Resistance and Resistance and Capacitance Capacitance Electrostatic Electrostatic Boundary value problems;Boundary value problems;

Sandra Cruz-Pol, Ph. D.Sandra Cruz-Pol, Ph. D.INEL 4151 INEL 4151 ch6ch6

Electromagnetics IElectromagnetics IECE UPRMECE UPRM

Mayagüez, PRMayagüez, PR

https://www.youtube.com/watch?v=JEIkB_8v7qk

Last Chapters: Last Chapters: we knew either V or we knew either V or charge distribution, to find E,Dcharge distribution, to find E,D..

NOW: we NOW: we only know values of V or only know values of V or Q at some places (boundaries). Q at some places (boundaries).

Some applicationsSome applications

ResistanceResistance CapacitorsCapacitors Microstrip lines capacitanceMicrostrip lines capacitance

To find E, we will use:To find E, we will use:

PoissonPoisson’’ss equation: equation:

LaplaceLaplace’’ss equation: equation: (if charge-free)(if charge-free)

They can be derived from They can be derived from GaussGauss’’s Laws Law

02 V

vV 2

VE

ED v

Depending Depending on the geometryon the geometry::

We use appropriate coordinates:We use appropriate coordinates:

Cartesian:Cartesian:

cylindrical:cylindrical:

spherical:spherical:

vV 2

v

z

V

y

V

x

V

2

2

2

2

2

2

v

z

VVV

2

2

2

2

2

11

vV

r

V

rr

Vr

rr

2

2

2222

2 sin

1sin

sin

11

(It’s a scalar)

1. Solve for Laplace or Poisson depending on value of v

2. Use Separation of Variables to solve for V3. Apply Boundary values to find unique answer for V

02 V

ResistanceResistance Defined as:Defined as:

The problem of finding the resistance of a conductor of nonuniform cross section can be solved by :1.Choose a suitable coordinate system2.Assume Vo a the potential difference between conductor’s terminals3.Solve Laplace’s eq. for V, then obtain E and I from and4. Finally obtain R as Vo/I

Resistance PE 6.8Resistance PE 6.8A disc of thicknessA disc of thickness tt has radius has radius bb and a central hole of radius a. Take and a central hole of radius a. Take = conductivity, find R = conductivity, find R

a.a.between hole and rim of the diskbetween hole and rim of the disk

b.b.Between the 2 fat sides of diskBetween the 2 fat sides of disk

P.E. 6.8 Part a) find Resistance of disk P.E. 6.8 Part a) find Resistance of disk of radius of radius bb and central hole of radius and central hole of radius aa..

oVbV

aVBC

)(

0)(:

aab

VV o

ln/ln

ˆd

dVVE

)/ln(

2

ab

tVSdEI o

S

ˆddzSd

a

t

BAV ln

b

01

V0

112

2

2

2

2

z

VVV

P.E. 6.8 Part b) find Resistance of disk P.E. 6.8 Part b) find Resistance of disk Between the 2 fat sides of diskBetween the 2 fat sides of disk

S

oo

SdE

V

I

VR

a

t

b

ResistenceResistence

Resist the flow of

electrons

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 CoulombGauss or Coulomb))

Assume V and find Q Assume V and find Q ((LaplaceLaplace))

And substitute E in the And substitute E in the equation.equation.

FaradsV

QC

FldE

SdE

V

QC

l

S

The (-) sign in V can be ignored because we want the absolute value of V

Two cases: CapacitanceTwo cases: Capacitance

1.1. Parallel plateParallel plate

2.2. Coaxial Coaxial

Parallel plate CapacitorParallel plate Capacitor Charge Charge QQ and – and –QQ

oror

Dielectric,

Plate area, S

S

Qs

d

S

V

QC

S

Qddx

S

QldEV

dd

00

SESdEQ x

Effect of Effect of rr on Capacitance on Capacitance

d

S

V

QC

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

Capacitors connectionCapacitors connection

SeriesSeries

ParallelParallel

21 CCC

21

111

CCC

How to tell if C is in:How to tell if C is in: Parallel:Parallel: when they have same voltage when they have same voltage

across their plates. across their plates. EE is || is || to interface. to interface.

SeriesSeries: when : when E & D are normal E & D are normal to the to the dielectric interface.dielectric interface.

21

1221

111

CCC

21 CCC

So In summary we obtainedSo In summary we obtained::Capacitor C R (not derived)

Parallel Plate

Coaxial

Spherical(not derived)

ba11

4

S

d

ab

L

ln

2d

S

Lab

2

ln

4

11

ba

ba

a CCC

CCC

2

2

FpC 3.14

They are connected in parallel and series

800,91

000,472

21mm

mm5

aC

2CbC

pFmmd

SC o

a 3.17)5.2(

)1000/5.0(800,9 2

Fmmd

SC o

b 7.8)5(

1000/)5.0(9800 2

pFmmd

SC o 3.8

)5.2(

1000/5.047000 2

2

Find the capacitance ofFind the capacitance of

Barium titanatepolymer