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