introduction to electrostatic
TRANSCRIPT
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Summary of electrostatic equations
in uniform field
2011년도 2 학기
서광석
신소재공학부 [email protected]
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Application of voltage ? Subjected to an E field ?
Charge, electric field, Coulombic forces
Charge in conducting spheres ?
Behavior of charge at electrodes and dielectrics under E field ?
Real charge vs. imaginary (compensation) charge
Measured currents vs. charge motion
Charge vs. breakdown
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Breakdown ?
materials have max. charge density
ex) air: breakdown strength: ~ 3.0 X 106 V/m ~ 2.64 X 10-5 C/m2
materials cannot possess higher charge density than max. charge density
excess charge should be dissipated
rate of charge dissipation:fast, sodden dissipation : breakdown
slow dissipation : no problems (no breakdown)
electron avalanche (전자사태)
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Summary of electrostatic equations in uniform field
uniform Fieldnon-uniform field
by assuming that electric field & charge density are uniform,equations of electrostatics become very simple
1. Coulomb Law
two charges, q, q’, distance d in the
medium of relative dielectric const. r dielectric constant
F = force btw charges,
24
'
d
qqF
r o
Found experimentally
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2. Electric field
Electric force due to a charge q’
qE F
24
'
d
q E
r o
d V E
direction
3. Capacitance① capacitance : ability to store charge
C qV V
qC / ,
q : charge, V : voltage, C : capacitance
** If a charge “feels” a(n electric) force, the charge is said to be in an electric field
+q
E
r+q
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② energy stored in a capacitance
C qqV CV U 22
2
1
2
1
2
1
③ slab,
: relative dielectric constant
r
area = A
d
d
AC r o
④ isolated conducting sphere
r
r
r oC 4
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⑤ parallel combination
1C
3C
2C
321 C C C C T use many layer to increase
film capacitor
T C
⑥ series combination
1C
2C
3C 321
1111
C C C C T
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4. Resistance
①
Conductor :
I V R
Insulator :
R
)(V f R
constant
② resistivity
A
d R
③ conductivity
1
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④ series combination
1 R
2 R
3 R
321 R R R R
⑤ parallel combination
1 R
2 R
3 R 321
1111
R R R R
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5. time constant for charge decay
- time that charge will take to decay away
37%
time
charge
initially fast
later slow
: time constant ( = RC )
time taken for the charge to reachapproximately 37% of its initial value
or resistivity of materials
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Example
two charges of 1 coulomb, 1 m apart
Force =2
4' d qq o
N 9121004.9)108.84(1
=
gravitational force on theobject weighing 109 kg
eq. says 1 coulomb can lift 109 kg unrealistic
so charge which can be built up in practice much less than 1 C.
practically possible
Lift a 5 mm diameter, PS sphere, 100 mg, 500 kV/m
mgqE
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charge density on surface : 25105.2~ mC
charge C 9
102
)106.11(19C e
Therefore,
atomic density
Charges on the surface monolayer of atomsonly 8 atoms/million atoms.
very low value electrostatic phenomena 가 왜 unpredictable 한지 설명함
14106.1 extra electrons/ m2
19102~ atoms/m2
1 surface atoms in 125000 atoms is charged, then
surface contamination levels of few ppm can significantlychange the charge on a surface
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Gauss Law
If an imaginary surface of area S is drawn round an uniform density of
charge, the component of the electric field at the surface which is
perpendicular to the surface is proportional to the total charge enclosed
mathematically
E r oS q
q
S
: charge (surface)
: area
sQS q : surface charge density
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sphere, w/ surface charge density (C/m2)in air
S Q
total charge enclosed S Qq S
from Gauss law
oS oS QS S Q E )()(
breakdown strength of air = (V/m) :6103 max. E
so max. charge density
25126
max 1064.2108.8103 mC Q
a few surface atomsper million atoms
above this, breakdown of air occurswhen charges dissipate fast
this is the max. charge density in air
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1 mm 간격, air, 1 eV 에너지를 갖기 위한 전압은?
diameter 1 cm2 area =23)105)((
= 251085.7 m
aird = 1 mm
air gap = 1 mm = m3101
C = capacitance =
d
Ao
= F 131095.6
1
mF / 1085.812
0
25
1085.7 m A
md 3
101
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Energy W
1813210242.61095.6
2
1)(
2
1
JouleCV W
eV J 1810242.61
2169095 2V
721061.4
2169095
1 V
471079.61061.4
Volt V
∴ 1eV 의 에너지를 갖기 위해서 정도가 필요함 4
1079.6 V
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Internal charge vs. surface charge
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