Emtl Note Unit II

8/13/2019 Emtl Note Unit II

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law is applicable in finding electric field due to an charge distribution$ !auss(s law is easier

to use when the distribution is smmetrical,

Coulomb's Law: Coulomb(s Law states that the force between two point

charges Q5and Q6is directl proportional to the product of the charges and in-ersel

proportional to the s%uare of the distance between them,

#oint charge is a hpothetical charge located at a single point in space, It is an idealised

model of a particle ha-ing an electric charge,

*athematicall$

$

where kis the proportionalit constant,In I units$ Q5and Q6are expressed in

Coulombs/C3 and Ris in meters, Force Fis in 0ewtons /N3 and

$

is called the permitti-it of free space,/.e are assuming the charges are in free space,

If the charges are an other dielectric medium$ we will use instead where is

called the relati-e permitti-it or the dielectric constant of the medium3,7herefore

,,,,,,,,,,,,,,,,,,,,,,,/6,53

"s shown in the Figure 6,5 let the position -ectors of the point charges Q5and Q6are gi-en

b and , Let represent the force on Q1due to charge Q6,


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Fig 2.1: Coulomb's Law

7he charges are separated b a distance of ,

.e define the unit -ectors as

and ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,63

can be defined as

,

imilarl the force on Q1due to charge Q2can be calculated and if represents this forcethen we can write

.hen we ha-e a number of point charges$ to determine the force on a particular

charge due to all othercharges$we appl principle of superposition.If we ha-eNnumber of

chargesQ1,Q2...QNlocatedrespecti-el at the points represented b the position -ectors

, , ...... , the force experienced b achargeQlocated at is gi-en b,

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,83


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Electric Field: 7he electric field intensit or the electric field strength at a point is

defined as the force per unit charge, 7hat is

or$ ,,,,,,,,,,,,,,,,,,,,,,,,/6,93

7he electric field intensitEat a pointr/obser-ation point3due a point chargeQlocated

at /source point3 is gi-en b:

,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,3

For a collection ofNpoint chargesQ1,Q2,.........QNlocated at , ,...... $ t he electric

field intensit atpoint is obtained as

,,,,,,,,,,,,,,,,,,,,,,/6,;3

7he expression /6,;3 can be modified suitabl to compute the electric filed due to

continuous distribution of charges,

In figure 6,6 we consider a continuous -olume distribution of charge (t) in the region

denoted as the source region,

For an elementar charge , i,e, considering this charge as point charge$ we

can write the field expression as:

,,,,,,,,,,,,,/6,


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Fig 2.2: Continuous Volume Distribution of Charge

.hen this expression is integrated o-er the source region$ we get the electric field at the

pointPdue to this distribution of charges, 7hus the expression for the electric field atPcan

be written as:

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,=3

imilar techni%ue can be adopted when the charge distribution is in the form of a line

charge densit or a surface charge densit,

,,,,,,,,,,,,,,,,,,,,,,,,/6,>3

,,,,,,,,,,,,,,,,/6,5?3


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Electric flu densit!:

"s stated earlier electric field intensit or simpl @Electric field( gi-es the strength of

the field at a particular point, 7he electric field depends on the material media in

which the field is being considered, 7he flux densit -ector is defined to be

independent of the material media /as we(ll see that it relates to the charge that is

producing it3,For a linear isotropic medium under considerationA the flux densit

-ector is defined as:

,,,,,,,,,,,,,,,,,,,,,/6,553

.e define the electric flux as

,,,,,,,,,,,/6,563

"auss's Law: !auss(s law is one of the fundamental laws of electromagnetism and it states

that the total electric flux through a closed surface is e%ual to the total charge enclosed b the

surface,

Fig 2.#: "auss's Law


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Let us consider a point charge Q located in an isotropic homogeneous medium of dielectric

constant, 7he flux densit at a distance ron a surface enclosing the charge is gi-en b

,,,,,,,,,,,,,,,,,,,,,,,/6,583

If we consider an elementar area ds$ the amount of flux passing through the elementar area is

gi-en b

,,,,,,,,,,,,,,,,,,,/6,593

'ut

$

is the elementar solid angle subtended b the area at the location of Q, 7herefore we can

write

For a closed surface enclosing the charge$ we can write

which can seen to be same as what we ha-e stated in the

definition of !auss(s Law,

$%%lication of "auss's Law

!auss(s law is particularl useful in computing or where the charge distribution has

some smmetr, .eshall illustrate the application of !auss(s Law with some examples,

1. $n infinite line charge: "s the first example of illustration of use of !auss(s law$ let

consider the problem of determination of the electric field produced b an infinite line charge


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of densitLC/m.Let us consider a line charge positioned along the z2axis as shown in

Fig, 6,9/a3 /next slide3, ince the line charge is assumed to be infinitel long$ the electric

field will be of the form as shown in Fig, 6,9/b3 /next slide3,If we consider a close cylindrical

surface as shown in Fig. 6,4(a, using !auss"s #heorm we can wri#e,

,,,,,,,,,,,,,,,,,,/6,53

Considering the fact that the unit normal -ector to areasS1andS$are perpendicular to the

electric field$ the surface integrals for the top and bottom surfaces e-aluates to 4ero, 1ence

we can write$

Fig 2.&: Infinite Line Charge

,,,,,,,,,,,,,,,,,,,,,,,,,,/6,5;3


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2. Infinite heet of Charge: "s a second example of application of !auss(s theorem$ we

consider an infinite charged sheet co-ering thex-zplane as shown in figure 6,, "ssuming

a surface charge densit of for the infinite surface charge$ if we consider a clindrical

-olumeha-ing sides placed smmetricall as shown in figure $ we can write:

,,,,,,,,,,,,,,/6,5


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#. (niforml! Charged %here:Let us consider a sphere of radius r%ha-ing a uniform

-olume charge densit of -CBm8, 7o determine e-erwhere$ inside and outside the

sphere$ we construct !aussian surfaces of radius r < r?and r > r?as shown in Fig, 6,; /a3

and Fig, 6,;/b3, For the region &the total enclosed charge will be

,,,,,,,,,,,,,,,,,,,,,,,,,/6,5=3

Fig 2.):(niforml! Charged %here

' appling !auss(s theorem$ ,,,,,,,/6,5>3

7herefore ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,6?3

For the region A the total enclosed charge will be ,,,,,,,,,,/6,653


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' appling !auss(s theorem$

,,,,,,,,,,,,,,,,,,,,,/6,663

Electrostatic *otential and E+ui%otential urfaces

In the pre-ious sections we ha-e seen how the electric field intensit

due to a charge or a charge distribution can be found using

Coulomb(s law or !auss(s law, ince a charge placed in the -icinit

of another charge /or in other words in the field of other charge3

experiences a force$ the mo-ement of the charge represents energ

exchange, Electrostatic potential is related to the wor done in

carring a charge from one point to the other in the presence of an

electric field,

Let us suppose that we wish to mo-e a positi-e test charge from

a pointPto another point Qas shown in the Fig, 6,=,

7he force at an point along its path would cause the particle to

accelerate and mo-e it out of the region if unconstrained, ince we

are dealing with an electrostatic case$ a force e%ual to the negati-e

of that acting on the charge is to be applied while mo-es

fromPto Q, 7he wor done b this external agent in mo-ing the

charge b a distance is gi-en b:

Fig 2.,: -oement of /est Charge in E

,,,,,,,,,,,,,,,,,,,,/6,683

7he negati-e sign accounts for the fact that wor is done on the sstem b the external

agent,

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,693

7he potential difference between two pointsPand Q$ VPQ$ is defined as the wor done perunit charge$ i,e,


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,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,63

It ma be noted that in mo-ing a charge from the initial point to the final point if the potential

difference is positi-e$ there is a gain in potential energ in the mo-ement$ external agent

performs the wor against the field, If the sign of the potential difference is negati-e$ wor is

done b the field,

.e will see that the electrostatic sstem is conser-ati-e in that no net energ is

exchanged if the test charge is mo-ed about a closed path$ i,e, returning to its initial

position, Further$ the potential difference between two points in an electrostatic field is a

point functionA it is independent of the path taen, 7he potential difference is measured in

oulesBCoulomb which is referred to as Volts,

Let us consider a point charge Qas shown in the Fig, 6,>,

Fig 2.0: Electrostatic *otential calculation for a %oint charge


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Further consider the two pointsAandBas shown in the Fig, 6,>, Considering the

mo-ement of a unit positi-e test charge fromBtoA$ we can write an expression for the

potential difference as:

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,6;3

It is customar to choose the potential to be 4ero at infinit, 7hus potential at an point

/ rA r3 due to a point charge G can be written as the amount of wor done in bringing a unit

positi-e charge from infinit to that point /i,e, rB ?3,

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,6


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,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,6>3

o far we ha-e considered the potential due to point charges onl, "s an other tpe of

charge distribution can be considered to be consisting of point charges$ the same basic

ideas now can be extended to other tpes of charge distribution also,

Let us first consider Npoint charges Q1, Q2,.....QNlocated at points with position -ectors ,

,...... , 7he potential at a point ha-ing position -ector can be written as:

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,8?a3

or, ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,8?b3

For continuous charge distribution$ we replace point charges Qnb corresponding charge

elements or or depending on whether the charge distribution is linear$

surface or a -olume charge distribution and the summation is replaced b an integral, .ith

these modifications we can write:

For line charge$ ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,853

For surface charge$ ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,863

For -olume charge$ ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,883

It ma be noted here that the primed coordinates represent the source coordinates and the

unprimed coordinates represent field point,


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Further$ in our discussion so far we ha-e used the reference or 4ero potential at infinit, If

an other point is chosen as reference$ we can write:

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,893

.here C is a constant, In the same manner when potential is computed from a nown

electric field we can write: ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,83

7he potential difference is howe-er independent of the choice of reference,

,,,,,,,,,,,,,,,,,,,,,,,/6,8;3

.e ha-e mentioned that electrostatic field is a conser-ati-e fieldA the wor done in mo-ing a charge from o

point to the other is independent of the path, Let us consider mo-ing a charge from point P1toP2in one pa

and then from pointP6bac toP5o-er a different path, If the wor done on the two paths were different$ a n

positi-e or negati-e amount of wor would ha-e been done when the bod returns to its original position

In a conser-ati-e field there is no mechanism for dissipating energ corresponding to an positi-e wo

neither an source is present from which energ could be absorbed in the case of negati-e wor, 1ence t

%uestion of different wors in two paths is untenable$ the wor must ha-e to be independent of path a

depends on the initial and final positions,

ince the potential difference is independent of the paths taen$ VAB 2 VBA$ and o-er a closed path$

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,8


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"n -ector field that satisfies is called an irrotational field,

From our definition of potential$ we can write

,,,,,,,,,,,,/6,9?3

from which we obtain$

,,,,,,,,,,,,,,,,,,,,/6,953

From the foregoing discussions we obser-e that the electric field strength at an point is the

negati-e of the potential gradient at an point$ negati-e sign shows that is directed from

higher to lower -alues of , 7his gi-es us another method of computing the electric field$ i,

e, if we now the potential function$ the electric field ma be computed, .e ma note here

that that one scalar function contain all the information that three components of carr$

the same is possible because of the fact that three components of are interrelated b the

relation ,

Eam%le: Electric Di%ole

"n electric dipole consists of two point charges of e%ual magnitude but of opposite sign and

separated b a small distance, "s shown in figure 6,55$ the dipole is formed b the two

point charges Qand -Qseparated b a distance d$ the charges being placed smmetricall

about the origin, Let us consider a pointPat a distance r$ where we are interested to find

the field,


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Fig 2.11: Electric Di%ole

7he potential at # due to the dipole can be written as:

,,,,,,,,,,,,,,,,,,,,,,,,,,/6,963

.hen r1and r6d$ we can write and ,

7herefore$

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,983

.e can write$

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,993

7he %uantit is called the di%ole moment of the electric dipole,

1ence the expression for the electric potential can now be written as:

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,93

It ma be noted that while potential of an isolated charge -aries with distance as 5B rthat of

an electric dipole -aries as 5Br6with distance,


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If the dipole is not centered at the origin$ but the dipole center lies at $ the expression for

the potential can be written as:

,,,,,,,,,,,,,,,,,,,,,,,,/6,9;3

7he electric field for the dipole centered at the origin can be computed as

,,,,,,,,,,,,,,/6,9


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Fig 2.12: E+ui%otential Lines for a *ositie *oint Charge

*ichael Farada as a wa of -isuali4ing electric fields introduced flux lines, It ma be seen

that the electric flux lines and the e%uipotential lines are normal to each other,

In order to plot the e%uipotential lines for an electric dipole$ we obser-e that for a

gi-enQandd, a constantV requires that is a constant, From this we can

write to be the e%uation for an e%uipotentialsurface and a famil of surfaces

can be generated for -arious -alues ofcv . .hen plotted in 62D this would gi-e e%uipotential

lines, 7o determine the e%uation for the electric field lines$ we note that field lines represent

the direction of in space, 7herefore$

, 'is a constant,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,9=3

,,,,,,,,,,,,,,,,,/6,9>3

For the dipole under consideration ? $ and therefore we can write$

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,?3


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Boundary conditions for Electrostatic fields: In our discussions so far we hae considered #he

e)is#ence of elec#ric field in #he homogeneous medium. *rac#ical elec#romagne#ic +rolems of#en

inole media wi#h differen# +hysical +ro+er#ies. -e#ermina#ion of elec#ric field for such +rolems

reuires #he 'nowledge of #he rela#ions of field uan#i#ies a# an in#erface e#ween #wo media. he

condi#ions #ha# #he fields mus# sa#isfy a# #he in#erface of #wo differen# media are referred #o

as !u"d#r$ c!"diti!"s.

In order #o discuss #he oundary condi#ions, we firs# consider #he field ehaior in some common

ma#erial media.

In general, ased on #he elec#ric +ro+er#ies, ma#erials can e classified in#o #hree ca#egories0

conduc#ors, semiconduc#ors and insula#ors (dielec#rics. In c!"duct!r , elec#rons in #he ou#ermos#

shells of #he a#oms are ery loosely held and #hey migra#e easily from one a#om #o #he o#her.

os# me#als elong #o #his grou+. he elec#rons in #he a#oms of i"su%#t!rs or die%ectrics remain

confined #o #heir ori#s and under normal circums#ances #hey are no# liera#ed under #he

influence of an e)#ernally a++lied field. he elec#rical +ro+er#ies ofse&ic!"duct!rs fall e#ween

#hose of conduc#ors and insula#ors since semiconduc#ors hae ery few numers of free charges.

he +arame#er c!"ductivit$ is used charac#eries #he macrosco+ic elec#rical +ro+er#y of a

ma#erial medium. he no#ion of conduc#ii#y is more im+or#an# in dealing wi#h #he curren# flow

and hence #he same will e considered in de#ail la#er on.

If some free charge is in#roduced inside a conduc#or, #he charges will e)+erience a force due #o

mu#ual re+ulsion and owing #o #he fac# #ha# #hey are free #o moe, #he charges will a++ear on #he

surface. he charges will redis#riu#e #hemseles in such a manner #ha# #he field wi#hin #he

conduc#or is ero. herefore, under s#eady condi#ion, inside a conduc#or ,


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From !auss(s theorem it follows that

? ,,,,,,,,,,,,,,,,,,,,,,,/6,53

7he surface charge distribution on a conductor depends on the shape of the conductor, 7he charges on

surface of the conductor will not be in e%uilibrium if there is a tangential component of the electric fie

present$ which would produce mo-ement of the charges, 1ence under static field conditions$ tange

component of the electric field on the conductor surface is 4ero, /he electric field on the surface of

conductor is normal eer!where to the surface , ince the tangential component of electric field is 4

the conductor surface is an e+ui%otential surface, "s ? inside the conductor$ the conductor

whole has the same potential, .e ma further note that charges re%uire a finite time to redistribute

conductor, 1owe-er$ this time is -er small sec for good conductor lie copper,

Let us now consider an inter

between a conductor and

space as shown in the figure 6

Fig 2.1&: 3oundar! Conditions for at the surface of a Conductor

Let us consider the closed path 'qrs'for which we can write$

,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,63

For and noting that inside the conductor is 4ero$ we can write

?,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,83

Etis the tangential component of the field, 7herefore we find that

Et ? ,,,,,,,,,,,,,,,,,,,,,,,,,/6,93


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In order to determine the normal componentE"$ the normal component of $ at the surface

of the conductor$ we consider a small clindrical !aussian surface as shown in the Fig,56,

Let represent the area of the top and bottom faces and represents the height of the

clinder, Hnce again$ as $ we approach the surface of the conductor, ince

? inside the conductor is 4ero$

,,,,,,,,,,,,,/6,3

,,,,,,,,,,,,,,,,,,/6,;3

7herefore$ we can summari4e the boundar conditions at the surface of a conductor as:

Et3 % ,,,,,,,,,,,,,,,,,,,,,,,,/6,


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water molecule2, In polar molecules the atoms do not arrange themsel-es to mae the

net dipole moment 4ero, 1owe-er$ in the absence of an external field$ the molecules

arrange themsel-es in a random manner so that net dipole moment o-er a -olume becomes

4ero, )nder the influence of an applied electric field$ these dipoles tend to align themsel-es

along the field as shown in figure 6,5, 7here are some materials that can exhibit net

permanent dipole moment e-en in the absence of applied field, 7hese materials are

called electrets that made b heating certain waxes or plastics in the presence of electric

field, 7he applied field aligns the polari4ed molecules when the material is in the heated

state and the are fro4en to their new position when after the temperature is brought down

to its normal temperatures, #ermanent polari4ation remains without an externall applied

field,

"s a measure of intensit of polari4ation$ polari4ation -ector (in C/m2 is defined

as: ,,,,,,,,,,,,,,,,/6,>3

"being the number of molecules per unit -olume i,e, is the dipole moment per unit

-olume, Let us now consider a dielectric material ha-ing polari4ation and compute the

potential at an external point H due to an elementar dipole dv*,

Fig 2.1): *otential at an Eternal *oint due to an Elementar! Di%ole dv'.


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.ith reference to the figure 6,5;$ we can write:

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,;?3

7herefore$

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,;53

,,,,,,,,/6,;63

where x$$4 represent the coordinates of the external point H and x($($4( are the coordinatesof the source point,

From the expression of+, we can -erif that

,,,,,,,,,,,,,,,,,,,,/6,;83

,,,,,,,,,,,,,,,,,,,/6,;93

)sing the -ector identit$ $ where fis a scalar %uantit $ weha-e$

,,,,,,,,,,,,,,,,,,,,,,,/6,;3

Con-erting the first -olume integral of the abo-e expression to surface integral$ we can write

,,,,,,,,,,,,,,,,,/6,;;3

.here is the outward normal from the surface element ds*of the dielectric, From theabo-e expression we find that the electric potential of a polari4ed dielectric ma be foundfrom the contribution of -olume and surface charge distributions ha-ing densities


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,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,;3

7he charge that remains inside the surface is

,,,,,,,,,,,,,,,,,,,,,,/6,


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7herefore the electric flux densit

.hen the dielectric properties of the medium are linear and isotropic$ polarisation is directl

proportional to the applied field strength and

,,,,,,,,,,,,,,,,,,,,,,/6,


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*edia exhibiting such characteristics are called biaial, Further$ if then the medium

is called uniaial, It ma be noted that for isotropic media$ ,

Loss dielectric materials are represented b a complex dielectric constant$ theimaginar part of which pro-ides the power loss in the medium and this is in general

dependant on fre%uenc,

"nother phenomenon is of importance is dielectric brea5down, .e obser-ed that the

applied electric field causes small displacement of bound charges in a dielectric material

that results into polari4ation, trong field can pull electrons completel out of the molecules,

7hese electrons being accelerated under influence of electric field will collide with molecular

lattice structure causing damage or distortion of material, For -er strong fields$ a-alanchebreadown ma also occur, 7he dielectric under such condition will become conducting,

7he maximum electric field intensit a dielectric can withstand without breadown is referred

to as the dielectric strengthof the material,

3oundar! Conditions for Electrostatic Fields: Let us consider the relationship among

the field components that exist at the interface between two dielectrics as shown in the

figure 6,5


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Fig 2.16: 3oundar! Conditions at the interface between two dielectrics

.e can express the electric field in terms of the tangential and normal components

,,,,,,,,,,/6,3

.hereEtandEnare the tangential and normal components of the electric field respecti-el,

Let us assume that the closed path is -er small so that o-er the elemental path length the

-ariation of E can be neglected, *oreo-er -er near to the interface$ , 7herefore

,,,,,,,,,,,,,,,,,,,,,,,/6,=?3

7hus$ we ha-e$

or i,e, the tangential com%onent of an electric field is continuousacross the interface.

For relating the flux densit -ectors on two sides of the interface we appl !ausss law to a

small pillbox -olume as shown in the figure, Hnce again as $ we can write

,,,,,,,,,,,,,,,,,,/6,=5a3


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i,e,$ ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,=5b3

7hus we find that the normal com%onent of the flu densit! ectorDis discontinuousacross an interface b! an amount of discontinuit! e+ual to the surface charge densit!at the interface.

Eam%le: 7wo further illustrate these pointsA let us consider an example$ which in-ol-es the

refraction of D or E at a charge free dielectric interface as shown in the figure 6,5=,)sing

the relationships we ha-e ust deri-ed$ we can write

,,,,,,,,,,,,,,,,,,,,,,,/6,=6a3

,,,,,,,,,,,,,,,,,,,,,,,/6,=6b3

In terms of flux densit -ectors$

,,,,,,,,,,,,,,,,,,,,,,,/6,=8a3

,,,,,,,,,,,,,,,,,,,,,,,/6,=8b3

7herefore$ ,,,,,,,,,,,,,,,,,,,,,,,/6,=93

Fig 2.1,: 7efraction of D or E at a Charge Free Dielectric Interface

Ca%acitance and Ca%acitors: .e ha-e alread stated that a conductor in an electrostaticfield is an E%uipotential bod and an charge gi-en to such conductor will distributethemsel-es in such a manner that electric field inside the conductor -anishes, If anadditional amount of charge is supplied to an isolated conductor at a gi-en potential$ this


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additional charge will increase the surface charge densit , ince the potential of theconductor is gi-en b

$ the potential of the conductor will also increase

maintaining the ratio same, 7hus we can write where the constant of

proportionalit Cis called the capacitance of the isolated conductor, I unit of capacitance is

CoulombB +olt also called Farad denoted b , It can It can be seen that if V5$ C Q, 7hus

capacit of an isolated conductor can also be defined as the amount of charge in Coulomb

re%uired to raise the potential of the conductor b 5 +olt,

Hf considerable interest in practice is a capacitor that consists of two /or more3 conductors

carring e%ual and opposite charges and separated b some dielectric media or free space,

7he conductors ma ha-e arbitrar shapes, " two2conductor capacitor is shown in

figure 6,5>,

Fig 2.10: Ca%acitance and Ca%acitors

.hen a d2c -oltage source is connected between the conductors$ a charge transfer occurs which resu

nto a positi-e charge on one conductor and negati-e charge on the other conductor, 7he condu

e%uipotential surfaces and the field lines are perpendicular to the conductor surface, If Vis the mea

difference between the conductors$ the capacitance is gi-en b , Capacitance of a capacitor d

the geometr

of the conductor and the permitti-it of the medium between them and does not depend on the charge


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potential difference between conductors, 7he capacitance can be computed b assuming Q/at the sam

time 2Qon the /other conductor3$ first determining using !ausss theorem and then determining

.e illustrate this procedure b taing the example of a parallel plate capacitor,

Example: #arallel plate capacitor

Fig 6,6?: #arallel #late CapacitorFor the parallel plate capacitor shown in the figure 6,6?$ let each plate has area " and a distance h separates

plates, " dielectric of permitti-it fills the region between the plates, 7he electric field lines are confined be

the plates, .e ignore the flux fringing at the edges of the plates and charges are assumed to be uniforml

distributed o-er the conducting plates with densities and 2 $ ,

' !ausss theorem we can write$ ,,,,,,,,,,,,,,,,,,,,,,,/6,=3

"s we ha-e assumed s to be uniform and fringing of field is neglected$ we see that E is

constant in the region between the plates and therefore$ we can writeA

-QE-V

== ,


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7hus$ for a parallel plate capacitor we ha-e

, ,,,,,,,,,,,,,,,,,,,,,,,/6,=;3

eries and %arallel Connection of ca%acitors: Capacitors are connected in -arious

manners in electrical circuitsA series and parallel connections are the two basic was of

connecting capacitors, .e compute the e%ui-alent capacitance for such connections,

eries Case: eries connection of two capacitors is shown in the figure 6,65, For thiscase we can write$

,,,,,,,,,,,,,,,,,,,,,,,/6,=


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therefore$ ,,,,,,,,,,,,,,,,,,,,,,,/6,==3

Electrostatic Energ! and Energ! Densit!:.e ha-e stated that the electric potential at a

point in an electric field is the amount of wor re%uired to bring a unit positi-e charge from

infinit /reference of 4ero potential3 to that point, 7o determine the energ that is present in

an assembl of charges$ let us first determine the amount of wor re%uired to assemble

them, Let us consider a number of discrete charges Q5$ Q6 $,,,,,,,$ Q0are brought from

infinit to their present position one b one, ince initiall there is no field present$ the

amount of wor done in bringing G5is 4ero, G6is brought in the presence of the field of G 5$

the wor done 5 Q6V65where V65is the potential at the location of G6due to G5,

#roceeding in this manner$ we can write$ the total wor done

,,,,,,,,,,,,,,,,,,,,/6,=>3

1ad the charges been brought in the re-erse order$

,,,,,,,/6,>?3

7herefore$

,,,,,/6,>53

1ere VIJrepresent -oltage at the/thcharge location due to0thcharge, 7herefore$

Hr$ ,,,,,,,,,,,,,,,,/6,>63

If instead of discrete charges$ we now ha-e a distribution of charges o-er a -olume vthenwe can write$

,,,,,,,,,,,,,,,,/6,>83


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.here v the -olume charge densit and + is represents the potential function,

ince$ 1v .= $ we can write

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,>93

)sing the -ector identit$

$ we can write

,,,,,,,,,,,,,,,,/6,>3

In the expression ds1Vs

.(2

1 $ for point charges$ since V -aries as

r

1and D -aries as

2

1

r$ the te

-aries as$

1

rwhile the area -aries as r6, 1ence the integral term -aries at least as

r

1and the as su

becomes large /i,e, r 3 the integral term tends to 4ero,

7hus the e%uation for . reduces to

, KK,,,,,,,,,,,,,,,/6,>;3

$ I s called the energ densit in the electrostatic field,

#oissons and Laplaces E%uations For electrostatic field$ we ha-e seen that

,,,,,,,,,,,,,,,,,,,,,,/6,>=3)sing -ector identit we can write$

,,,,,,,,,,,,,,,,/6,>>3


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For a simple homogeneous medium$ is constant and %= , 7herefore$

,,,,,,,,,,,,,,,,/6,5??3

7his e%uation is nown as #oissons e%uation, 1ere we ha-e introduced a new operator$ / del s

called the Laplacian operator, In Cartesian coordinates$

,,,,,,,,,/6,5?53

7herefore$ in Cartesian coordinates$ #oisson e%uation can be written as:

,,,,,,,,,,,,,,,/6,5?63In clindrical coordinates$

,,,,,,,,,,,,,,,/6,5?83

In spherical polar coordinate sstem$

,,,,,,,,,/6,5?93

"t points in simple media$ where no free charge is present$ #oissons e%uation reduces to

,,,,,,,,,,,,/6,5?3 which is nown as Laplaces e%uation,

Laplaces and #oissons e%uation are -er useful for sol-ing man practical electrostatic

field problems where onl the electrostatic conditions /potential and charge3 at some

boundaries are nown and solution of electric field and potential is to be found throughout

the -olume, .e shall consider such applications in the section where we deal with boundar

-alue problems,


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(ni+ueness /heorem:olution of Laplaces and #oissons E%uation can be obtained in a

number of was, For a gi-en set of boundar conditions$ if we can find a solution to

#oissons e%uation /Laplaces e%uation is a special case3$ we first establish the fact that the

solution is a uni%ue solution regardless of the method used to obtain the solution,

)ni%ueness theorem thus can stated as

,,,,,,,,,,,,,,,,,,,,,,,,/6,5?;3

and ,,,,,,,,,,,,,,,/6,5?3

olution of an electrostatic problem specifing its boundar condition is the onl possible so

irrespecti-e of the method b which this solution is obtained, 7o pro-e this theorem$ as shown in figure

we consider a -olume Vr and a closed surface Srencloses this -olume, Sr is such that it ma also

surface at infinit, Inside the closed surface Sr$ there are charged conducting bodies with surfaces

S3$,,,,and these charged bodies are at specified potentials,

If possible$ let there be two solutions of #oissons e%uation in Vr, Each of these solutions Vand V2sa#oisson e%uation as well as the boundar conditions on S$ S2$,,,,S"and Sr, ince Vand V2are assum

be different$ let Vd V2V2be a different potential function such that$

Figure:


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)sing the fact that $we can write

,,,,,,,,,,,,,,,,,,,,,,,,,,/6,55?3

1ere the surface Sconsists of Sras well as S$ S2$ ,,,,,S3,

.e note that Vd? o-er the conducting boundaries$ therefore the contribution to the surface

integral from these surfaces are 4ero, For large surface Sr$ we can thin the surface to be a

spherical surface of radius r , In the surface integral$ the term dd VV -aries as $1

r

whereas the area increases asr2$ hence the surface integral decreases asr

1, For r $

the integral -anishes, 7herefore$ %442 = dvV

xV

d , ince244

dV is non2negati-e

e-erwhere$ the -olume integral can be 4ero onl if %44 = dV $ which means V V2, 7hat

is$ our assumption that two solutions are different does not hold, o if a solution exists for

#oissons /and Laplaces3 e%uation for a gi-en set of boundar conditions$ this solution is a

uni%ue solution,

*oreo-er$ in the context of uni%ueness of solution$ let us loo into the role of the boundar

condition, .e obser-e that$

$

which$ can be written as ,

)ni%ueness of solution is guaranteed e-erwhere if

or ? on S,

7he condition %=dV means that SVVV == 21 on , 7his specifies the potential function

on and is called Dirichlet boundar! condition, Hn the other hand$


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means

on , pecification of the normal deri-ati-e of the potential function is called the 8eumann

boundar! conditionand corresponds to specification of normal electric field strength or

charge densit, If both the Dirichlet and the 0eumann boundar conditions are specified

o-er part of then the problem is said to be o-er specified or improperl posed,

-ethod of Images: Form uni%ueness theorem$ we ha-e seen that in a gi-en region if the

distribution of charge and the boundar conditions are specified properl$ we can ha-e a

uni%ue solution for the electric potential, 1owe-er$ obtaining this solution calls for sol-ing

#oisson /or Laplace3 e%uation, " conse%uence of the uni%ueness theorem is that for a

gi-en electrostatics problem$ we can replace the original problem b another problem at the

same time retaining the same charges and boundar conditions, 7his is the basis for the

method of images, *ethod of images is particularl useful for e-aluating potential and field

%uantities due to charges in the presence of conductors without actuall sol-ing for

#oissons /or Laplaces3 e%uation, )tili4ing the fact that a conducting surface is an

e%uipotential$ charge configurations near perfect conducting plane can be replaced b the

charge itself and its image so as to produce an e%uipotential in the place of the conducting

plane, 7o ha-e insight into how this method wors$ we consider the case of point

charge Qat a distance dabo-e a large grounded conducting plane as shown in figure 6,69,

Fig 2.2&

For this case$ the presence of the positi-e charge MQwill induce negati-e charges on the surf

conducting plane and the electric field lines will be normal to the conductor, For the time being

ha-e the nowledge of the induced charge densit ,


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Fig 2.29

7o appl the method of images$ if we place an image charge 2G as shown in the figure 6,6;$

the sstem of two charges /essentiall a dipole3 will produce 4ero potential at the location of

the conducting plane in the original problem, olution of field and potential in the

region plane /conducting plane in the original problem3 will remain the same e-en the

solution is obtained b sol-ing the problem with the image charge as the charge and the

boundar condition remains unchanged in the region %

z ,

Fig 2.2)


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.e find that at the pointP/x$$$z3$

,,,,,,,,,,,,,,,/6,5553

imilarl$ the electric field at # can be computed as:

,,,,,,,,,,,,,,,/6,5563

7he induced surface charge densit on the conductor can be computed as:

,,,,,,,,,,,,,,,/6,5583

7he total induced charge can be computed as follows:

,,,,,,,,,,,,,,,/6,5593

)sing change of -ariables$

,,,,,,,,,,,,,,,/6,553

7hus$ as expected$ we find that an e%ual amount of charge ha-ing opposite sign is induced

on the conductor,

*ethod of images can be effecti-el used to sol-e a -ariet of problems$ some of which are

shown in the next slide,

5, #oint charge between semi2infinite conducting planes,


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Fig 2.26 :*oint charge between two%er%endicular semiinfinite conducting %lanes

Fig 2.2,: E+uialent Image charge arrang

In general$ for a sstem consisting of a point charge between two semi2infinite conducting planes in

an angle /in degrees3$ the number of image charges is gi-en b 1$5%( =

N ,

6, Infinite line charge /densit CBm3 located at a distance dand parallel to an infiniteconducting clinder of radius #,

Fig 2.20: Infinite line charge and %arallel conducting c!linder

Fig 2.#: Line charge and its image


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7he image line charge is located at and has densit%

,

8, " point charge in the presence of a spherical conductor can similarl be represented in

terms of the original charge and its image,

Electrostatic boundar! alue %roblem: In this section we consider the solution for field

and potential in a region where the electrostatic conditions are nown onl at the

boundaries, Finding solution to such problems re%uires sol-ing Laplaces or #oissons

e%uation satisfing the specified boundar condition, 7hese tpes of problems are usuall

referred to as boundar -alue problem, 'oundar -alue problems for potential functions can

be classified as:

Dirichlet problem where the potential is specified e-erwhere in the boundar

0eumann problem in which the normal deri-ati-es of the potential function are specified

e-erwhere in the boundar

*ixed boundar -alue problem where the potential is specified o-er some boundaries and

the normal deri-ati-e of the potential is specified o-er the remaining ones,

)suall the boundar -alue problems are sol-ed using the method of separation of-ariables$ which is illustrated$ for different tpes of coordinate sstems,

3oundar! Value *roblems In Cartesian Coordinates: Laplace(s e%uation in cartesiancoordinate can be written as$

,,,,,,,,,,,,,,,/6,55;3

"ssuming that V/x,$,z3 can be expressed as V/x,$,z3 4/x35/$36/z3 where4/x3$ 5/$3 and6/z3

are functions ofx$ yand zrespecti-el$ we can write:

,,,,,,,,,,,,,,/6,55


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,

In the abo-e e%uation$ all terms are function of one coordinate -ariabl onl$ hence we canwrite:

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,55=3

kx$ k$and kzare seperation constant satisfing

"ll the second order differential e%uations abo-e has the same form$ therefore we discuss

the possible form of solution for the first one onl,

Depending upon the -alue of kx $ we can ha-e different possible solutions,

,,,,,,,,,,,,,,,,,,,,,,,,,/6,55>3

A$B$Cand1are constants,

imilarl we ha-e the possible solutions for other two differential e%uations, 7he particular

solutions for4/x3$ 5/$3 and6/z3 to be used will be chosen based on the nature of the

problems and the constantsA$Betc are e-aluated from the boundar conditions, .e

illustrate the method discussed abo-e with some examples,


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Eam%le 1:"s shown in the figure 6,85$ let us consider two electrodes$ one grounded and

the other maintained at a potential V, 7he region between the plates is filled up with two

dielectric laers, Further$ we assume that there is no -ariation of the field along xor$and

there is no free charge at the interface,

Fig 2.#1

7he Laplace e%uation can be written as:

For which we can write the solution in general form as V7 A z 8 B.

Considering the two regions$

,,,,,,,,,,,,,,,,,,,,,,,/6,56?3

"ppling the boundar conditions V? forz? and VV?forz d$ we can write

,,,,,,,,,,,,,,,,,,/6,5653

'oth the solution should gi-e the same potential at z#, 7herefore$

,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,566a3

or$ ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,566b3


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"nother e%uation relating toAandA2is re%uired to sol-e for the two constants AandA2.

.e note that at the dielectric boundar at 4 a $ the normal component of electric flux

densit -ector is constant, 0oting that $ we can write

,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,568a3

or$ ,,,,,,,,,,,,,,,,,,/6,568b3

ol-ing forAandA2$ we find that

,,,,,,,,,,,,,,,,,,/6,569a3

,,,,,,,,,,,,,,,,,/6,569b3

)sing the abo-e expressions forAandA2$ we can find V and E ,

Eam%le2:In the earlier example$ we considered a situation where the potential function was a

function of onl one coordinate, In the figure 6,86 we consider a problem where the potential is a fu

of two coordinate -ariables, .e consider two number of grounded semi2infinite parallel plate electr

separated b a distance d, " third electrode maintained at potential V9and insulated from the groun

conductors is placed as shown in the figure 6,86, 7he potential function is to be determined in the re

enclosed b the electrodes,

Fig 2.#2: *otential Function Enclosed b! the Electrodes


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"s discussed abo-e$ we can write V(x,$,z)4(x) 5($) 6(z)..e now construct the solution for Vsuchsatisfies the stated conditions at the boundar,


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.e find that V is independent ofx, 7herefore$4(x) C5$ where C5is a constant,

7herefore$

,,,,,,,,,,,,,,,,,,/6,563

.e obser-e that the potential has to be e%ual to V9at$ ? and as the potential

function becomes 4ero, 7herefore$ the$dependence can be expressed as an exponential

function, imilarl$ since V ? for z ? andz d$ thezdependence will be a sinusoidal

function with kz7 k$ kbeing a real number,

7herefore$ ,,,,,,,,,,,,,,,/6,56;3

,,,,,,,,,,,,,,,/6,563

)sing orthogonalit conditions and e-aluating the integrals we can show that


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,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,58?3

7he general solution for the potential function can therefore be written as

,,,,,,,,,,,,,,,,,,,,,,,,,/6,5853

n the region$ ? and ?NzN d,

3oundar! Value *roblem in C!lindrical and %herical *olar Coordinates

"s in the case of Cartesian coordinates$ method of separation of -ariables can be used to

obtain the general solution for boundar -alue problems in the clindrical and spherical

polar coordinates also, 1ere we illustrate the case of solution in clindrical coordinates with

a simple example where the potential is a function of one coordinate -ariable onl,

Eam%le #: *otential distribution in a coaial conductor

Let us consider a -er long coaxial conductor as shown in the figure 6,88$ the inner

conductor of which is maintained at potential V9and outer conductor is grounded,

Fig 2.##: *otential Distribution in a Coaial Conductor

From the smmetr of the problem$ we obser-e that the potential is independent

of O -ariation, imilarl$ we assume the potential is not a function of z, 7hus the Laplaces

e%uation can be written as


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,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,5863

Integrating the abo-e e%uation two times we can write

,,,,,,,,,,,,,,,,,,,,,,,,,/6,5883

where Cand C2are constants$ which can be e-aluated from the boundar conditions

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,5893

E-aluating these constants we can write the expression for the potential as

,,,,,,,,,,,,,,,,,,,,,,,/6,583

Furthure$ ,,,,,,,,,,,,,,,,,,,,,,,/6,58;3

7herefore$ ,,,,,,,,,,,,,,,,,,,,,,,,/6,58


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Considering the smmetr$ we find that the potential is function

of r onl, 7he Laplaces e%uation can be written as:

,,,,,,,,,,,,/6,58>3

From the abo-e e%uation$ b direct integration we can write$

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,59?3"ppling boundar conditions

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,5953we can find Cand C2and the expression for the potential can

be written as : ,,,,,,,,,,,,,,,,,,,,,,,,,,,/6,5963

7he electric field and the capacitor can be obtained using thesame procedure described in Example 28,

Fig 2.#&: Concentric conducting s%her

Emtl Note Unit II

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