electric fields in material space sandra cruz-pol, ph. d. inel 4151 ch 5 electromagnetics i ece uprm...
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Electric fields in Material Space
Sandra Cruz-Pol, Ph. D.INEL 4151 ch 5Electromagnetics IECE UPRMMayagüez, PR
Last Chapter: free space
NOW: different materials
Some applications superconductors High permittivity dielectrics Transistors Electromagnets
We will study Electric charges: Conductors or Insulators
Depends on Frequency and Temperature… Boundary conditions
Conductors
(metals)
Insulators
(dielectrics)Semiconducto
rs
Material @ 20oCLow frequency
Conductivity (S/m)
Silver 6.1 x 107
Copper 5.8 x 107
Gold 4.1 x 107
Aluminum 3.5 x 107
Carbon 3 x 104
Sea water 4
Silicon 4.4 x 10-4
Pure water 10-4
Dry Earth 10-5
Glass, Quartz 10-12, 10-17
Colder metals conduct better.
(superconductivity)
Insulators at most lower frequencies.
Conductors- have many free electrons available.
semiconductor
Appendix B
Current Units: Amperes [A]
Definition: is the electric charge passing through an area per unit time.
Current Density, [A/m2]Is the current thru a perpendicular
surface:
dt
dQI
S
IJ n
S
SdJI
Depending on how I is produced:
There are different types of currents. Convection- I flows thru isolator: liquid,
gas, vacuum. Doesn’t involve conductors, doesn’t satisfies Ohm’s Law
Conduction- flows thru a conductor Displacement (ch9)
Current in a filament
Convection current, [A]
Convection density, A/m2
uSt
lS
t
QI v
v
uS
IJ v
S
v
u
l
Conduction Current Requires free electrons, it’s inside conductor.
Suffers collisions, drifts from atom to atom
Conduction current density is:
EeF
city drift velo
collisionsbetween time
electron of mass
u
m
Eeum
EEm
neuJ v
2
Newton’s Law
where v=ne
A Perfect conductor
Has many charges that are free to move. Therefore it can’t have an E field inside which
would not let the charges move freely. So, inside a conductor
0
0
0
ab
v
V
E
Charges move to the surface to make E=0
Resistance If you force a Voltage across a conductor: Then E is not 0 The e encounter resistance to move
S
l
S
l
I
VR
ESIJ
lVE
c
/
/
I
E
V+ -
S
l
c= resistivity of the material
Power in Watts
=Rate of change of energy or force x velocity
dvuEuEdv vv
dvJEP
SL
dSJdlEP
VIP
Joule’s Law
PE 5.1 Find the current thru the cylindrical surface
For the current density ]/[ˆsin10 22 mAazJ
mz 51,2
dzdzdSJIS
2
2
0
25
1
sin10
2
04
2sin
22
)125(20
I
A
I
754
240
PE 5.2 In a Van de Graaff generator, w=0.1m, u=10m/s and the leakage paths have resistance 1014 .
If the belt carries charge 0.5 C/m2, find the potential difference between the dome and the base.
MVIRV 50)10(10)5(. 146
)1)(.10(105.0 6 uwI S
w= width of the belt
u= speed of the belt
PE 5.3 The free charge density in Cu is 1.81 x 1010C/m3..
For a current density of 8 x 106 A/m2, find the electric field intensity and the drift velocity.
EuJ v
mVJ
E /138.108.5
1087
6
smJ
uv
/1042.41081.1
108 410
6
Polarization in dielectrics
The effect of polarization on a dielectric is to have a surface bound charge of:
and leave within it an accumulation of volume bound charge:
EP
PED
oe
o
P
aP
dvQ
pv
nps
v
pvb
ˆ
ps and pv are the polarization (bounded) surface and volume charge densities
Permittivity and Strength Not really a constant!
Ro
oeoo EEPEED
oer
1
Dielectric properties Linear = doesn’t change with E Isotropic= doesn’t change with direction Homogeneous= doesn’t change from point
to point. Coulomb’s Law for any material:
12221
12 ˆ4
aR
QQF
ro
PE 5.6.A parallel plate capacitor with plate separation of 2mm has a 1kV voltage applied to its plane.
If the space between its plates is filled with polystyrene, find E and P.55.2r
mVakd
VE x /ˆ500
002.
1000
2512 /ˆ86.61051085.8)55.1(
55.11
mCaVEP xoe
re
PE 5.7.In a dielectric material, Ex= 5V/m and
Find:
3/ˆ4ˆˆ310
1mnCaaaP zyx
EP eo
DandEe
,,
zyxeo
aaaP
E ˆ67.6ˆ67.1ˆ5
16.2xo
xe E
P
zyxe
rro aaa
PED ˆ186ˆ477ˆ140
Continuity Equation
Charge is conserved.
dvJSdJIout
v
vin dvdt
d
dt
dQI
dt
dJ v
For steady currents: Change= output current –input current = 0
0
0
J
dt
d v
Substituting in:
dt
dEE vv
EJ
vD
dt
dJ v
where Tr=is called the Relaxation time
rTtvov
vv
e
dt
d
/
0
What is Relaxation Time? [s]
What is Relaxation Time? [s]
Is the time it takes a charge placed in the interior of a material to drop to e-1 of its initial value.
Find Tr for silver
Find Tr for rubber:
sT or
197
1045.1101.6
1
hrs
sT or
6.7
435,2710
1.315
We have two materials How the fields behave @ interface?
Boundary Conditions
Boundary Conditions We have two
materials How do the fields
behave @ interface?
S
enc
l
QdSD
dlE
sMaxwell
0
:' Evaluate
We look at the tangential and the perpendicular component of the fields.
nt EEE
Cases for Boundary Conditions:
1. Dielectric- dielectric2. Conductor- Dielectric3. Conductor-Free Space
Dielectric-dielectric B.C. Consider the figure
below:nt
nt
EEE
EEE
222
111
E1
E2
E1t
E1n
E2t
E2n
a b
cd w
h
22220
0
122211
hE
hEwE
hE
hEwE
dlE
nntnnt
l
continuous
EE tt 21
ousdiscontinu
DD tt
2
2
1
1
1
2
1
Dielectric-dielectric B.C. Consider the figure
below:
D1
D2
D1t
D1n
D2t
D2n
h
snn DD 21
nn EE 2211
:charges free no if
SDSD
dSDSQ
nn
S
s
21
S
2
1
.continuous is
:charges free no if
21 nn DD S
Dielectric-Dielectric B.C.
E1
D2
E1t
E1n
D2t
D2n
h
222111
21
coscos
:charges free no if
EE
DD nn
2211
21
sinsin EE
EE tt
2
1
2
1
tan
tan
:charges free no if
2
1
In summary:
1
Conductor-dielectric B.C. Consider the figure
below:
20
222000
0
hhEwE
hE
hw
dlE
ntn
l
E
Et
En
a b
cd
w
h
0tE1
2=∞
E2=0
1dielectric
conductor
0tD
Conductor-dielectric B.C. Consider the figure
below:
E
Et
En
h
SnD
1
2=∞
E2=0
1dielectric
conductor
SDdSDSQ n
S
s
S
S
Conductor-Free Space B.C. Consider the figure
below:
E
Et
En
a b
cd
w
h
Snon ED
1
2=∞
E2=0
oFree space
conductor
0 tot ED
PE 5.9 A homogeneous dielectric (r=2.5) fills region 1 (x<0), while region 2(x>0) is free space.
Find 2122 nC/m ˆ4ˆ10ˆ12 if and zyx aaaDD
xn aa ˆˆ xn
zyt
aD
aaD
ˆ12
ˆ4ˆ10
1
1
xnn aDD ˆ1212
tt EE 12
zyzyt
t aaaaD
D ˆ6.1ˆ4ˆ4ˆ105.2
1
1
122
zyxtn aaaDDD ˆ6.1ˆ4ˆ1222
o
n
t
D
D
75.19
tan
2
2
22
5.29 Lightning strikes a dielectric sphere of radius 2-mm for which r=2.5, =5x10-6 S/m and deposits uniformly a charge of 1C.
Determine the initial volume charge density and the volume charge density 2s later.
425.4/2
3
/
6
12
341
425.4105
)1085.8(5.2
er
Ce
sT
rTtvov
r
Answer: 29.84KC/m3, 18.98 kC/m3