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Capacitance and Capacitance and ResistanceResistance
Sandra Cruz-Pol, Ph. D.Sandra Cruz-Pol, Ph. D.INEL 4151 ch6INEL 4151 ch6
Electromagnetics IElectromagnetics IECE UPRMECE UPRM
Mayagüez, PRMayagüez, PR
Resistance and CapacitanceResistance and Capacitance
To find E, we will use:To find E, we will use:
Poisson’sPoisson’s equation: equation:
Laplace’sLaplace’s equation: equation: (if charge-free)(if charge-free)
They can be derived from They can be derived from Gauss’s LawGauss’s Law
02 V
vV 2
VE
ED v
ResistanceResistance
If the cross section of a conductor is not If the cross section of a conductor is not uniform we need to integrate:uniform we need to integrate:
Solve for VSolve for V Then find E from its differentialThen find E from its differential And substitute in the above equationAnd substitute in the above equation
S
l
SdE
ldE
I
VR
P.E. 6.8 find Resistance of disk of P.E. 6.8 find Resistance of disk of radius radius bb and central hole of radius and central hole of radius aa..
S
oo
SdE
V
I
VR
oVbV
aVBC
)(
0)(:
aab
VV o
ln/ln
ˆd
dVVE
ˆ/ln ab
Vo
)/ln(
2
ab
tVSdEI o
S
ˆddzSd
t
ab
o2
)/ln(
a
t
BAV ln
b
01
V0
112
2
2
2
2
z
VVV
vV 2
ResistenceResistence
Las resistencias reducen o resisten el paso de electrones
0 Negro
1 Marrón
2 Rojo
3 Naranja
4 Amarillo
5 Verde
6 Azul
7 Violeta
8 Gris
99 BlancoBlanco
CapacitanceCapacitance Is defined as the ratio of Is defined as the ratio of
the charge on one of the the charge on one of the plates to the potential plates to the potential difference between the difference between the plates:plates:
Assume Q and find V Assume Q and find V (Gauss or Coulomb)(Gauss or Coulomb)
Assume V and find Q Assume V and find Q (Laplace)(Laplace)
And substitute E in the And substitute E in the equation.equation.
FaradsldE
SdE
V
QC
l
S
CapacitanceCapacitance
1.1. Parallel plateParallel plate
2.2. Coaxial Coaxial
3.3. Spherical Spherical
Parallel plate CapacitorParallel plate Capacitor Charge Charge QQ and – and –QQ
oror
xs
xsn
aE
aD
ˆ
ˆ
Dielectric,
Plate area, S
S
Qs
d
S
V
QC
S
Qddx
S
QldEV
dd
00
SESdEQ x
Coaxial CapacitorCoaxial Capacitor
Charge +Charge +QQ & & -Q-Q
LESdEQ 2
abL
V
QC
ln
2
Dielectric,
Plate area, S
S
Qd
S
QdSEV
dd
00
++
+
+
+
-
-
-
-
--
-
-
-
c
a
b
L
Qd
L
QldEV
a
b
ln2
ˆˆ2
Spherical CapacitorSpherical Capacitor
Charge +Q & -QCharge +Q & -Q
24 rEdSEQ r
baV
QC
114
ba
Qrdrr
r
QldEV
a
b
11
4ˆˆ
4 2
What is the Earth's What is the Earth's charge?charge?
The electrical resistivity of the atmosphere decreases with The electrical resistivity of the atmosphere decreases with height to an altitude of about 48 kilometres (km), where the height to an altitude of about 48 kilometres (km), where the resistivity becomes more-or-less constant. This region is resistivity becomes more-or-less constant. This region is known as the electrosphere. known as the electrosphere. There is about a 300 000 volt There is about a 300 000 volt (V) potential difference between the Earth's surface and the (V) potential difference between the Earth's surface and the electrosphere,electrosphere, which gives an average electric field strength which gives an average electric field strength of about 6 V/metre (m) throughout the atmosphere. Near of about 6 V/metre (m) throughout the atmosphere. Near the surface, the fine-weather electric field strength is about the surface, the fine-weather electric field strength is about 100 V/m. 100 V/m.
The Earth is electrically charged and The Earth is electrically charged and acts as a spherical capacitor. The acts as a spherical capacitor. The EarthEarth has a has a net net negative charge negative charge of of about a million coulombsabout a million coulombs, while an , while an equal and equal and positive charge positive charge resides in resides in the the atmosphereatmosphere..
Capacitors connectionCapacitors connection
SeriesSeries
ParallelParallel
21 CCC
21
111
CCC
ResistanceResistance
S
SdE
ldE
I
VR
ldE
SdE
V
QC S
Recall that:Recall that:
Multiplying, we obtain the Relaxation Time:Multiplying, we obtain the Relaxation Time:
Solving for R, we obtain it in terms of C:Solving for R, we obtain it in terms of C:
RC
CR
So in summary we obtainedSo in summary we obtained::Capacitor C R=C
Parallel Plate
Coaxial
Spherical
ba11
4
S
d
ab
L
ln
2d
S
Lab
2
ln
4
11
ba
P.E. 6.9P.E. 6.9 A coaxial cable contains an insulating material of A coaxial cable contains an insulating material of 11 in in
its upper half and another material with its upper half and another material with 22 in its lower in its lower
half. Radius of central wire is half. Radius of central wire is aa and of the sheath is and of the sheath is b.b. Find the leakage resistance of length Find the leakage resistance of length L.L.
1
2
Lab
RLab
R2
21
1
lnln
21
2121 ||
RR
RRRRR
21
1ln
Lab
R
They are connected in parallel because voltage across them is =
P.E. 6.10P.E. 6.10aa Two concentric spherical capacitors with Two concentric spherical capacitors with 1r1r=2.5 in its =2.5 in its
outer half and another material with outer half and another material with 2r2r=3.5 in its =3.5 in its
inner half. The inner radius is inner half. The inner radius is a=1mma=1mm, , b=3mm b=3mm and and c=2mm .c=2mm . Find their C Find their C.. 1
pFC 53.0
2
c
We have two capacitors in series:
21
21
CC
CCC
ba
C
ba
C oror
114
114 2
21
1
P.E. 6.10bP.E. 6.10b Two spherical capacitors with Two spherical capacitors with 1r1r=2.5 in its upper =2.5 in its upper
half and another material with half and another material with 2r2r=3.5 in its lower =3.5 in its lower
half. Inner radius is half. Inner radius is a=1mma=1mm and and b=3mm.b=3mm. Find their Find their CC..1
2
2121 || CCCCC
ddErdSEQ sin2
We have two capacitors in parallel:
22 ErQ
ba
VQC oro 11
2/ 1
1
pF
ba
C 5.011
2 21
ba
Qrdrr
r
QldEV
a
b
11
2ˆˆ
2 2