for for
S
A
r
For the charge enclosed we have
Hence, from Gauss's law we have
0RHS encl sQ A
0 0
2 2s s
z
AD
A
so
14
0
0
ˆ2
sE z
Therefore
LHS RHS
Example
15
x
x = h
x = 0 0As
0Bs
(a) x > h
(c) x < 0
(b) 0 < x < h
0 0
0 0
ˆ2 2
A Bs sE x
0 0
0 0
ˆ2 2
A Bs sE x
0 0
0 0
ˆ2 2
A Bs sE x
From superposition:
Example (cont.)
(a) x > h
(c) x < 0
(b) 0 < x < h
Choose: 0 0 0 0,A Bs s s s
16
0 0
0 0
ˆ 02 2
A Bs sE x E
0 0 0
0 0 0
ˆ ˆ2 2
A Bs s sE x E x
0 0
0 0
ˆ 02 2
A Bs sE x E
x
x = h
x = 0 0s
0s
0As
0Bs
Example (cont.)
0
0
ˆ sE x
0 < x < h
Ideal parallel-plate capacitor
17
Metal plates
0s
0s
xh
Example
Infinite slab of uniform volume charge density
ˆ x
x x
E x E x
E x E x
Assume
(since Ex (x) is a continuous function)
y
x
3
0 C/m[ ]vd
r
Find the electric field vector everywhere
18
0 0xE
Example (cont.)(a) x > d / 2
0
0
ˆ ˆ
ˆ ˆ ˆ ˆ
0 ( / 2)
/ 2
top bottom
x encl
S
x x encl
S S
x x encl v
x v
D x n dS Q
D x x dS D x x dS Q
D x A D A Q A d
D x d
0
0
ˆ V/m , ( / 2)2
v dE x x d
A
S x
xr
3
0 C/m[ ]v
d
19
Alternative choice: Another choice of Gaussian surface would be a symmetrical surface, symmetrical about x = 0 (as was done for the sheet of charge).
Example (cont.)
Note: If we define 0effs v d
Q
seff
Q
v0
(sheet formula) then
d
A A
0
V/mˆ2
effsE x
0
0
: effv s
effs v
Q Ad A
d
Note
so
20
Example (cont.)
(b) 0 < x < d / 2
0
0
0x x encl v
x v
D x A D A Q A x
D x
0
0
ˆ V/m , 0 / 2v xE x x d
d
y
x
x = 0
x = xS
r3
0 C/m[ ]v
21
Example (cont.)
d / 2
v0 d / (20 )
x
Ex
- d / 2
0
0
V/mˆ , / 2 / 2v xE x d x d
0
0
V/mˆ , ( / 2)2
v dE x x d
Summary
y
x
3
0 C/m[ ]vd
0
0
V/mˆ , ( / 2)2
v dE x x d
22
Note: In the second formula we had to introduce a minus sign,
while in the third one we did not.
