prof. david r. jackson ece dept. fall 2014 notes 11 ece 2317 applied electricity and magnetism 1
TRANSCRIPT
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.