Lecture 3: Electrostatic Fieldswebpages.iust.ac.ir/nayyeri/Courses/BEE/Lecture_1.pdf1. Electrostatic...

Lecture 1: Electrostatic Fields

Instructor:

Dr. Vahid Nayyeri

Contact:

[email protected]

Class web site:

http://webpages.iust.ac.

ir/nayyeri/courses/BEE

mailto:[email protected]

http://webpages.iust.ac.ir/nayyeri/courses/BEE

1. Electrostatic Fields

1.1. Coulomb’s Law

Something known from the ancient time (here comes amber): two charged

particles exert a force on each other…

1 2

2

0

[ ]4

R

Q QF u N

R (2.1)

where Q1 and Q2 are charges,

R –distance between particles,

uR – the unit-vector

12 9

0

18.854 10 10 /

36F m the permittivity of free space

In this notation, negative force means attraction, positive – repelling.

Electrostatic

(Coulomb’s)

force:


1.1. Coulomb’s Law (Example)

Find the magnitude of the Coulomb force that exists between an electron

and a proton in a hydrogen atom. Compare the Coulomb force and the

gravitational force between the two particles. The two particles are

separated approximately by 1 Ångström 1Å 10-10 m.

219

81 2

22

9 100

1.602 102.3 10

144 10 10

36

C

QQF N

R

31 31

11 47

22 10

9.11 10 1836 9.11 106.67 10 1.02 10

10

e p

G

m mF G N

R

39: 2.27 10C

G

FRatio times

F

This is why chemical bounds are so strong!


1.2. Electric (electrostatic) Field

Electrostatic field due to the

charge Q:

2

04

N V

C mR

F QE u

q R

For a system of two charges:

(4.1)

An “alternative definition”:

0lim

q q q

q

F FdFE

dq q

What’s wrong with it?

(4.2)


1.3. Superposition

For several charges placed at different locations in space, the total electric field

at the particular location would be a superposition (vector summation) of

individual electric fields:

1

N

tot n

n

E E

a vector sum!

(5.1)

(Example): find the EF at P

1 1 1

1 2 3

, , , ,

31 2

2 2 2

0 1 0 2 0 3

3 2 2

0 0 0

4 4 4

1(3 4 ) 2 3

4 5 4 4 4 3

tot P Q P Q P Q P

R R R

x y y x

E E E E

QQ Qu u u

R R R

u u u u

Q1 = +1C, Q1 = +2C, Q3 = -3C


1.3. Superposition (cont)

3[ / ]v

QC m

v Volume charge density:

2[ / ]s

QC m

s Surface charge density:

[ / ]l

QC m

l Linear charge density:

(6.1)

(6.2)

(6.3)

2

0

10, '

4v

i R

v

if v number of volumes u dvR

(6.4)

There is a differential electric field directed radially from each differential charges


1.3. Superposition (Example)

Calculate the electric field from a

finite charge uniformly distributed

along a finite line.

Linear charge density: (z’)l

2 2

':

'

z

R

z u uThe unit vector u

z

We assume a symmetry along z with respect to the observation point. Therefore, it

will be a charge element at –z’ for every charge element at +z’. As a result, fields

in z direction will cancel each other:

0iz

i

due toE symmetry


1.3. Superposition (Example, cont)

The radial component:2 2

cos'

dE dE dE dER z

Combining (3.4.1) and (3.6.3), we arrive to:

2 2

0

'

4 ( ' )

ldzdE

z

which, combined with (3.8.1) and integrated leads to:

3 2 2

2 20 0

1'

4 2'

a

l l

a

aE dz

az

0

22

la E

(8.1)


1.3. Superposition (Example 2)

Calculate the electric field

from an infinite plane charged

with s and consisting of an

infinite number of parallel

charged lines.

Utilize (8.1) and that . The linear charge density:2 2R x y l sdx

Symmetry leads to cancellation

of tangent components.

1

2 22 2 2 20 00 0

cos tan2 2 22 '

s s sy

sy y xE dE dx dx dx

x y yx y x y


1.4. Gauss’s Law

A charge Q is uniformly distributed within a

sphere of radius a.

We can assume first that the charge is

located at the center. Than, by (4.1):

2

04r

QE u

a (10.1)

By evaluating surface integrals of both sides 2

04r

QE ds u ds

a (10.2)

At the surface of the sphere, the unit-vector associated with the differential surface

area ds points in the radial direction. Therefore, and the closed

surface integral is1r ru u

24 a


0

enclQE ds

(11.1)

1.4. Gauss’s Law (cont)

Here Qencl is the charge enclosed within the closed surface.

By using divergence theorem and volume charge density concept:

0

v

v

s v

dv

E ds Edv

0

vE

Therefore, the

integral form:

Differential form: (11.2)


1.4. Gauss’s Law (cont 2)

For the charge Q uniformly distributed within the spherical

volume 34 / 3v a

The volume charge density:34

3

v

Q Q

v a

0 0

v

v encl

dvQ

The total charge

enclosed:

a) Outside the sphere: r > a, Qencl = Q2

0 0

4

v

vr

s

dvQ

E ds r E

2

04r

QE

r (12.1)

Gauss Law


a) Inside the sphere: r < a

1.4. Gauss’s Law (cont 3)

33

3

0 0 0 0

1 4

4 3

3

v

enc v

dvQ Q r Q r

a a

3

2

0 0

4

v

vr

s

dvQ r

E ds r Ea

Gauss Law

3

04r

QrE

a (13.1)


1.5. Gaussian Surface

A Gaussian surface is a closed two-dimensional surface through which a flux or

electric field is calculated. The surface is used in conjunction with Gauss's law (a

consequence of the divergence theorem), allowing to calculate the total enclosed

electric charge by exploiting a symmetry while performing a surface integral.

Commonly used are:

a) Spherical surface for

• A point charge;

• A uniformly distributed spherical shell of charge;

• Other charge distribution with a spherical symmetry

b) Cylindrical surface for

• A long, straight wire with a uniformly distributed charge;

• Any long, straight cylinder or cylindrical shell with uniform charge distribution.


1.6. Potential Energy and Electric Potential

A charged particle will gain a certain amount of potential energy as the

particle is moved against an electric field.

[ ]b b

e

a a

W F dl Q E dl J (15.1)


1.6. Potential Energy and Electric Potential (cont)

Imaginary experiment: compute a total work required to bring three charged

particles from - to the shaded region. No electric field exists at - and there

are no friction, no gravity, and no other forces.

I

There are no forces here, therefore, no work is required! W1 = 0;


II

1.6. Potential Energy and Electric Potential (cont 2)

We need to overcome the Coulomb’s force, therefore, some work is required.

1 2 1 22 2 12

0 04 ( ) 4

bx

a a

QQ QQW dx Q V

x x x x

since both charges are positive

1 11 2

0 04 ( ) 4

bx

a a

Q QV dx

x x x x

V1 is an absolute electric potential caused by the charge Q1

xaxb

(17.1)

(17.2)



III

We need again to overcome the Coulomb’s force, therefore, some work is required.

3 3

1 3 2 3 1 3 2 33 3 13 3 232 2

0 1 0 2 0 3 1 0 3 24 ( ) 4 ( ) 4 4

x xQQ Q Q QQ Q Q

W dx dx QV QVx x x x x x x x

Totally, for the three particles: 1 2 3 2 12 3 13 230 ( )totW W W W QV Q V V

Note:04

i j

i ji j ji

i j

QQQV Q V

x x

Note: the total work in our case is equal to the total electrostatic energy

stored in the shaded region.



(19.3)

The electrostatic energy can also be evaluated as

1[ ]

2e v

v

W Vdv J

Or, for N particles:1 1, 0 1

1

2 4

1

2

N Ni j

tot

i j j i ij

N

i ii

QQW

xQV

1, 04

Ni

i

j j i ij

QV

x

Here xij is the distance between charges i and j;

Note: the total work (and the total energy) do not depend on the order, in

which particles are brought.

(19.1)

(19.2)


(20.1)

(20.2)


We can express the electric potential difference or voltage as:

a

aV E dl

[ ]

b

ab

a

J

C

VV E dl

Electric potential difference between points a and b is the work required to

move the charge from point a to point b divided by that charge.

1 1b b

ab

a a

V F dl Q E dlQ Q

aab bV V V b

bV E dl


1.6. Potential Energy and Electric Potential (Example)

1’

2’

3’

4’

5’

6’

Evaluate the work (charge times potential

difference) required to move a charge q from

a radius b to a radius a.

The electric field is2

04r

QE u

r

The potential difference between the two

spherical surfaces is

2

0 0 0

1 1

4 4 4

bb

ab r r

a a

Q Q QV u u dr

r r a b

The potential at r = is assumed to be 0 and is called a ground potential.

The electric potential defined with respect to the ground potential is called

an absolute potential.

(21.1)


1.6. Potential Energy and Electric Potential (Example, cont)

Considering the path 1-2-3-4, we notice that there are only potential differences

while going 1 2 and 3 4. Therefore, these are the only paths where some

work is required. When moving 2 3, the potential is constant, therefore no

work is required.

A surface that has the same potential is called an equipotential surface.

If the separation between two equipotential surfaces and the voltage between

them are small:

x y z

V V VdV E dl E dx E dy E dz dx dy dz

x y z

: x y z

V

m

V V VE u u u

x yThe electric fi

zeld

(22.1)


(23.1)

(23.2)


We can modify equation (22.1) as following:

E V

Note that when a charged particle is moved along a closed contour,

no work is required

0

0s

E dl dV

E dl E ds E

Electrostatic field is conservative and irrotational.



(24.1)

(24.2)

The potential energy would be

0 00

1 1( )

2 2 2 2e v

v v v v

W Vdv E Vdv VE E V dv VE ds

0

2v

E Vdv

for R

200

2 2v

e

v

W E dEdv vE

2

00

2 vv V

mS Vince E V

Poisson’s equation, Laplace’s eqn. when v = 0

2 2 2

0

( ', ', ')1( , , ) ' ' '

4 ' ' '

v

v

x y zV x y z dx dy dz

x x y y z z


1.7. Dielectric materials

A material can be considered as a collection of randomly (in general) oriented

small electric dipoles.

If an external electric

field is applied, the

dipoles may orient

themselves.


1.7. Dielectric materials (cont)

We may suggest that an external electric

field causes a “thin layer of charge” of the

opposite sign at either edge of the material.

This charge is called a polarization charge.

The density of the polarization charge:

p P

where P is the polarization field:0

1

1lim

N

jv

j

P pv

Here pj = Qdud is the dipole moment of individual dipole, N – number of atoms

(dipoles)

(26.1)

(26.2)


1.7. Dielectric materials (cont 2)

Let us add the polarization charge density to the real charge density. The

Gauss’s Law will take a form:

0

v pE

which leads to

vD

where20 ( )

C

m

electric displacemenD E P t flux density

(27.1)

(27.2)

(27.3)

The total flux that passes through the surface [ ]e

s

D ds C (27.4)


1.7. Dielectric materials (cont 3)

Integrating (27.2) over a volume, leads to

enc

s

D ds Q (28.1)

Dielectric materials are susceptible to polarization. Usually, polarization is linearly

proportional to the applied (small) electric field. Then 0 eP E (28.2)

where e is the electric susceptibility

0 0(1 )e rD E E E

r is the relative dielectric constant

(28.3)

for linear and isotropic materials

We consider only linear materials here.


1.8. Capacitance

[ ]Q

C FV

(29.1)

A parallel-plate capacitor

Area A w z (29.2)

Assume A >> d


1.8. Capacitance (cont)

0 0 0

22

enc s s

s

Q AE ds EA E between plates

b

ab

a

V E dl Ed

0

0

,s

s

A AQC or in case of dielectric

V dC

dd

A

222

0 0

2 2 2e

v

V CVW E dv Ad

d

Stored energy:

(30.1)

(30.2)

(30.3)

(30.4)

Assumed uniform field in the capacitor and uniform distribution of charge on plates


1.8. Capacitance (Example)

Calculate the mutual capacitance of a coax cable

with dielectric r inside…

From the Gauss’s Law:

2enc lD ds Q D L L

ln2 2

b b

l lab

a a

bV E dl d

a

Potential difference:

The total charge: lQ L

n

2

lnl2

l

lab

LQ

bV

a

C Lb

a

(31.1)

(31.2)

(31.3)

(31.4)


(32.1)

(32.2)

(32.3)

1.9. Electric currents

Let’s consider a wire…

v drift v mJ v E E

where is the electron volume charge density, is an average

electron drift velocity, is the mobility of the material.v driftv

m

1/v m R theconductivity

The total current that passes through the wire

A

I J ds


(33.1)

(33.2)

(33.3)

1.9. Electric currents (cont)

2I J a

In our case f the current is distributed uniformly in a cylindrical wire:

The power density (density of power dissipating within a conductor):

2W mp J E

The total power absorbed within the volume:

[ ]v

P pdv W


1.9. Electric currents (Examples)

a) Calculate the current flowing through the wire of radius a; current density:

0 zJ I ua

2 2

0 0

0 0

23

a

z z

aI I u d d u I

a

b) Calculate the power dissipated within a resistor with a uniform conductivity .

The voltage across the resistor is V, a current passing through is I.

2

2

0 0 0

L a

v z

I VJ E dv d d dz

LP VI

(34.1)

Lecture 3: Electrostatic Fieldswebpages.iust.ac.ir/nayyeri/Courses/BEE/Lecture_1.pdf1. Electrostatic...

Documents

Transcript of Lecture 3: Electrostatic Fieldswebpages.iust.ac.ir/nayyeri/Courses/BEE/Lecture_1.pdf1. Electrostatic...