Lect04

(1)

EEE 498/598

Overview of Electrical

Engineering

Engineering Lecture 4:

Lecture 4:

Electrostatics: Electrostatic

Shielding; Poisson’s and

Laplace’s Equations;

Capacitance; Dielectric

(2)

Lecture 4 Objectives Lecture 4 Objectives

 To continue our study of To continue our study of

electrostatics with electrostatic electrostatics with electrostatic

shielding; Poisson’s and shielding; Poisson’s and

Laplace’s equations; Laplace’s equations;

capacitance; and dielectric capacitance; and dielectric

materials and permittivity. materials and permittivity.

(3)

Ungrounded Spherical Ungrounded Spherical

Metallic Shell Metallic Shell

 Consider a point charge at the Consider a point charge at the

center of a spherical metallic shell:

Q

a

b

Electrically neutral

(4)

Ungrounded Spherical Ungrounded Spherical Metallic Shell (Cont’d) Metallic Shell (Cont’d)

 The applied electric field is given byThe applied electric field is given by

2 0

4 ˆ

r Q a

E _app _r





(5)

Ungrounded Spherical Ungrounded Spherical Metallic Shell (Cont’d) Metallic Shell (Cont’d) The total electric field can be obtained using Gauss’s law together _{The total electric field can be obtained using Gauss’s law together}

with our knowledge of how fields behave in a conductor. with our knowledge of how fields behave in a conductor.

(6)

(1) Assume from symmetry the form

of the field

(2) Construct a family of Gaussian

surfaces

 

r D

a

D  ˆ_r _r

spheres of radius r where





r

(7)

Ungrounded Spherical Ungrounded Spherical Metallic Shell (Cont’d) Metallic Shell (Cont’d)  Here, we shall need to treat Here, we shall need to treat

separately 3 sub-families of

Gaussian surfaces:

a r  

0 1)

b r

a  

2)

b r 

3)

0 )

(r  D

a b

(8)

(3) Evaluate the total charge within the volume (3) Evaluate the total charge within the volume

enclosed by each Gaussian surface enclosed by each Gaussian surface





V

ev

encl q dv

(9)

Gaussian surfaces for which

a r 



0

b r

a  

b r 

(10)

 ForFor

 For For

Q Q_encl  a

r  

0

Q Q_encl  b

r 

Shell is electrically neutral:

(11)

 ForFor

since the electric field is zero since the electric field is zero inside conductor.

inside conductor.

 A surface charge must exist on A surface charge must exist on the inner surface and be given the inner surface and be given

by by

b r

a   Q_encl  0

2

4 a Q q_esa



 

(12)

 Since the conducting shell is Since the conducting shell is initially neutral, a surface

initially neutral, a surface

charge must also exist on the charge must also exist on the

outer surface and be given by outer surface and be given by

2

4 b Q q_esb



(13)

(4) For each Gaussian surface,

evaluate the integral

DS s

d D

S

 



 _r ₄ _r2

D s

d

D _r

S

 





magnitude of D on Gaussian

surface.

surface area of Gaussian

(14)

(5) Solve for

(5) Solve for DD on each Gaussian surface on each Gaussian surface

S Q D  encl

(6) Evaluate E as

0

 D E 

(15)

              b r r Q a b r a a r r Q a E r r , 4 ˆ , 0 0 , 4 ˆ 2 0 2 0   Ungrounded Spherical Ungrounded Spherical Metallic Shell (Cont’d) Metallic Shell (Cont’d)

(16)

Ungrounded Spherical Ungrounded Spherical Metallic Shell (Cont’d) Metallic Shell (Cont’d)  The induced field is given byThe induced field is given by

                b r b r a r Q a a r E E

E_ind _app _r , 0 , 4 ˆ 0 , 0 2 0 

(17)

r

E

E_ind

total electric

field Eapp

(18)

Ungrounded Spherical Ungrounded Spherical Metallic Shell (Cont’d) Metallic Shell (Cont’d)  The electrostatic potential is The electrostatic potential is

obtained by taking the line integral

of

of EE. To do this correctly, we must _{. To do this correctly, we must}

start at infinity (the reference point

or

or groundground) and “move in” back toward ) and “move in” back toward the point charge.

the point charge.

 For For r > br > b

 

r Q dr

E r

V

r

0

4





 

(19)

 Since the conductor is an Since the conductor is an equipotential body

equipotential body (and potential is _{(and potential is} a continuous function), we have a continuous function), we have

for

for _a _ _r _ _b

   

b Q b

V r

V

0

4 

(20)

 For For 0  r  a

   

  

 _ _









a r

b Q

dr E

b V r

V

r a

r

1 1

1 4₀

(21)

r

V

No metallic shell

(22)

Grounded Spherical Grounded Spherical

 When the conducting sphere is grounded, When the conducting sphere is grounded,

we can consider it and ground to be one

huge conducting body at ground (zero)

potential.

 Electrons migrate from the ground, so Electrons migrate from the ground, so

that the conducting sphere now has an

excess charge exactly equal to

excess charge exactly equal to --QQ. This _{. This}

charge appears in the form of a surface

charge density on the inner surface of the

sphere.

(23)

Grounded Spherical Grounded Spherical

 There is no longer a surface charge on the There is no longer a surface charge on the

outer surface of the sphere.

 The total field outside the sphere is zero.The total field outside the sphere is zero.

 The electrostatic potential of the sphere is zero.The electrostatic potential of the sphere is zero.

Q

a

b

-

(24)

-Grounded Spherical Grounded Spherical Metallic Shell (Cont’d) Metallic Shell (Cont’d)

R

E

E_ind

total electric

field Eapp

(25)

Grounded Spherical Grounded Spherical Metallic Shell (Cont’d) Metallic Shell (Cont’d)

R

V

a b

Grounded

metallic shell acts as a

(26)

The Need for Poisson’s The Need for Poisson’s and Laplace’s Equations and Laplace’s Equations

 So far, we have studied two So far, we have studied two approaches for finding the approaches for finding the

electric field and electrostatic electric field and electrostatic

potential due to a given charge potential due to a given charge

distribution. distribution.

(27)

(Cont’d) (Cont’d)

 Method 1Method 1: given the position of all : given the position of all

the charges, find the electric field

and electrostatic potential using

 (A)(A)

      _          P V ev l d E r V R v d R r q r E ₃ 0 4

(28)

 (B)(B)

     r V  r E

R v d r

q r

V

ev

 

 

 

 40

(29)

The Need for Poisson’s The Need for Poisson’s

and Laplace’s and Laplace’s

Equations (Cont’d) Equations (Cont’d)

 Method 2Method 2: Find the electric field : Find the electric field

and electrostatic potential using

  _  



 

  

P V

ev S

l d E r

V

dv q

s d D

 Method 2 works only for symmetric charge distributions, but we can have materials other than free space present.

(30)

Equations (Cont’d) Equations (Cont’d)  Consider the following problem:Consider the following problem:

Conducting bodies

 r



 What are E and V in the region?

2

V V 

1

V V 

Neither Method 1 nor Method 2 can be used!

(31)

Equations (Cont’d) Equations (Cont’d)

 Poisson’s equation_{Poisson’s equation} is a differential equation for is a differential equation for

the electrostatic potential

the electrostatic potential VV. Poisson’s . Poisson’s equation and the boundary conditions

equation and the boundary conditions

applicable to the particular geometry form a

boundary-value problem that can be solved

either analytically for some geometries or

numerically for any geometry.

 After the electrostatic potential is evaluated, After the electrostatic potential is evaluated,

the electric field is obtained using

 

r V

 

r

(32)

Derivation of Poisson’s Derivation of Poisson’s

Equation Equation

 For now, we shall assume the For now, we shall assume the only materials present are free only materials present are free

space and conductors on which space and conductors on which

the electrostatic potential is the electrostatic potential is

specified. However, Poisson’s specified. However, Poisson’s

equation can be generalized for equation can be generalized for

other materials (dielectric and other materials (dielectric and

magnetic as well). magnetic as well).

(33)

Equation (Cont’d) Equation (Cont’d) 0 0   ev ev ev q V V E q E q D                V 2 

(34)

Equation (Cont’d) Equation (Cont’d)

0 2



ev

q

V







Poisson’s

equation

 2 is the Laplacian operator. The Laplacian of a scalar

function is a scalar function equal to the divergence of the gradient of the original scalar function.

(35)

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 sin 1 sin sin 1 1 1 1                                                                   V r V r r V r r r V z V V V V z V y V x V V

Laplacian Operator in

Cartesian, Cylindrical, and

Spherical Coordinates

(36)

Laplace’s Equation Laplace’s Equation

 Laplace’s equation_{Laplace’s equation} is the homogeneous form is the homogeneous form of

of Poisson’s equationPoisson’s equation._.

 We use Laplace’s equation to solve We use Laplace’s equation to solve

problems where potentials are specified

on conducting bodies, but no charge

exists in the free space region.

0

2



(37)

Uniqueness Theorem Uniqueness Theorem

 A solution to Poisson’s or A solution to Poisson’s or

Laplace’s equation that satisfies Laplace’s equation that satisfies

the given boundary conditions is the given boundary conditions is

the

the uniqueunique (i.e., the one and only (i.e., the one and only correct) solution to the problem. correct) solution to the problem.

(38)

Potential Between Coaxial

Cylinders Using Laplace’s

Equation

 Two conducting coaxial cylinders exist such thatTwo conducting coaxial cylinders exist such that

 

  0

0

 

b V

V a

V

 

a b

x y

+ V₀

(39)

Equation (Cont’d)

 Assume from symmetry thatAssume from symmetry that



 



V V 

0 1

2 _

   



 

 

 



 d

d d

d V

(40)

 Two successive integrations yieldTwo successive integrations yield

 The two constants are obtained from The two constants are obtained from

the two BCs:

 

C₁ ln

 

C₂

V    









00 ₁ 1ln ₂ 2 ln

C b

V

C a

C V

a V

 

 

 

 

 

(41)

 Solving for Solving for CC₁₁ and and CC₂₂, we obtain:, we obtain:

 The potential isThe potential is

 a b V

C

/ ln

0

1   

a b b V C / ln ln 0

2  

 

_

_

       b b a V

V  ln 

/ ln

(42)

 The electric field between the The electric field between the

plates is given by:

 The surface charge densities on The surface charge densities on

the inner and outer conductors

are given by

are given by  

 

  _ _ a b b V b E a q a b a V a E a q esb esa / ln ˆ / ln ˆ 0 0 0 0 0 0                  

b a

V a d dV a V E / ln ˆ ˆ 0         

(43)

Capacitance of a Two Capacitance of a Two

Conductor System Conductor System

 The The capacitancecapacitance of a two conductor system is of a two conductor system is

the ratio of the total charge on one of the

conductors to the potential difference

between that conductor and the other

conductor.

+

V₁₂= V₂-V₁

V₂ _V

1

+

-12

V Q C 

(44)

 Capacitance_Capacitance is a positive quantity is a positive quantity

measured in units of

measured in units of FaradsFarads._.

 CapacitanceCapacitance is a measure of the ability is a measure of the ability

of a conductor configuration to store

charge.

(45)

V Q C 

 The capacitance of an isolated The capacitance of an isolated conductor can be considered to conductor can be considered to

be equal to the capacitance of a be equal to the capacitance of a

two conductor system where the two conductor system where the

second conductor is an infinite second conductor is an infinite

distance away from the first and distance away from the first and

at ground potential. at ground potential.

(46)

Capacitors Capacitors

 A A capacitorcapacitor is an electrical device consisting is an electrical device consisting

of two conductors separated by free space

or another conducting medium.

 To evaluate the capacitance of a two To evaluate the capacitance of a two

conductor system, we must find either the

charge on each conductor in terms of an

assumed potential difference between the

conductors, or the potential difference

between the conductors for an assumed

charge on the conductors.

(47)

Capacitors (Cont’d) Capacitors (Cont’d)

 The former method is the more The former method is the more

general but requires solution of

Laplace’s equation.

 The latter method is useful in The latter method is useful in

cases where the symmetry of the

problem allows us to use Gauss’s

law to find the electric field from

a given charge distribution.

(48)

Parallel-Plate Capacitor Parallel-Plate Capacitor

 Determine an approximate expression Determine an approximate expression

for the capacitance of a parallel-plate

capacitor by

capacitor by neglecting fringingneglecting fringing._.

d _A

Conductor 1 Conductor 2

(49)

 ““Neglecting fringingNeglecting fringing” means to assume that ” means to assume that

the field that exists in the real problem

is the same as for the infinite problem.

V = V₁₂ V = 0

z

z = 0 z = d

(50)

 Determine the potential between Determine the potential between

the plates by solving Laplace’s

equation.









₁₂

2 2 2

0 0

0

V d

z V

dz V d

V

 

(51)

Parallel-Plate Capacitor Parallel-Plate Capacitor (Cont’d) (Cont’d)

 









 

z

d V z V d V c d c V d z V c z V c z c z V dz V d 12 12 1 1 12 2 2 1 2 2 0 0 0             

(52)

d V a

dz dV a

V

E     _ˆ_z   _ˆ_z 12

(53)





d V d

V a

a E

a

q_es _n _z _z 12 0 12

0 0

2 ˆ ˆ ˆ

 

       

• Evaluate the surface charge on conductor 2

• Evaluate the total charge on conductor 2

d

A V

A q

Q _es 0 12

2



 

(54)

• Evaluate the capacitance

d A V

Q

C 0

12



 

(55)

Dielectric Materials Dielectric Materials

 A A dielectric_dielectric (insulator) is a medium which (insulator) is a medium which

possess no (or very few) free electrons to

provide currents due to an impressed

electric field.

 Although there is no macroscopic migration Although there is no macroscopic migration

of charge when a dielectric is placed in an

electric field, microscopic displacements (on

the order of the size of atoms or molecules)

of charge occur resulting in the appearance

of induced electric dipoles.

(56)

 A A dielectric_dielectric is said to be is said to be polarized_polarized when when induced electric dipoles are present. induced electric dipoles are present.

 Although all substances are Although all substances are polarizable_polarizable to to some extent, the effects of polarization some extent, the effects of polarization

become important only for insulating become important only for insulating

materials. materials.

 The presence of induced electric dipoles The presence of induced electric dipoles within the dielectric causes the electric within the dielectric causes the electric

field both inside and outside the material field both inside and outside the material

to be modified. to be modified.

(57)

Polarizability Polarizability

 PolarizabilityPolarizability is a measure of the is a measure of the ability of a material to become ability of a material to become

polarized in the presence of an polarized in the presence of an

applied electric field. applied electric field.

 Polarization occurs in both Polarization occurs in both polarpolar and

(58)

Electronic Polarizability Electronic Polarizability

 In the absence of an In the absence of an

applied electric field,

the positively

charged nucleus is

surrounded by a

spherical electron

cloud with equal and

opposite charge.

 Outside the atom, Outside the atom,

the electric field is

zero.

electron

(59)

Electronic Polarizability

(Cont’d)

 In the presence of In the presence of an applied electric

an applied electric

field, the electron

cloud is distorted

such that it is

displaced in a

direction (w.r.t. the

nucleus) opposite to

that of the applied

electric field.

(60)

Electronic Polarizability Electronic Polarizability

 The net effect is The net effect is

that each atom

becomes a small

charge dipole

which affects

the total electric

field both inside

and outside the

material.

e

 e

loc e E

p  

dipole moment

(61)

Ionic Polarizability Ionic Polarizability

 In the absence of In the absence of

an applied electric

field, the ionic

molecules are

randomly oriented

such that the net

dipole moment

within any small

volume is zero.

negative

(62)

 In the presence In the presence

of an applied

electric field,

the dipoles tend

to align

themselves with

the applied

electric field.

(63)

 The net effect is that The net effect is that each ionic molecule

each ionic molecule

is a small charge

dipole which aligns

with the applied

electric field and

influences the total

electric field both

inside and outside

the material.

e

 e

loc i E

p  

(C-m)

polarizability (F-m2₎

(64)

Orientational Orientational

Polarizability Polarizability

 In the absence of In the absence of

an applied electric

field, the polar

molecules are

randomly oriented

such that the net

dipole moment

within any small

volume is zero.

(65)

Orientational Polarizability

(Cont’d)

 In the presence In the presence

of an applied

electric field,

the dipoles tend

to align

themselves with

the applied

electric field.

(66)

Orientational Polarizability

(Cont’d)

 The net effect is that The net effect is that each polar molecule

each polar molecule

is a small charge

dipole which aligns

with the applied

electric field and

influences the total

electric field both

inside and outside

the material.

e

 e

loc o E

p  

(C-m)

polarizability (F-m2₎

(67)

Polarization Per Unit Polarization Per Unit

Volume Volume

 The total polarization of a given The total polarization of a given material may arise from a

material may arise from a

combination of electronic, ionic, combination of electronic, ionic,

and orientational polarizability. and orientational polarizability.  The The polarization per unit volumepolarization per unit volume is is

given by given by

loc T

E

N

p

N

(68)

Polarization Per Unit Volume

(Cont’d)

 PP is the polarization per unit volume. is the polarization per unit volume.

(C/m

(C/m22))

 NN is the number of dipoles per unit is the number of dipoles per unit

volume. (m

volume. (m-3-3))

 pp is the average dipole moment of the is the average dipole moment of the

dipoles in the medium. (C-m)

 _T_T is the average polarizability of the is the average polarizability of the

dipoles in the medium. (F-m

dipoles in the medium. (F-m22))

o i

e

T   

(69)

(Cont’d)

 EE_loc_loc is the total electric field that is the total electric field that actually exists at each dipole

actually exists at each dipole location.

location.

 For gases For gases EE_loc_loc = = EE where where EE is the is the total macroscopic field.

total macroscopic field.  For solids For solids

1

0

3 1

    

 

 

 _T

loc

N E

(70)

(Cont’d)

 From the macroscopic point of From the macroscopic point of view, it suffices to use

view, it suffices to use

E

P





₀



_e

electron susceptibility (dimensionless)

(71)

 The effect of an applied electric field The effect of an applied electric field

on a dielectric material is to create a

net dipole moment per unit volume

net dipole moment per unit volume P_P..

 The dipole moment distribution sets The dipole moment distribution sets

up induced secondary fields:

ind app E

E

(72)

Volume and Surface Volume and Surface

Bound Charge Densities Bound Charge Densities

 A volume distribution of dipoles may be A volume distribution of dipoles may be

represented as an equivalent volume (

represented as an equivalent volume (qqevb_evb) )

and surface

and surface ((qqesb_esb)) distribution of distribution of boundbound

charge.

 These charge distributions are related to These charge distributions are related to

the dipole moment distribution:

n P

q

P q

esb evb

ˆ

 

  

(73)

Gauss’s Law in Gauss’s Law in

Dielectrics Dielectrics

 Gauss’s law in differential form in free space:Gauss’s law in differential form in free space:

 Gauss’s law in differential form in dielectric:Gauss’s law in differential form in dielectric:

ev q E 

 

0



evb ev q q

E  

 

0

(74)

Displacement Flux Displacement Flux

Density Density



E Pev



qevb_ev ev

P q

q q

E

 

 

   

 

 

0 0

 

• Hence, the displacement flux density

vector is given by

P E

(75)

General Forms of General Forms of

Gauss’s Law Gauss’s Law

_{Gauss’s law in differential form:}_{Gauss’s law in differential form:} Gauss’s law in integral form:_{Gauss’s law in integral form:}

ev

q

D







encl S

Q s

d

D  

(76)

Permittivity Concept Permittivity Concept

 Assuming thatAssuming that

we have_{we have}

 The parameter The parameter  is the is the electric electric

permittivity

permittivity or the _{or the}dielectric constantdielectric constant

of the material. of the material.

E P   ₀_e





E E

(77)

 The concepts of polarizability and dipole The concepts of polarizability and dipole

moment distribution are introduced to

relate microscopic phenomena to the

macroscopic fields.

 The introduction of The introduction of permittivity_permittivity eliminates the eliminates the

need for us to explicitly consider

microscopic effects.

 Knowing theKnowing the permittivity_permittivity of a dielectric tells us of a dielectric tells us

all we need to know from the point of view

of macroscopic electromagnetics.

(78)

 For the most part in macroscopic For the most part in macroscopic

electromagnetics, we specify the

permittivity of the material and if

necessary calculate the dipole

moment distribution within the

medium by using





E

D

(79)

Relative Permittivity Relative Permittivity

 The The relative permittivityrelative permittivity of a of a

dielectric is the ratio of the dielectric is the ratio of the

permittivity of the dielectric to permittivity of the dielectric to

the permittivity of free space the permittivity of free space

0