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