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Chapter 5
Electric Field in
Material Space
Part 4
3
Boundary Conditions
Boundary conditions are helpful in determining
the field on one side if the field on the other side
is known.
Dielectric and dielectric
Conductor and dielectric
Conductor and free space
E=Et+En
D1n-D2n=ρs
4
(1) Dielectric and dielectric
E1=E1t+E1n
E2=E2t+E2n
5
6
7
8
9
10
11
12
Substitute in boundary conditions:
14
15
(2) Conductor and free space
ρs
16
ρs
17Note: σf=ρs , Qf = Qen
ρs
Qen
18
ρs
19
ε =εo
21
(3) Conductor and Dielectric
22
(4) Dielectric and free space
As Dielectric- Dielectric but with ε1=εo
Dielectric
free space
23
E1=10ax-6ay+12az V/m , find
(a) P1 (b) E2 and the angle E2 makes with y-axis
(c) The energy density in each region.
1 2
1 2
1 1 1 1
2
( )
( 1) (2)
0.177 0.10624 0.2125 nC/m
t t
n n
o r o
x y z
a
E E
D D
P E E
a a a
Di - Di
24
1 2
1
1 1 1
1 2
2 2
2
2 2 2
( )
10 12
6
3 ( 6 ) 18
18
184
4.5
10 4 12 V/m
t x z t
n y
n n o y o y
n n
o y n
o y
n y
o
t n x y z
b
E a a E
E a
D E a a
D D
a E
aE a
E E E a a a
25
22
2
1
2
2 2 3
1 1
2 2 3
2 2
| | 100 144tan 4
| | 16
tan (4) 75.6367
( )
1 1(3 )( 100 36 144) 3.718 nJ/m
2 2
1 1(4.5 )( 100 16 144) 5.18 nJ/m
2 2
t
n
o
E o
E o
E
E
c
W E
W E
26
(a) D1=12ax-10ay+4az n C/m2 , find
D2 and the angle E2 makes with x-axis
1 2
1 2
1 2 2 2 2
1 1 1 1 2
2 2 2
2
2 2 2
2 2 2
1212
10 4 4 1.610 4
2.5
4 1.6( ) 4 1.6
12 4 1.6 nC/m
1(12 4
t t
n n
xn x n n n
o
y z y z
t y z t t t
o o
y z
t t o y z
o
n t x y z
n t x
o
E E
D D
aD a D E E
a a a aD a a E E E
a aD E a a
D D D a a a
E E E a a
Di - free
1.6 ) nV/my za
27
2 2
22
22
1
2
| | 4 1.6tan 0.36
| | 12
tan (0.36) 19.7487
t
n
o
E
E
28
1t 2t
1n 2n
E = E
D = D
2
22
2
2 1
2
1 1 2 2
21 2
1
11 1
1
11 1
1
| | 12
| |sin
| |
| (12)sin 60 10.4 | |
| | (12)cos 60 6
1(6) 2.4
2.5
| | 10.4tan 77
| | 2.4
| | 10.4sin | |
| | sin 77
t
t t
n
n n
n n
ot
n
t
o
E
E
E
E E
E
E E
E E
E
E
EE
E
1n 2nD = D
10.6735 V / m
(b) If E2 =12 V/m and θ2=60o , Find E1 and θ1
29
(a) E=60ax+20ay-30az m V/m
find D and ρs
2 2 2 2
0.531 0.177 0.265
| | (0.531) (0.177) (0.265) 0.62 pc/m
n o
x y z
s n
D D E
a a a
D
30
1 2 1 2
( )
,
=0= ,
=
(3) (8.5) =(1)
3( )(2000) 706 V/m8.5
3( )(2000) 6000 V/m1
t t n n
ot gt at
on gn an
on gn an
gn
an
a
E E D D
E E E
D D D
E E E
E
E
Di - Di
31
_
_
_
( )
sin
(2000)sin(75) 1931.85
(2000)cos(75) 517.638
_______________________________
oil t
oil
oil
oil t
oil n
b
E
E
E
E
for oil(1)
32
_ _
_ _
_ _
_
2 2
_ _
2 2
_
_
1931.85
(8.5) (3)
3(517.638) 182.7
8.5
( ) ( )
(182.7) (1931.85) 1940.47 V/m
| |tan
|
glass t oil t
glass n oil n
glass n oil n
glass n
glass glass n glass t
glass t
glass
glass n
E E
D D
E E
E
E E E
E
E
Di(oil) - Di(glass)
1931.85
| 182.7
84.6o
glass
33
_ _
_ _
_ _
_
2 2
_ _
2 2
_
_
1931.85
(1) (8.5)
8.5(182.7) 1552.95
1
( ) ( )
(1552.95) (1931.85) 2478.6484 V/m
| | 1tan
| |
air t glass t
air n glass n
air n glass n
air n
air air n air t
air t
air
air n
E E
D D
E E
E
E E E
E
E
Di(glass) - freespace(air)
931.85
1552.95
51.2o
glass
34
2
92
2 2
2
2
r2 2 2
2 2
2
.
(4 )
12 10381.97 nC/m
4 4 (0.05)
for r>a .
(4 )
12 3a nC/m
4 4
energy:
1 1 1. | | | |
2 2 2
1 3| |
2
E o
o
E
o
D ds Qen
D a Qen
QenD
a
D ds Qen
D r Qen
QenD
r r r
electrostatic
W D Edv E dv D dv
Wr
(a)
(b)
(c)
2
2 2 6
4
5 0 0
3 1sin 13 10
2 o r
dv r d d dz Jr
A silver-coated sphere of radius 5 cm carries a total charge of 12 nC uniformly
distributed on its surface in free space. Calculate:
(a) |D| on the surface of the sphere,
(b) D external to the sphere,
(c) the total energy stored in the field.
