Static Electric Fields

8/22/2019 Static Electric Fields

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STATIC ELECTRIC FIELDS


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Static Electric Fields 2

Introduction

A field is a spatial distribution of a scalar orvector quantity, which may or may not be a

function of time.

A scalarexample is the altitude of a location ona mountain relative to the sea level.

It is not function of time if long term erosion and

earthquake effects are neglected.

Various locations on the mountain have different

altitudes, constituting an altitude field.


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Introduction

The gradient of altitude is a vector that givesboth the direction and the magnitude of themaximum rate of increase (the upward slope) ofaltitude.

On a flat mountaintop or flat land the altitude isconstant, and its gradient vanishes.

The gravitational field of the earth, representing theforce of gravity on a unit mass, is a vector field

directed toward the center of the earth, having amagnitude depending on the altitude of the mass.

Electric and Magnetic field intensities are vectorfields.


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Introduction

In Electrostatics, electric charges (the sources)are at rest, and electric fields do not change withtime.

Many natural phenomena such as lightning,corona and grain explosion are based onelectrostatics.

Some important industrial applications such as

oscilloscope, ink-jet printer, xerography, andelectric microscope are also base onelectrostatics.


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Introduction

Coulombs law states that the force between

two charged bodies q1

and q2, that are very

small compared to distance between them, R12,

is proportional to the product of the charges and

inversely proportional to the square of the

distance between them. Coulomb found that unlike charges attracts and

like charges repel each other.

Using vector notation, coulombs law can

mathematically be written as:

,2

12

2112 12 R

qqkaF R


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Introduction

F12 is the vector force exerted by q1 on q2. If q1 and q2are of same sign, F12 is +ve (repulsive); and if theseare of opposite signs, F12 isve (attractive).

aR12 is a unit vector in the direction from q1 to q2.

K is the proportionality constant depending on themedium and the system of units.

R12 is the distance between the charges.

Electrostatics can proceed from Coulombs law

to define electric field intensity E, electric scalarpotential V, electric flux density D, and then leadto Gausss law and other relations


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Introduction

Coulombs law itself is a postulate and its exact

equation is a law of nature discovered and

assumed by Coulomb on the basis of his

experiments of limited accuracy. From Helmholtzs theorem, a vector field is

determined if its divergence and curl are

specified. Here Gausss law and coulombs law

are derived from divergence and curl relations.The concept of scalar potential follows naturally

from a vector identity.


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Electrostatics in Free Space Electric Field intensity is defined as the force per

unit charge that a very small stationary testcharge experiences when it is placed in a region

where an electric field exists. That is:

Electric Field Intensity is then proportional to and

in the direction of the force F. Its unit is N/C or

V/m.

The test charge q cannot be zero and cannot beless that the charge of an electron. It is so small

that E does not differ from its calculated value.

)/(0lim mV

q

FqE


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An inverse relation of previous equation gives

the force F on an stationary charge q in an

electric field of intensity E.

Two fundamental postulates of electrostatics in

free space specify the divergence and curl of E.

and

Electrostatics in Free Space

(N)qEF

0 E


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First of above equations implies that a static

electric field is not solenoidal unless =0.

Second of above equations asserts that Static

electric fields are irrotational. Above equations are point relations. That is,

they hold at every point in space. They areb

referred to as the differential form of the

postulates of electrostatics, since both

divergence and curl operations involve spatial

derivatives.



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In practical applications we are usually

interested in the total field of an aggregate or a

distribution of charges. This can be obtained by

an integral form of first equation.

According to divergence theorem:

In view of above two equations:


V SdsAAdv


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Where Q is the total charge contained in volume

V bounded by surface S.

Above equation is of the form of Gausss law,

which states that: The total outward flux of the electric field

intensity over any closed surface in free space is

equal to the total charge enclosed in the surface

divided by .


S


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Now according to Stokess theorem:

Combining above equation with last equation on

page 9, we get:

The line integral is performed over a closed

contour C bounding an arbitrary surface; hence

C is itself arbitrary.

Above equation asserts that ,the scalar lineintegral of the static electric field intensity around

any closed path vanishes.

CS dlAdsA

C

dlE 0


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The scalar product E dl integrated over any pathis the voltage along that path.

Above equation is an expression of Kirchhoffs

voltage law in circuit theory that, The algebraic

sum of voltage drops around any closed circuit iszero.

This expression is another way of saying that E

is irrotational (conservative). Refer to figure onnext page, we see that if the scalar line integral

of E over the arbitrary closed contour C1C2 is

zero, then:


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Above equation says that the scalar line integralof the irrotational E field is independent of the

path; it depends only on the end points.

It is the work done by the electric field in moving

a unit charge from P1 to P2. Hence it shows the

conservation of work or energy in an

electrostatic field.

2

21

2

11

1

22

2

11

21

CAlongCAlong

CAlongCAlong

0

P

P

P

P

P

P

P

P

CC

dlEdlEor

dlEdlEor

dlEdlE

P2P1

C2

C1


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Coulombs Law

Consider a point charge q, at rest in a boundless

free space. Draw a hypothetical spherical

surface of radius R centered at q. Electric field

due to point charge must be everywhere radialand has the same intensity at all points on the

spherical surface.

Applying last equation

On slide 11 to figure a,

We have:


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Coulombs Law

Or

Therefore

The above equation tells us that, The electric field

intensity of a +ve point charge is in the outward

radial direction and has a magnitude proportional to

the charge and inversely proportional to the square

of the distance from the charge.


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Coulombs Law

If the charge q is not located at the origin of thechosen coordinate system then, let the position

vector of q be R and that of a field point P be R,

as shown in figure b on slide 16 Then from

previous equation:

Where aqP is the unit vector drawn from q to P.

Since

We have


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Coulombs Law

Example: Determine the electric field intensity at

P(-0.2,0,-2.3) due to a point charge of +5 (nC) at

Q(0.2,0.1,-2.5) in air. All dimensions are in

meters. Solution: The position vector for the field point P:

The position vector for the point charge Q is:

The difference is:

.3.22.0 zx aaOPR

.5.21.02.0' zyx aaaOQR

,2.01.04.0' zyx aaaRR


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Coulombs Law

which has the magnitude:

substituting in equation on page 18, we obtain:

The quantity within the parentheses is the unitvector aQP, and EP has the magnitude of 214.5(V/m).

).(458.02.01.04.0' 21

222mRR


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Example: A total charge Q is put on a thin sphericalshell of radius b. Determine the electric fieldintensity at an arbitrary point inside the shell.

Solution: We shall solve

this problem in two ways:a) At any point, such as P,

inside the hollow shell, an

arbitrary hypothetical closed

surface (a Gaussian surface)may be drawn, over which we

apply Gausss Law.


Coulombs Law


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Since no charge exists inside the shell and the

surface is arbitrary, we conclude easily that E=0

everywhere inside the shell.

b) Draw a pair of elementary cones of solid angled with vertex at an arbitrary point P. The cones

extend in both directions, intersecting the shell in

areas ds1 and ds2 at distances r1 and r2,

respectively, from the point P. Since charge Qdistributes uniformly over the spherical shell,

there is a uniform surface charge density.


Coulombs Law


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The magnitude of the electric field intensity at P

due to charges on the elementary surfaces ds1

and ds2 is from equation on the page 17.

But the solid angle d equals:


Coulombs Law

24 b

QS

.coscos2

2

2

2

1

1 r

ds

r

dsd


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Combining the expressions of dE and d, we

find that:

Since the above result applies to every pair ofelementary cones, we conclude that E=0

everywhere inside the conducting shell as

before.


Coulombs Law

C l b L


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The electrostatic deflection system of a cathoderay oscilloscope is depicted in figure. Electronsfrom a heated cathode are given an initialvelocity u0=azu0 by a +vely charged anode. Theelectrons enter at z=0 into a region of deflectionplates where a uniform electric field Ed=-ayEd is

maintained over a width w. Ignoring gravitationaleffects, find the

vertical deflection

of the electrons

on the fluorescent

screen at z=L.


Coulombs Law


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Solution: Since there is no force in the z-

direction in the z>0 region, the horizontal

velocity u0 is maintained. The field Ed exerts a

force on the electrons each carrying a chargee, causing a deflection in the y-direction:

From Newtons second law of motion in the

vertical direction we have:


Coulombs Law

dyd eEaEeF

,d

yeE

dt

dum


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Where m is the mass of an electron. Integratingboth sides, we obtain:

Where the constant of integration is set to zerobecause uy=0 at t=0. Integrating again, we have:

The constant of integration is again zero

because y=0 at t=0. Note that the electrons havea parabolic trajectory between the deflectionplates.


Coulombs Law

tEm

e

dt

dyu dy

2

2tE

m

ey d


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At the exit from the deflection plates, t=w/u0,

When the electrons reach the screen, they have

travelled a further horizontal distance of (L-w)which takes (L-w)/u0 seconds. During that time

there is an additional vertical deflection.


Coulombs Law

00

2

0

1

1

2

u

w

m

eE

u

wtuu

and

u

w

m

eEd

dyy

d


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Hence the deflection at the screen is:

Ink-jet printers used in computer output, like

cathode ray oscilloscopes, are devices based onthe principle of electrostatic deflection of a

stream of charged particles.


Coulombs Law

2

00

12u

wLw

m

eE

u

wLud dy

.220

210

wLw

mu

eEddd d


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Minute droplets of ink are forced throughvibrating nozzle controlled by a piezoelectrictransducer.

The output of the computer imparts variableamounts of charges on the ink droplets, whichthen pass through a pair of deflection plateswhere a uniform static electric field exists.

The amount of droplet deflection depends on thecharge it carries, causing the ink jet to strike theprint surface and form an image as the printhead moves in a horizontal direction.


Coulombs Law

C l b L


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Coulombs LawElectric Field due to a System of Discrete

Charges Let a electrostatic field is created by a group of n

discrete point charges q1, q2, ,qn located at

different positions.

Since electric field intensity is a linear function of(proportional to) aRq/R

2, the principle of

superposition applies, and the total E field at a

point is the vector sum of fields caused by all the

individual charges.

From last equation on page 18 we can write the

E at a field point whose position vector is R as


C l b L


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Above equation is somewhat inconvenient to use

because of the need to add vectors of differentmagnitudes and directions.

Let us consider the simple case of an electric dipole

that consist of a pair of equal and opposite charges

+q and q, separated by a small distance d asshown in figure. Let the centre of the dipole coincide

with the origin of a spherical coordinate system.



Charges

C l b L


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Then the E at the point P is the sum of the

contributions due to +q andq. Thus:

The first term on the right side

of above equation can besimplified if d


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Charges

C l b L


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Where the binomial expansion has been used

and all terms containing the second and higher

powers of (d/R) have been neglected.

Similarly for the second term on the right side ofequation on page 33; we have:

Substituting these in equation on page 33:



Charges

.

Co lombs La


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We define the product of the charge q and the

vector d (going from q to +q) as the electric

dipole moment, p:

p=qd Hence equation on the end of page 34 can be

written as:

If the dipole lies along the z-axis as in previous

figure, then:



Charges

Coulombs Law


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And equation on the end of page 35 becomes:

Above equation gives the electric field intensity of anelectric dipole in spherical coordinates. We see that E ofa dipole is inversely proportional to the cube of the

distance R. This is reasonable because as R increases,the fields due to the closely spaced +q and q tend tocancel each other more completely, thus decreasingmore rapidly than that of a single point charge.



Charges

Coulombs Law


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The electric field caused by a continuous

distribution of charge can be obtained by

integrating the contribution of an element of

charge over the charge distribution. Volume charge distribution is

shown in figure.

The volume charge density

(C/m3) is a Function of the

coordinates.


Coulomb s LawElectric Field due to a Continuous

Distribution of Charges

Coulombs Law


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Since a differential element of charge behave

like a point charge, the contribution of the

charge dv in a differential volume element dv

to the electric field intensity at the field point P is:

We have:

Or, since aR=R/R,




Coulombs Law


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Except for some especially simple cases, the

vector triple integral in above equations is

difficult to carry out because in general all three

quantities in the integrand (aR, , and R) changewith the location of the differential volume dv.

If the charge is distributed on a surface with a

surface charge density S (C/m2), then the

integration is to be carried out over the surface.Thus:




Coulombs Law


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For a line charge we have:

Where l (C/m) is the line charge density, and L

the line (not necessarily straight) along which

the charge is distributed.




Coulombs Law


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Example: Determine the electric field intensity of

an infinitely long, straight, line charge of a

uniform density l in air.

Solution: Let us assume thatthe line charge lies along the z

-axis as shown ( We are free to

do this as the field obviouslydoes not depend on how we

designate the line).




Coulombs Law


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It is an accepted convention to use primedcoordinates for source points and unprimedcoordinates for field points when there is apossibility of confusion.

We are asked to find the electric field intensity at apoint P, which is at a distance r from the line.

Since the problem has a cylindrical symmetry (ie,the electric field is independent of the azimuth angle

), it would be most convenient to work withcylindrical coordinates. We write the previousequation as




Coulombs Law


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For this problem l is a constant and the line

element dl=dz is chosen to be at an arbitrary

distance z from the origin. It is important to note

that R is the distance vector directed from thesource to the field point, not the other way

around. We have:

The electric field dE, due to the differential line

charge element l dl=l dz is:




'zaraR zr

Coulombs Law


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where:

and

In last equation on page 44, we have decomposed

dEinto its components in the ar and az directions. It

is easy to say that for every ldz at +z there is acharge element ldzatz,which will produce a dEwith components dEranddEz.




Coulombs Law


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Hence the az components will cancel in theintegration process, and we only need to integratethe dEr in 2

nd equation on page 45:

or:

This equation is an important result for an infiniteline charge. Of course, no physical line charge is

infinitely long; nevertheless above equation givesthe approximate E field of a long straight line chargeat a point close to the line charge.





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Gausss Law and Applications

Gausss law follows directly from the divergence

postulate of electrostatics (2nd last equation on

page 9) by the application of the divergence

theorem. It was derived earlier in last equationon page 11.

Gausss law asserts that the total outward flux of

the E-field over any closed surface in free spaceis equal to the total charge enclosed in the

surface divided by .



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Gausss law is useful in determining the E-field ofcharge distributions with some symmetry conditions,such that the normal component of the electric fieldintensity is constant over an enclosed surface.

On the other hand, when symmetry conditions donot exist, Gausss law would not be of much help.

The essence of applying Gausss law lies first in therecognition of symmetry conditions and second in

the suitable choice of a surface over which thenormal component of E resulting from a givencharge distribution is a constant.



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Such a surface is referred to as a Gaussian

surface. This basic principle was used to obtain

last equation on page 17, for a point charge that

possesses spherical symmetry; consequently aproper Gaussian surface is the surface of a

sphere centered at the point charge.



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Example: Use Gausss law to determine theelectric field intensity of an infinitely long,straight, line charge of a uniform density l inair.

Solution: Since the line charge is

infinitely long, the resultant E field

must be radial and perpendicular

to the line charge (E=arEr) and acomponent of E along the line can

not exist.



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With the obvious cylindrical symmetry weconstruct a cylindrical Gaussian surface of aradius r and an arbitrary length L with the linecharge as its axis, as shown. On this surface Er

is constant, and ds= arrddz. We have:

There is no contribution from the top or thebottom face of the cylinder because on the topface ds=azrdrd but E has no z-componentthere, making E.ds=0. Similarly for the bottomface


L

rrS

rLEdzrdEdsE0

2

0.2


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The total charge enclosed in the cylinder is Q=lL.Substituting in equation:

We have:

or

We note that the length L of the cylindrical Gaussiansurface does not appear in the final expression;hence we could have chosen a cylinder of a unitlength.




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Example: Determine the electric field intensity of

an infinite planar charge with a uniform surface

charge density s.

Solution: It is clear that the E field caused by acharged sheet of an infinite extent is normal to

the sheet. Gausss law can be used to find E.

We chose as the Gaussian surface a rectangular

box with top and bottom faces of an arbitraryarea A equidistant from the planer charge, asshown in figure.




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Gausss Law and Applications The sides of the box are

perpendicular to the charged

sheet.

If the charged sheet

coincides with the xy-plane,then on the top face,

AEdsEds

dsEdsaEads

dsEdsaEads

zAS

z

zzzz

zzzz

22

:havewefaces,sidethefromoncontributinoisthereSince

face,bottomOn the

E

E

E


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The total charge enclosed in the box is Q=sA.

Therefore:

from which we obtain:

and:

Of course, the charged sheet may not coincide

with the xy-plane, but the E field always points




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away from the sheet if s is +ve.

It is obvious that the Gaussian surface couldhave been a pillbox of any shape, notnecessarily rectangular.

The lighting scheme of an office or a classroommay consist of incandescent bulbs, longfluorescent tubes, or ceiling panel lights. Thesecorrespond roughly to point sources, linesources, and planner sources, respectively.

From last equation on page 17, last equation on




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page 46, and last two equations on page 55 we

can estimate that light intensity will fall off rapidly

as the square of the distance from the source in

the case of incandescent bulbs, less rapidly asthe first power of the distance for long

fluorescent tubes, and not at all for ceiling panel

lights.




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Example: Determine the E field caused by a

spherical cloud of electrons with a volume

charge density = -0 for 0 R b (both 0 and

b are +ve) and = 0 for R > b.

Solution: First we recognize that the given

source condition has spherical symmetry. The

proper Gaussian surfaces must therefore beconcentric spherical surfaces. We must find the

E field in two regions. See figure on next page:




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a) 0 R b

A hypothetical

spherical Gaussian

surface Si with R< bis constructed within

the electron cloud.

On this surface, E is

radial and has aconstant magnitude:

E=aRER, ds=aRds.




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The total outward Eflux is:

Substituting in equation on page 47 yields:



.3

4

:issurfaceGaussianewithin thenclosedchargetotalThe

4

3

00

2

Rdv

dvQ

REdsEd

V

V

SRR

S ii


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We see within the uniform electron cloud, the E

field is directed towards the center and has a

magnitude proportional to the distance from

the center.b) Rb

For this case we construct a spherical

Gaussian surface S0 with R > b outside the

electron cloud. We obtain the same expressionfor as in case (a). The total charge

enclosed is:



0S

ds

Ga sss La and Applications


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Consequently:

which follows the inverse square law and could

have been obtained directly from last equationon page 17.

We observe that outside the charged cloud theE field is exactly the same as though the total

charge is concentrated on a single pointcharge at the center. This is true in general, fora spherically symmetrical charged region eventhough is a function of R.


Gausss Law and Applications3

03

4bQ


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In connection with Null identity wenoted that a curl-free vector field could alwaysbe expressed as the gradient of a scalar field.This induces us to define a scalar electric

potential V such that:

Electric potential does have physicalsignificance, and it is related to the work donein carrying a charge from one point to another.Previously we have defined the electric fieldintensity as a force acting on a unit test charge.


Electric Potential

0 V

V


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Therefore in moving a unit charge from P1 toP2 in an electric field, work must be doneagainst the field(cause of ve sign.) and isequal to:

Many paths may be followed in going from P1to P2. Two such paths are drawn in figure onthe next page. Since the path between P

1

andP2 is not specified in above equation, thequestion naturally arises, how does the workdepend on the path taken?


Electric Potential

2

1 .Vor

P

P CJ

dlEq

W


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A little thought will lead

us to conclude that W/q

in above equation sould

not depend on the path,


Electric Potential

If it did, one will be able to go from P1 to P2 along

a path for which W is smaller and then to come

back to P1 along another path, achieving a net

gain in work or energy. This would be contrary tothe principle of conservation of energy. We

have already alluded to the path independant


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nature of the scalar line integral of theirrotational (conservative) E field .

Analogous to concept of potential energy in

mechanics, above equation represents thedifference in electric potential energy of a unit

charge between point P2 and P1. Denoting the

electric potential energy per unit charge by V,

the electr ic po tent ial, we have:


Electric Potential

C dl 0

VdlVVP

P

2

112 E


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Above equation can be obtained by substitutingequation on page 63 in equation on page 64.

Equation on the end of page 66 specifies a potentialdifference (electrostatic voltage) between points P2and P1.

It makes no sense to talk about the absolutepotential of a point, absolute phase of a phasor, orabsolute altitude of a geographic location.


Electric Potential

2

1

2

1

2

1

12P

P

P

Pl

P

P

VVdV

dlaVdl


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A reference zero potential point, a referencezero phase (usually at t=0), or a reference zeroaltitude (usually at sea level) must first bespecified.

In most cases zero-potential point is taken atinfinity. When it is not at infinity, it should bespecifically stated.

The inclusion ofve sign in equation on page 63is necessary in order to conform with theconvention that in going against the E field theelectric potential V increases.


Electric Potential


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When a DC battery of a voltage Vo is connectedbetween two parallel conducting plates, as in fig.

+ve and ve charges comulate on the top and

bottom plates respectively. The E field is directed

from +ve to ve

charges, while the

potential increases in

the opposite direction.


Electric Potential


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When we defined the gradient of scalar field, thedirection of V is normal to the surfaces of

constant V. Hence if we used directed field linesor streamlines to indicate the direction of the Efield, they are everywhere perpendicular to

equipotential lines and equipotential surfaces.


Electric Potential

Electric Potential due to a


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The electric potential of a point at a distance Rfrom a point charge q referred to that at infinity

can be obtained readily from equation:

This is a scalar quantity and depends on,

besides q, only the distance R.



Charge Distribution

VdlVVP

P

2

112

E



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The potential difference between any two pointsP2 and P1 at distances R2 and R1, respectively

from q is:



Charge Distribution

As shown in fig, the concentriccircles (Spheres) passing

through P2 and P1 are

equipotential lines (surfaces),

and VP2-VP1 is the same as VP2-

VP3.

From the point of view of



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equation on top of page 71 we can chose thepath of integration from P1 to P3 and then from

P3 to P2.

No work is done from P1 to P3 because F isperpendicular to dl=aR1d along the circular

path (E dl = 0).t

The electric potential at R due to a system of n

discrete point charges q1, q2, , qn located atR1, R2, , Rn is by superposition, the sum of

potentials due to the individual charges:



Charge Distribution



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Since this is a scalar sum, it is in general, easier

to determine E by taking the ve gradient of Vthan from the vector sum in equation on page 32

directly.

As an example, let us consider an electric dipole

consisting of charges +q and q with a small

separation d. The distances from the charges to

a field point P are designated R+ and R- asStatic Electric Fields 74


Charge Distribution



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Shown in figure. The potential at P can bewritten down directly:



Charge Distribution

If d


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And:

Substituting above two equations in equation on

the top of page 75 gives:

or

Where p = qd.


cos2

1cos2

1 11

R

dR

dR

R


Charge Distribution



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The E field can be obtained from - V. Inspherical coordinates we have:

Above equation is same as last equation on

page 36 but has been obtained by a simplerprocedure without manipulating position vectors.



Charge Distribution



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Example: Make a two-dimensional sketch of theequipotential lines and the electric field lines foran electric dipole.

Solution: The equation of an equipotential

surface of a charge distribution is obtained bysetting the expression for V to equal a constant.

Since q, d, and o in above equation for anelectric dipole are fixed quantities, a constant V



Charge Distribution



79/175

requires a constant ratio (cos/R2

). Hence theequation for an equipotential surface is:

Where Cv is a constant. By plotting R versus

for various values of Cv we draw the solidequipotential lines as shown in figure on nextpage.

In the range 0 /2, V is positive; R is

maximum at =0 and zero at =90. A mirrorimage is obtained in the range /2 whereV isve.



Charge Distribution

cosVCR



80/175



Charge Distribution



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The electric field lines or streamlines representthe direction of the E field in space. We set:

Where k is a constant. In spherical coordinatesabove equation becomes:



Charge Distribution

Ekdl

E

dR

E

Rd

E

dR

EaEaEakdRaRdadRa

R

RRR

sin

writtenbecanwhich

),(sin



82/175

For the electric dipole in previous figure there isno E component, and:



Charge Distribution

2sin

getweequation,abovegIntegratinsin

)(sin2

sincos2

ECR

d

R

dR

or

RddR



83/175

Where CE is a constant. The electric field linesare drawn as dashed lines in previous figure.

They are rotationally symmetrical about the z-

axis (independent of ) and are everywhere

normal to the equipotential lines.

The electric potential due to a continuous

distribution of charge confined in a given region

is obtained by integrating the contribution of anelement of charge over the charged region.

We have, for a volume charge distribution:



Charge Distribution



84/175

For a surface charge distribution:

and for a line charge:



Charge Distribution



85/175

Example: Obtain a formula for the electric fieldintensity on the axis of a circular disk of radius b

that carries a uniform surface charge density s.



Charge Distribution

Solution: Although the diskhas circular symmetry, we

cannot visualize a surface

around it over which the

normal component of E hasa constant magnitude;

hence Gausss law is not



86/175

Useful for the solution of this problem. We usesurface charge equation on page 84, working

with cylindrical coordinates as in figure, we have:

The electric potential at the point P(0, 0, z)

referring to the point at infinity is:


ect c ote t a due to a

Charge Distribution

22 '

''''

rzR

and

ddrrds



87/175

Therefore:

The determination ofE field at an off-axis point

would be a much more difficult problem.


Charge Distribution



88/175

For very large z, it is convenient to expand thesecond term in above two equations into a

binomial series and neglect the second and all

higher powers of the ratio (b2/z2). We have:

Substituting above into two equation on last

page, we obtain:


Charge Distribution



89/175

Where Q is the total charge on the disk. Hence

when the point of observation is very far away

from the charged disk, the E field approximately

follows the inverse square law as if the total

charge were concentrated at a point.


Charge Distribution



90/175

Example: Obtain a formula for the electric fieldintensity along the axis of a uniform line charge

of length L. The uniform line charge density is l.

Solution: For an infinity long linecharge, the E field can

determined readily by applying

gausss law. However for a line

charge of finite length, as in figure

we can not construct a Gaussian


Charge Distribution



91/175

surface over which E ds is constant. Gausss law istherefore not useful here.

Instead we use last equation on page 84 by takingan element of charge dl=dz at z. The distance R

from the charge element to the point P(0, 0, z) alongthe axis of the line charge is:

Here it is extremely important to distinguish the

position of the field point(un-primed coordinates)from the position of the source point (primedcoordinates). We integrate over the source region:


Charge Distribution

2

,'L

zzzR



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The E field at P is the ve gradient of V with

respect to the unprimed field coordinates. For

this problem:


Charge Distribution

Conductors in Static Electric


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In general we classify materials according totheir electrical properties into three types: Conductors

Semiconductors

Insulators (or Dielectrics)

In terms of the crude atomic model of an atomconsisting of a +vely charged nucleus withorbiting electrons, the electrons in the outermost

shells of the atoms of conductors are veryloosely held and migrate easily from one atom toanother. Most metals belong to this group.


Field



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The electrons in the atoms of insulators ordielectrics, however are confined to their orbits;they cannot be librated in normal circumstances,even by the application of an external electric

field. The electrical properties of semiconductors fall

between those of conductors and insulators inthat they possess a relatively small number of

freely movable charges. In terms of the band theory of solids we find that

there are allowed energy bands consisting of


Field



95/175

many closely spaced, discrete energy states.Between these energy bands there may beforbidden regions or gaps where no electron ofthe solid atom can reside.

Conductors have an upper energy band partiallyfilled with electrons or an upper pair ofoverlapping bands that are partially filled so thatthe electrons in these bands can move from one

to another with only a small change in energy. Insulators or dielectrics are materials with a

completely filled upper band, so conduction


Field



96/175

could not normally occur because of theexistence of a large energy gap to the next

higher band.

If the energy gap of the forbidden region isrelatively small, small amounts of external

energy may be sufficient to excite the electrons

in the filled upper band to jump into the next

band, causing conduction. Such materials aresemiconductors.

Assume for the present that some +ve (orve)


Field



97/175

charges are introduced in the interior of aconductor. An electric field will be set up in the

conductor, the field exerting a force on the

charges and making them move away from one

another. This movement will continue until all thecharges reach the conductor surface and

redistribute themselves in such a way that both

the charge and the field inside vanish. Hence:


Field

Inside a Conductor(Under Static Conditions)

= 0

E = 0



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When there is no charge in the interior of aconductor (=0), E must be zero because

according to Gausss law, the total outward

electric flux through any closed surface

constructed inside the conductor must vanish.

The charge distribution on the surface of a

conductor depends on the shape of the surface.

Obviously the charges would not be in a state ofequilibrium if there were a tangential component

of the electric field intensity that produces a


Field



99/175

Tangential force and moves the charge.

Therefore under stat ic cond i t ions the E f ield

on a conductor sur face is everywhere norm al

to the su r face. In other words, the surface of a conducto r is

an equipotent ia l surface under stat ic

condi t ions.

Since E=0 everywhere inside a conductor, thewhole conductor has the same electrostatic

potential.Static Electric Fields 99

Field



100/175

A finite time is required for the charges toredistribute on a conductor surface and reach

the equilibrium state. This time depends on the

conductivity of the material.


Field

Figure shows an interface

between a conductor and

free space. Consider the

contourabcda, which haswidth ab=cd=w and

height=bc=da=h. Sides



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ab and cd are parallel to the interface. Applyinglast equation on page 13, letting h0, and

noting that E in the conductor is zero, we obtain

immediately:

Which says that the tangential component of the

E field on a conductor surface is zero. In otherwords to find En, the normal component of E at

the surface of the conductor, we construct aStatic Electric Fields 101

Field

0

0

t

abcda t

E

orwEdlE



102/175

Gaussian surface in the form of a thin pillbox withthe top face in free space and the bottom face in theconductor where E=0. Using last equation on page11, we obtain:

or

Hence the normal component of E field at a

conductor/free space boundary is equal to thesurface charge density on the conductor divided bythe permittivity of free space.


Field



103/175

Summarizing the boundary conditions at theconductor surface, we have:

When an uncharged conductor is placed in a

static electric field, the external field will causeloosely held electrons inside the conductor tomove in a direction opposite to that of the field


Field

Boundary Conditions at a

Conductor/Free Space

InterfaceEt = 0

En = s/o



104/175

And cause net +ve charges to move in thedirection of the field. These induced free

charges will distribute on the conductor surface

and create an induced field in such a way that

they cancel the external field both inside the

conductor and tangent to its surface.

When the surface char distribution reaches an

equilibrium, all four equations on page 97 to 102will hold, and the conductor is again an

equipotential body.


Field



105/175

Example: A +ve point charge Q is at the centerof a spherical conducting shell of an inner radius

Ri and an outer radius Ro. Determine E and V

functions of the radial distance R.

Solution: The geometry of the problem is shown

in figure on next slide. Since there is spherical

symmetry, it is simplest to use Gausss law to

determine E and find V by integration. There arethree distinct regions: (a) R>Ro, (b) Ri


106/175


Field



107/175

Surfaces will be constructed in these regions.Obviously, E=aRER in all three regions.

a) R>R0 (Gaussian surface S1)

or

The E field is the same as that of a point charge Qwithout the presence of the shell. The potential

referring to the point at infinity is:


Field



108/175

b) R1


109/175

The conducting shell is an equipotential body. Hence:

c) R


110/175

Where the integration constant C is determined byrequiring V3 at R=Ri to equal V2 in equation on top of page

110, we have:

And

The variation of ER and V versus R in all three regions are

plotted in figures b and c on page 106. Note that while the

electric intensity has discontinuous jumps, the potential

remains continuous. A discontinuous jump in potential

would mean an infinite electric field intensity.


Field

Dielectrics in Static Electric


111/175

Field.

Ideal dielectrics do not contain free charges. When a dielectric body is placed in an external

electric field, there are no induced free charges

that move to the surface and make the interiorcharge density and electric field vanish, as with

conductors.

However since dielectric contains bound

charges, we cannot conclude that they have noeffect on the electric field in which they are

placed.




112/175

Field.

Although the molecules of dielectrics are neutral,the presence of an external electric field causes

a force to be exerted on each charged particle

and results in small displacements of +ve and

ve charges in opposite directions. These

displacements, though small in comparison to

atomic dimensions, nevertheless polarize a

dielectric material and create electric dipole asshown in figure on next page.

Inasmuch as electric dipoles do have




113/175

Field.

nonvanishing electric potential and electric fieldintensity, we expect that the induced electric

dipoles will modify the electric field both inside

and outside the dielectric material.


The molecules of some dielectrics

possess permanent dipole

moments, even in the absence of

an external polarizing field. Suchmolecules usually consist of two

or more dissimilar atoms and are

called polar molecules, in contrast



114/175

to nonpolar molecules, which do not havepermanent dipole moments.

An example is the water molecule H2O. The

atoms do not arrange themselves in a manner

that makes the molecule have a zero dipole

moment; i.e. the hydrogen atoms do not lie

exactly on diametrically opposite sides of the

oxygen atom. The dipole moments of polar molecules are of

the order of 10-30 (C.m). When there is no


Field.



115/175

external field, the individual dipoles in a polardielectric are randomly oriented, producing no

net dipole moment.

An applied electric field will exert a torque on the

individual dipoles and tend to align them with the

field in a manner similar to that shown in

previous figure.

Some dielectric materials called electrets, canexhibit a permanent dipole moment even in the

absence of an externally applied electric field.


Field.



116/175

Electrets can be made by heating certain waxesor plastics and placing them in an electric field.The polarized molecules in these materials tendto align with the applied field and to be frozen in

their new positions after they return to normaltemperatures.

Permanent polarization remains without anexternal electric field. Electrets are the electrical

equivalents of permanent magnets; they havefound important applications in high fidelityelectret microphones.


Field.

Equivalent Charge Distributions

f


117/175

To analyze the macroscopic effect of induceddipoles we define a polarization vector P as:

Where n is the number of molecules per unitvolume and the numerator represents the vector

sum of the induced dipole moments contained in

a very small volumev.

The vector P, a smoothed point function, is thevolume density of electric dipole moment. Thedipole moment dp of an elemental volume dv is


of Polarized Dielectrics


f P l i d Di l i


118/175

dp = Pdv, which produces an electrostaticpotential (see last equation on page 76):

Integrating over the volume V of the dielectric,

we obtain the potential due to the polarizeddielectric.

Where R is the distance from the elementalvolume dv to a fixed field point. In Cartesiancoordinates:




f P l i d Di l t i


119/175

and it is readily verified that the gradient of 1/R

with respect to the primed coordinates is:

Hence last equation on previous page can be

written as:

Recalling the vector identity:





120/175

Letting A=P and f=1/R, we can rewrite 2nd lastequation on previous page as:

First volume integral on the R.H.S of aboveequation can be converted into a closed surface

integral by the divergence theorem. We have:

Where an is the outward normal from the

surface element ds of the dielectric.





121/175

Comparison of two integrals on the R.H.S ofabove equation with 2nd and 1st equation on

page 84 respectively reveals that the electric

potential due to a polarized dielectric may be

calculated from the contributions of surface andvolume charge distributions having respectively,

densities:

and





122/175

These are referred to as polarization chargedensities or bound-charge densities. In other

words , a polarized dielectric may be replaced

by an equivalent polarization surface charge

density ps and an equivalent polarizationvolume charge density p for field calculations:

The sketch in previous figure clearly indicates

that charges from the ends of similarly orientedStatic Electric Fields 122




123/175

dipoles exist on surfaces not parallel to thedirection of polarization.

Consider an imaginary elemental surface S of

a nonpolar dielectric. The application of an

external electric field normal to S causes a

separation d of the bound charges. +ve charges

+q move a distance d/2 in the direction of the

field, and ve charges q move an equaldistance againist the direction of the field. The

net total chargeQ that crosses the surfaceS





124/175

in the direction of the field is nqd(s), where n isthe number of molecules per unit volume. If the

external field is not normal tos, the separation

of the bound charges in the direction of an will be

d an and:

But nqd, the dipole moment per unit volume, is

by definition the polarization vector P. We have:

And





125/175

Remember that an is always the outwardnormal. This relation correctly gives a +ve

surface charge on the right hand surface as in

previous figure and ave surface charge on the

left hand surface.

For a surface S bounding a volume V, the net

total charge flowing out of V as a result of

polarization is obtained by integrating 2nd

lastequation on previous page. The net charge

remaining within the volume V is theve of this





126/175

integral:

which leads to the expression for the volumecharge density in last equation on page 121.

Hence when the divergence of P does not

vanish, the bulk of the polarized dielectric

appears to be charged. However since westarted with an electrically neutral dielectric body,





127/175

the total charge of the body after polarizationmust remain zero. This can be radily verified by

noting that:

where the divergence theorem has again been

applied.



Electric Flux Density and

Di l t i C t t


128/175

Because a polarized dielectric gives rise to anequivalent volume charge density p, we expect

the electric field intensity due to a given source

distribution in a dielectric to be different from that

in free space. In particular, the divergencepostulated in second last equation on page 9

must be modified to include the effect ofp; that

is:

Using last equation on page 121, we have:


Dielectrics Constant




129/175

We now define a new fundamental field quantity,

the electr ic f lux densi ty, or electr ic

disp lacement, D, such that:

The use of vector D enables us to write a

divergence relation between the electric field

and the distribution of free charges in anymedium without the necessity of dealing

explicitly with the polarization vector P or theStatic Electric Fields 129





130/175

Polarization charge density p. Combiningprevious two equations, we obtain:

Where is the volume density of free charges.Above equation and last equation on page 9 are

the two fundamental governing differential

equations for electrostatics in any medium.

The corresponding integral form of aboveequation is obtained by taking the volume

integral of both sides. We have:Static Electric Fields 130





131/175

Or

Above equation is another form ofGausss law,which states that, The total outward flux of the

electric displacement over any closed surface is

equal to the total free charge enclosed in the

surface. When the dielectric properties of the medium are

linear and isotropic, the polarization is directlyStatic Electric Fields 131





132/175

proportional to the electric field intensity, and theproportionality constant is independent of the

direction of the field. We write:

where is a dimensionless quantity called

electric susceptibility. A dielectric medium is

linear if is independent of E and homogeneous

if is independent of space coordinates.Substituting above equation in the last equation

on page 129 yields:






133/175

where

is a dimensionless quantity known as the

relat ive perm itt iv i tyor the dielectr ic cons tant

of the medium. The coefficient is the

absolute perm it t iv i ty of the medium and ismeasured in farads per meter (F/m).






134/175

Note that the can be a function of spacecoordinates. If is independent of position, the

medium is said to be homogenous. A linear,

homogenous, and isotropic medium is called a

s imp le medium. The relative permittivity of asimple medium is a constant.

Foranisotropicmaterials the dielectric constant

is different for different directions of the electricfield, and D and E vectors generally have

different directions; permittivity is a tensor.






135/175

In matrix form we may write:

For crystals the reference coordinates can bechosen to be along the principal axis of the

crystal so that the off-diagonal terms of above

matrix are zero. We have:






136/175

Media having the property represented by abovematrix are said to be biaxial. We may write:

If further, , then the medium is said to be

uniaxial. Of course, if ,we have an

isotropic medium.






137/175

Example: A +ve point charge Q is at the centerof a spherical dielectric shell of an inner radius

Ri and an outer radius Ro. The dielectric

constant of the shell is . Determine E, V, D,

and P as functions of the radial distance R.

Solution: Because of the spherical symmetry, we

apply Gausss law to find E and D in threeregions. (a) R>Ro; (b) Ri


138/175

E, and polarization P is determined by therelation:

The E, D, and P vectors have only radialcomponents. Refer to figure on next page:

a)R>Ro

We have from equations on page 107 and 108:






139/175




Di l t i C t t


140/175

From equations on top of page 133 and 138, weobtain:

b) Ri


141/175

Note that DR2 has the same expression as DR1and that both ER and PR have a discontinuity at

R=Ro. In this region:




Di l t i C t t


142/175

c) RRo, the application of

Gausss law yields the same expression for ER,

DR, and PR in both regions:




Di l t i C t t


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To find V3, we must add to V2 at R=Ri the ve

line integral of ER3:

The variations of and DR

versus R are

plotted in previous figure b. The difference

is PR and is shown in figure c. The plot for V in




Di l t i C t t


144/175

figure d is a composite graph for V1, V2, and V3in the three regions. We note that DR is a

continuous curve exhibiting no sudden changes

in going from one medium to another and that

PR exists only in the dielectric region.

It is instructive to compare figure b and d with

figure b and c on page 106 respectively. From

equations on page 121 we find:




Di l t i C t t


145/175

on the inner shell surface,

on the outer shell surface, and:

Previous three equations indicate that there is nonet polarization volume charge inside the

dielectric shell. Howeverve polarization surface




Di l t i C t t


146/175

charges exist on the inner surface and +vepolarization surface charges on the outer surface.

These surface charges produce an electric field

intensity that is directed radially inward, thus

reducing the E field in region 2 due to the point

charge +Q at the center.



Dielectric Strength


147/175

An electric field causes small displacements ofthe bound charges in a dielectric material,

resulting in polarization.

If the electric field is very strong, it will pull

electrons completely out of the molecules. The

electrons will accelerate under the influence of

the electric field, collide violently with the

molecular lattice structure, and causepermanent dislocation and damage in the

material.


Dielectric Strength


148/175

Avalanche effect of ionization due to collisionmay occur. The material will become conducting,

and large currents may result. This phenomenon

is called dielectr ic breakdown.

The maximum electric field intensity that a

dielectric material can withstand without

breakdown is the dielectr ic strength of the

material. The dielectric strength of a material must not be

confused with its dielectric constant.


Dielectric Strength


149/175

The dielectric strength of air at the atmosphericpressure is 3kV/mm. When the electric field

intensity exceeds this value, air breaks down.

Massive ionization takes place, and sparking

(corona discharge) follows. Charge tends toconcentrate at sharp points.

In view of last equation on page 103, the electric

field intensity in the immediate vicinity of sharppoints is much higher than that at points on a

relatively flat surface with a small curvature.


Dielectric Strength


150/175

This is the principle upon which a lightningarrester with a sharp metal lightning rod on top

of tall buildings works.

When a cloud containing an abundance of

electric charges approaches a tall building

equipped with a lightning rod connected to the

ground, charges of an opposite sign are

attracted from the ground to the tip of the rod,where the electric field intensity is the strongest.

As the electric field intensity exceeds the

dielectricStatic Electric Fields 150

Dielectric Strength


151/175

dielectric strength of the wet air, breakdownoccurs, and the air near the tip is ionized and

becomes conducting. The electric charges in the

cloud are then discharged safely to the ground

through the conducting path.


Dielectric Strength


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Example: Consider two spherical conductorswith radii b1 and b2 (b2>b1) that are connected bya conducting wire. The distance of separationbetween the conductors is assumed to be very

large in comparison to b2 so that the charges onthe spherical conductors may be considered asuniformly distributed. A total charge Q isdeposited on the spheres. Find:

a)The charges on the two spheres.b)The electric field intensities at the spheres

surfaces.


g


153/175


Dielectric Strength


154/175

a) Since the spherical conductors are at the samepotential, we have:

or

Hence the charges on the spheres are directly

proportional to their radii. But since:


g

Dielectric Strength


155/175

we find that:

b) The electric field intensities at the surfaces of the

two conducting spheres are:

so

The electric field intensities are therefore inversely

proportional to the radii, being higher at the surface

of the smaller sphere which has a large curvature.


Boundary Conditions forElectrostatic Fields


156/175

Electrostatic Fields

Electromagnetic problems often involve mediawith different physical properties and require theknowledge of the relations of the field quantitiesat an interface between two media.

For instance, we may wish to determine how theE and D vectors change in crossing an interface.Boundary conditions must be satisfied at theconductor/free space interface. These conditions

have been given in equations on page 103. Letus consider an interface between two generalmedia as shown in figure on next page.




157/175





158/175

Let us construct a small path abcda with sidesab and cd in media 1 and 2 respectively, bothbeing parallel to the interface and equal tow.

Last equation on page 13 is applied to this path.

If we let sides bc=da= h approach zero, theircontribution to the line integral of E around the

path can be neglected. We have:





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Therefore:

Which states that the tangential component ofan E field is continuous across an interface.

Above equation simplifies to Et=0 if one of the

media is a conductor. If media 1 and 2 are

dielectrics with permittivities and respectively,

we have:





160/175

In order to find a relation between the normalcomponents of the fields at a boundary, we

construct a small pillbox with its top face in

medium 1 and bottom face in medium 2, as

shown in figure. The faces have an areaS, andthe height of the pillbox h is vanishingly small.

Applying Gausss law (2nd equation on page

121) to the pillbox, we have:





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where we have used the relation an2 = -an1. Unitvectors an1 and an2 are respectively, outward unit

normals from media 1 and 2. From aboveequation we obtain:

or

where the reference unit normal is outward frommedium 2.





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Above equation states that the normalcomponent ofDfield is discontinuous across aninterface where a surface charge exists - theamount of discontinuity being equal to the

surface charge density. If medium 2 is aconductor, D2=0 and above equation becomes:

which simplifies to 2nd

equation on page 103when medium 1 is free space.

When two dielectrics are in contact with no freeStatic Electric Fields 162




163/175

charges at the interface, s=0, we have:

or

We find that the boundary conditions that mustbe satisfied for static electric fields are as

follows:





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Example: A lucite sheet ( =3.2) is introducedperpendicularly in a uniform electric field

Eo=axEo in free space. Determine Ei, Di, and Pi

inside the lucite.




165/175

Solution: We assume that the introduction of thelucite sheet does not disturb the original uniform

electric field Eo. Since the interfaces are

perpendicular to the electric field, only the

normal field components need to be considered.No free charges exist

Boundary condition equation on top of page 163

at the left interface gives:

orStatic Electric Fields 165



166/175

There is no change in electric flux density acrossthe interface. The electric field intensity inside

the lucite sheet is:

The polarization vector is zero outside the lucite

sheet(Po=0). Inside the sheet:




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Example: Two dielectric media with permittivitiesand are separated by a charge free boundary

as shown in figure. The electric field intensity in

medium 1 at the point P1 has a magnitude Ei

and makes an angle 1 with the normal.Determine the magnitude and direction of the

electric field intensity at point P2 in medium 2.




168/175

Solution: Two equationsare needed to solve fortwo unknowns E2t andE2n. After E2t and E2n

have been found E2and 2 will followdirectly. Usingequations on the top ofpage 159 and on thetop of page 163, wehave:




169/175

and:

Division of above two equations gives:

The magnitude of E2 is:

or:Static Electric Fields 169



170/175

Example: When a coaxial cable is used to carryelectric power, the radius of the inner conductoris determined by the load current, and theoverall size by the voltage and the type of

insulating material used. Assume that the radiusof the inner conductor is 0.4 cm and thatconcentric layers of rubber ( =3.2) andpolystyrene ( =2.6) are used as insulatingmaterials. Design a cable that is to work at avoltage rating of 20 kV. In order to avoidbreakdown due to voltage surges caused by




171/175

lightning and other abnormal external conditions,the maximum electric field intensities in the

insulating materials are not to exceed 25% of

their dielectric strengths.

We find the dielectric strengths (from the table)of rubber and polystyrene to be 25x106 (V/m)

and 20x106 (V/m) respectively. Using last

equation on page 46 for specified 25% of

dielectric strengths, we have the following.

In rubber:




172/175

In polystyrene: Combination of above two equations yields:

Above equation indicates that the insulatinglayer of polystyrene should be placed outside of

that of rubber, as shown in figure on next page.

The cable is to work at a potential difference of

20,000 (V) between the inner and outerconductors. We set:




173/175




174/175

where both Ep and Er have the form given in last

equation on page 46. The above relation leads

to:

or

Since ri=0.4 cm is given, ro can be determinedby finding the factor from last equation




175/175

on page 171 and then using it in aboveequation. We obtain = 8x104, and

ro=2.08ri=0.832 cm.

In figures b and c are plotted the variations of

the radial electric field intensity E and thepotential V referred to that of the outer sheath.

Note that E has discontinuous jumps, while the

V curve is continuous.

Documents

Static Electric Fields