Field Theory Compiled Basvaraj

8/8/2019 Field Theory Compiled Basvaraj

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e-Notes by Prof.T.Basavaraj Sri Revanna Siddeshwara Institute of Technology, Bangalore

FIELD THEORY

Sub Code : EC44 IA Marks : 25Hrs/Week : 04 Exam Hrs. : 03

Total Hrs. : 52 Exam Marks : 100

1. Electric Fields 18 hours

a. Coulombs law and Electric field intensity

b. Electric flux density, Gauss law and divergence

c. Energy and potential

d. Conductors, dielectrics and capacitance

e. Poissons and Laplaces equations

2. Magnetic fields 14 hoursa. The steady magnetic field

b. Magnetic forces, materials and inductance

3. Time varying fields and Maxwells equations 5 hours

4. Electromagnetic waves 15 hours

Text Books :

William H Hayt Jr and John A Buck, Engineering Electromagnetics, Tata McGraw-Hill,

6th Edition, 2001

Reference books :

John Krauss and Daniel A Fleisch, Electromagnetics with Application, McGraw-Hill,

5th Edition, 1999

Guru and Hiziroglu, Electromagnetics Field theory fundamentals, Thomson Asia Pvt. Ltd

I Edition, 2001

Joseph Edminster, Electromagnetics, Schaum Outline Series, McGraw-Hill

Edward C Jordan and Keith G Balmain, Electromagnetic Waves and Radiating Systems,

Prentice-Hall of India, II Edition, 1968, Reprint 2002.

David K Cheng, Field and Wave Electromagnetics, Pearson Education Ais II Edition, 1989,

Indian Repr-01

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Introduction to Field Theory

The behavior of a physical device subjected to electric field can be studied either by Field

approach or by Circuit approach. The Circuit approach uses discrete circuit parameters like

RLCM, voltage and current sources. At higher frequencies (MHz or GHz) parameters would no

longer be discrete. They may become non linear also depending on material property andstrength of v and i associated. This makes circuit approach to be difficult and may not give very

accurate results.

Thus at high frequencies, Field approach is necessary to get a better understanding of

performance of the device.

FIELD THEORY

The Vector approach provides better insight into the various aspects of Electromagnetic

phenomenon. Vector analysis is therefore an essential tool for the study of Field Theory.

The Vector Analysis comprises of Vector Algebra and Vector Calculus.

Any physical quantity may be Scalar quantity or Vector quantity. A Scalar quantity is

specified by magnitude only while for a Vector quantity requires both magnitude and direction

to be specified.

Examples :

Scalar quantity : Mass, Time, Charge, Density, Potential, Energy etc.,

Represented by alphabets A, B, q, t etc

Vector quantity : Electric field, force, velocity, acceleration, weight etc., represented by alphabets

with arrow on top.

etc.,B,E,B,A

Vector algebra : If C,B,A are vectors and m, n are scalars then

(1) Addition

AssociativC)BA()CB(A

CommutativABBA++=++

+=+

(2) Subtraction

)B(-AB-A +=

(3) Multiplication by a scalar

Am)BA(m

nAmAn)(m

)A(mn)A(nm

mAAm

A vector is represented graphically by a directed line segment.

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A Unit vector is a vector of unit magnitude and directed along that vector.

Aa is a Unit vector along the direction of A .

Thus, the graphical representation of A and Aa are

AaAorA/AaAlso

ctorUnit veAVector

A

AA

==

Product of two or more vectors :

(1) Dot Product ( . )

0,B}COSA{ORCOSB(AB.A =

B B

CosA

A CosB A

A . B = B . A (A Scalar quantity)

(2) CROSS PRODUCT (X)

C = A x B = nSINBA

xA)CB(xA

AxB-BxA

BAsuchthatdirected

perpendicurunitvectoisnand

betweenangleis''whereEx.,

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CO-ORDINATE SYSTEMS :

For an explicit representation of a vector quantity, a co-ordinate system is essential.

Different systems used :

Sl.No. System Co-ordinate variables Unit vectors

1. Rectangular x, y, z ax , ay , az2. Cylindrical , , z a , a , az3. Spherical r, , ar , a , a

These are ORTHOGONAL i.e., unit vectors in such system of co-ordinates are mutually

perpendicular in the right circular way.

r,z,zyxi.e.,

RECTANGULAR CO-ORDINATE SYSTEM :

Z

x=0 plane

azp

y=0 Y

plane ay

ax z=0 plane

X

yxz

xzy

zyx

xzzyyx

aaxa

aaxa

aaxa

0a.aa.aa.a

=

=

=

===

az is in direction of advance of a right circular screw as it is turned from a x to ay

Co-ordinate variable x is intersection of planes OYX and OXZ i.e, z = 0 & y = 0

Location of point P :

If the point P is at a distance of r from O, then

If the components of r along X, Y, Z are x, y, z then

arazayaxrrzyx =++=

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Equation of Vector AB :

componentsareB&B,Band careA&A,Awhere

ABorBABA

aBBOBand

AaAAOAIf

zys zys

xx

xx

+

=

=

Dot and Cross Products :

wegrouping,andby termtermproducts'Cross'Taking

aBa(Bx)aAaAa(ABxA

aBa(B.)aAaAa(AB.A

yxxzzyyxx

yyxxzzyyxx

+++=

+++=

zyx

zyx

zyx

BBB

AAA

aaa

BxA =

ABlengthVector

where

AB

ABa

alongVectorUnit

represents)CxB(.A(ii)

then0BxA

Costhen0B.A(i)

nonareCandB,AIf

CBA)CxB(.A

AB

x

x

x

Differential length, surface and volume elements in rectangular co-ordinate systems

zyx

zyx

adzadyadxrd

z

rdy

rdx

x

rrd

azayaxr

+=

+

=

++=

y

Differential length 1-----]d zd yd x[rd 1 /2222 ++=

Differential surface element, sd

1. zadxdy:ztor

2. zadxdy:ztor ------ 2

3. zadxdy:ztor

Differential Volume element

dv = dx dy dz ------ 3

z

dx p

5

B

B AB

0 A A


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p dz

dy

rrdr+

0 y

x

Other Co-ordinate systems :-

Depending on the geometry of problem it is easier if we use the appropriate co-ordinate system

than to use the Cartesian co-ordinate system always. For problems having cylindrical symmetry

cylindrical co-ordinate system is to be used while for applications having spherical symmetry

spherical co-ordinate system is preferred.

Cylindrical Co-ordiante systems :-z

P(, , z) x = Cos y = Sin

az r z = z

ap r yzz

y / xtan

yx

1-

22

==

+=

x

ar

aSin-r SinaCosr rdrrd SiaCosr azayaxr

z

x

x

xyx

z

Thus unit vectors in (, , z) systems can be expressed in (x,y,z) system as

2222

z

zzz

y

xyx

(dz))d(drdand

adzadadrd,Further

orthogonalareaanda,a;aa

SinaaCosaSin-a

CosaaSinaCosa

++=

++=

=

=+=

=+=

yx

Differential areas :

zz

adz)(dads

3-------a.)d((dz)ads

a.)d()(dads

=

=

=

Differential volume :

6

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4-----dzdddor

(dz))d()(dd

==

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Spherical Co-ordinate Systems :-

Z X = r Sin Cos

Y = r Sin Sin z p Z = r Cos R

r

0 y Y

x r Sin

X

dddrSinrvd

ddrrSd

ddrSinrSd

ddSinrSd

adSinradradrRd

dR

dR

drr

RRd

aCosaSin-R

/R

a

aSinaSinCosaCosCosR

/R

a

aCosaSinSinaCosSinr

R/

r

Ra

aCosraSinSinraCosSinrR

2

2

2

r

r

yx

zyx

zyxr

zyx

=

=

=

=

++=

+

+

=

+=

=

+=

=

++=

=

++=

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General Orthogonal Curvilinear Co-ordinates :-

z u1 a3 u3

a1 u2a2

y

x

Co-ordinate Variables : (u1 , u2, u3) ;

Here

u1 is Intersection of surfaces u2 = C & u3 = C

u2 is Intersection of surfaces u1 = C & u3 = Cu3 is Intersection of surfaces u1 = C & u2 = C

2

1

1

321

1

1

x

321

h,u

Rh

h,h,hwhere

duh

u

RRdthen

yaxRIfOrthogonalisSystem ubnitvectarea,a,a

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Co-ordinate Variables, unit Vectors and Scale factors in different systems

Systems Co-ordinate Variables Unit Vector Scale factors

General u1 u2 u3 a1 a2 a3 h1 h2 h3

Rectangular x y z ax ay az 1 1 1

Cylindrical z a a az 1 1

Spherical r ar a a 1 r r sin

Transformation equations (x,y,z interms of cylindrical and spherical co-ordinate system

variables)

Cylindrical : x = Cos , y = Sin , z = z ; 0, 0 2 - < z <

Spherical

x = r Sin Cos , y = r Sin Cos , z = r Sin r 0 , 0 , 0 2

10

)u,u,(uAAand)u,u,(uA

)u,u,(uAAwherefieldVectoraisaAaAaAA

fieldScalara)u,u,u(VVere

AhAhAhuuu

ahahah

hhh

1Ax

)Ah(hu

)Ah(hu

)Ah(hu

hhh

1A

au

v

h

1a

u

v

h

1a

u

v

h

1V

3213332122

32111332211

321

332211

321

332211

321

321

3

231

2

132

1321

3

33

2

22

1

11

==

=++=

=

=

+

+

=

+

+

=


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Vector Transformation from Rectangular to Spherical :

11

= ++=+=++=

z

y

x

zyx

zyx

rzryrxr

zyxr

rr

RrrRS

zzyyxxR

A

A

A

a.aa.aa.a

a.aa.aa.a

a.aa.aa.a

A

A

A

asA,A,AtorelatedareA,A,Awe

aAaAaA

a)a.A(a)aA(a)aA(A:cal

aAaAaAA:rRangula

R


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Field Theory

A field is a region where any object experiences a force. The study of performance in the

presence of Electric field )E( , Magnetic field ( ) is the essence of EM Theory.

P1 : Obtain the equation for the line between the points P(1,2,3) and Q (2,-2,1)

zyx a2-a4-aPQ =

P2 : Obtain unit vector from the origin to G (2, -2, 1)

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Problems on Vector Analysis

Examples :-

1. Obtain the vector equation for the line PQ between the points P (1,2,3)m and Q (2,

-2, 1)mZ

PQ P (1,2,3)

Q(2,-2,-1)

0

Y

X

)a2-a4-a(a3)-(-1a2)-(-2a1)-(2

a)z-(za)y-(ya)x-(xPQvectorThe

zyx

zyx

zpqypqxpq

=

++=

++=

2. Obtain unit vector from origin to G (2,-2,-1)

G

G

0

a0.667-a(0.667a

(-2)2G

G

Ga,orunit vectThe

a2-a(2

a)0-(xGvectorThe

yxg

2

g

x

xg

=

=

=

3. Given

zyx

zyx

a5a2-a4-B

aa3-a2A

==

BxA(2)andB.A(1)find

Solution :

)a5a2-a(-4.)aa3-a(2B.A(1)zyxzyx

++=

= - 8 + 6 + 5 = 3

Since ax . ax = ay . ay = az . az = 0 and ax ay = ay az = az ax = 0

(2)524

132

aaa

BxA

zyx

=

= (-13 ax -14 ay - 16 az)

4. Find the distance between A( 2, /6, 0) and B = ( 1, /2,2)Soln : The points are given in Cylindrical Co-ordinate (, , z). To find the distance betweentwo points, the co-ordinates are to be in Cartesian (rectangular). The corresponding

rectangular co-ordinates are ( Cos , Sin , z)

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1.73AB)(

a1.73-

a1.73-

)A-(BAB

a6

CosB&

a6

Cos2A

2

x

x

xx

x

5. Find the distance between A( 1, /4, 0) and B = ( 1, 3 /4, )Soln : The specified co-ordinates (r, , ) are spherical. Writing in rectangular, they are

(r Sin Cos , r Sin Sin , r Cos ).Therefore, A & B in rectangular co-ordinates,

0.5(2 AB.AB(AB a1.414- )A-(BAB a0.707( Cos43Sin(B a0.707( Cos4Sin(1A

x

xx

xx

6. Find a unit vector along AB in Problem 5 above.

AB

ABa AB = = [ - 1.414 ax + (-0.707) ay + (-0.707) az]

1.732

1

= )a408.0a0.408-a0.816-( zyx

7. Transform ordinates.-ColCylindricainFinto)a6a8-a(10F zyx +=

Soln :

a)a.F(a)a.F(a)a.F(F zzppCyl ++=

yx

yx

yx

Siny

Cosx

Sin8-Cos(10

a8-a(10[a8-a10([ a8-a[(10

)a6a(12.8

a6a38.66)](-Cos8-38.66)(-Sin10[-a]38.66)-(Sin8-38.66)(-Cos10[F

z

zpCyl

+=

++=

8. Transform azax-ayB zyx += into Cylindrical Co-ordinates.

zxxCyl

x

aza- CosSin[ -aSin([ -aSin([ (B.a)a(B.B

Cos-aSinB

Siny,Cosx

9. Transform xa5 into Spherical Co-ordinates.

SinCos5aCosSin5

)]aCosaSin(-.a5[

CosaCos(Cos.a5[

SinaCos(Sin.a5[

a(A.a)a(A.a)a(A.A

r

yxx

xx

xx

rrSph

+=+++= ++=

10. Transform to Cylindrical Co-ordinates z),,(Qata4x)-(y-ay)x(2G yx +=

Soln :

14

x

y


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Siny,Cosx

Siny)x(2-[

-Cosy)2x([

(y-ay)x(2[

-(y-ay)x(2[G

(G.a)a(G.G

x

xCyl

Cyl

=

a)CosSin3-SinCos4(

a)Sin-CosSin5Cos2(

a]Cos4CosSin-Sin-CosSin2-[

a]CosSin4Sin-CosSinCos2[

a]Cos)Cos4-Sin(-Sin)SinCos2(-[

a]Sin)Cos4-Sin(-Cos)SinCos2([G

22

22

22

22

Cyl

+

+=

++

++=

+++=

11. Find a unit vector from ( 10, 3 /4, /6) to (5, /4, )Soln :

A(r, , ) expressed in rectangular co-ordinates

ABBAa 9.65AB A-BAB .a6.12A 4Sin5B Sin10A osSinrOA

AB

x

12. Transform a6a8-a10F zyx += into F in Spherical Co-orindates.

)a0.783a5.38a(11.529F

a0.781)x8-0.625-x(-10

a(0.625))x0.42x8-0.781x0.42x10(

a(-0.625))x0.9x8-0.781x0.9x(10F

0.781(-38.66)CosCos0.4264.69CosCos

0.625-(-38.66)SinSin0.964.69SinSin

38.66-10

8-tan

64.89200

6Cos

r

zCos;2006810r

a)Cos8-Sin10(-

a)Sin6-SinCos8-CosCos(10a)Cos6SinSin8-CosSin(10

a)a.F(a)a.F(a)a.F(F

aCosaSin-a

aSin-aSinCosaCosCosa

aCosaSinSinaCosSina

r

r

01-

1-1-222

r

rrSph

yx

zyx

zyxr

++=

+

+

=

====

====

==

====++=

+

+ +=

++=

+=

+=

++=

Line Integrals

In general orthogonal Curvilinear Co-ordinate system

++=++=

++=

C

333

C

222

C C

111

332211

333222111

duFhduFhduFhdl.F

aFaFaFF

aduhaduhaduhdl

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Conservative Field A field is said to be conservative if it is such that 0dl.C

=

contour.closedaaroundtakenisitifzeroisandaandbbetweenpotentialtherepresentdl.

andintensityfieldelectrictherepresent-E

thenflux,ticelectrostaisIf).!pathon thedependnot(does(a)-b)(ddl.

b

a

b

a

=

==

= 0dl.i.e.,Therefore ES flux field is Conservative.

EXAMPLES :

13. Evaluate line integral = dl.aI

where yx ax)-(yay)(xa ++=

(4,2)Bto(1,1)Afromxyalong

2 =

Soln : adyadxdl yx +=

++===

++=

2

1

2

1

22

2

dy)y-(ydy2yy)(ydl.a

dxdydy2orxy

dy)x-(ydxy)(xdl.a

++=2

1

223dy)y-yy2y(2

++=2

1

23dyy)yy(2

3

111

3

11-

3

212

3

4-2

3

88

2

1

3

1

2

1-

2

2

3

2

2

2

2

y

3

y

24

y2

234

2

1

234

==

++=

++

++=

++

//

=

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14. Evaluate the Integral =S

ds.EI where aradiusofhunisphereisSandaxE x=

Soln:

If S is hemisphere of radius a, then S is defined by

3

a2x

3

2xadCosdSinads.E

20,2/0

ddCosSinads.E

ddSina.a)CosSin(ads.Eds.E

CosSinax;aCosSinxE

a)a(E.a)a(E.a)a.(EE

addSinads

ad)Sin(a)d(ads;0z,azyx

3/2

0

2

0

3233

23

2

r

2

r

rr

rr

r

2

r

2222

===


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where r, r1 , r2 .. rm are the vector distances of q, q1 , qm from origin, 0.

mr-r is distance between charge qm and q.

ma is unit vector in the direction of line joining qm to q.

Electric field is the region or vicinity of a charged body where a test charge experiences a

force. It is expressed as a scalar function of co-ordinates variables. This can be illustrated by

drawing force lines and these may be termed as Electric Flux represented by and unit iscoulomb (C).

Electric Flux Density )D(

is the measure of cluster of electric lines of force. It is the

number of lines of force per unit area of cross section.

i.e.,

==S

2surfacetonormalorunit vectisnwhereCdsnDorc/m

A

D

Electric Field Intensity )E( at any point is the electric force on a unit +ve charge at that

point.

i.e., c/Nar4

q

q

FE 12

10

1

==

CEDorc/ND

c/Na

r4

q

1 0

0

12

1

1

0

=

=

=

in

vacuum

In any medium other than vacuum, the field Intensity at a point distant r m from + Q C is

Car4

QDorCEDand

m)/Vor(c/Nar4

QE

r20

r2

0

==

=

r

r

Thus D is independent of medium, while E depends on the property of medium.

r E+QC q = 1 C (Test Charge)

Source charge

E

E

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0 r , m

Electric Field Intensity E for different charge configurations

1. E due to Array of Discrete charges

Let Q, Q1 , Q2 , Qn be +ve charges at P, P1 , P2 , .. Pn . It is required to find E

at P.

Q1 1r nE

P1

Q2 2r P 2E

1E

P2 1rQn 0

Pr nr

m/Var-r

Q

4

1E m2

m

m

0

r =

2. E due to continuous volume charge distribution

Ra R P

v C / m3

The charge is uniformly distributed within in a closed surface with a volume charge density

of v C / m3 i.e,

vd

QdanddvQ

V

V

V ==

C/Na)r-(r4

)(rE

aR4

V

aR4

Q

E

R210

1

r

R20

R20

1

=

=

=

V

V

V

Ra is unit vector directed from source to filed point.

3. Electric field intensity E due to a line charge of infinite length with a line charge

densityof l C / m Ra

P

R

dl l C / m

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L

C/NaR

dl

4

1E R

L

2

0

p =l

4. E due to a surface charge with density of S C / m2

Ra

P (Field point)

ds R

(Source charge)

C/NaR

ds

4

1E R

S

2

0

p =S

Electrical Potential (V) The work done in moving a unit +ve charge from Infinity to that is

called the Electric Potential at that point. Its unit is volt (V).

Electric Potential Difference (V12) is the work done in moving a unit +ve charge from one

point to (1) another (2) in an electric field.

Relation between E and V

If the electric potential at a point is expressed as a Scalar function of co-ordinate variables

(say x,y,z) then V = V(x,y,z)

V-E(2)and(1)From

----dl.VdV

dyy

Vdx

x

VdVAlso,

----dl.E-dlq

f-dV

== +===

Determination of electric potential V at a point P due to a point charge of + Q C

la

dRR+

0 + Q R P Ra

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At point P, C/NaR4

QE R2

0=

Therefore, the force f on a unit charge at P.

NaR4

QEx1f R2

0

p

==

The work done in moving a unit charge over a distance dl in the electric field is

scalar(aVoltR4

QV

-)a.a(R4

dlQ-V

dl.E-dl.f-dV

2

0

P

lR

R

2

0

p

====

Electric Potential Difference between two points P & Q distant Rp and Rq from 0 is

voltR

1-

R

1

4

Q)V-(VV

qp0

qppq

==

Electric Potential at a point due to different charge configurations.

1. Discretecharges

. Q1

. PQ2 Rm

Qm

VR

Q

4

1V

n

1 m

m

0

1P

=

2. Linecharge

x P Vdl

R

l

4

1V

l

l

0

2P

=

l C / m

3. Surfacecharge

x P VR

ds

4

1V

S

S

0

3P =

s C / m2

4. Volumecharge

x P

21

v

C/ m3


22/37

R VR

dv

4

1V

V

V

0

4P =

5. Combination of above V5P = V1P + V2P + V3P + V4P

Equipotential Surface : All the points in space at which the potential has same value lie on a

surface called as Equipotential Surface.

Thus for a point change Q at origin the spherical surface with the centre of sphere at the

origin, is the equipotential surface.

Sphere of

Radius , R

R

P

equipotential surfaces

Q

V

0 R

Potential at every point on the spherical surface is

potentialsurfaceialequipotenttwopotentialofdifferenceisV

voltR4

QV

PQ

0

R

=

Gausss law : The surface integral of normal component of D emerging from a closed

surface is equal to the charge contained in the space bounded by the surface.

i.e., =S

(1)CQdsn.D

where S is called the Gaussian Surface.

By Divergence Theorem,

=S V

dvD.dsn.D ----------- (2)

Also, =V

V dvQ ---------- (3)

From 1, 2 & 3,D. = ----------- (4) is point form (or differential form) of Gausss law while

equation (1) is Integral form of Gauss law.

Poissons equation and Laplace equation

22

0

+Q


23/37

In equation 4, ED 0=

equationLaplace0V0,If

equationPoisson-V

/V)(-.or/E.

2

0

2

00

==

==

Till now, we have discussed (1) Colulombs law (2) Gauss law and (3) Laplace equation.

The determination of E and V can be carried out by using any one of the above relations.

However, the method of Coulombs law is fundamental in approach while the other two use

the physical concepts involved in the problem.

(1) Coulombs law : Here E is found as force f per unit charge. Thus for the simple case

of point charge of Q C,

=

=

l

20

VoltdlEV

MV/R

Q

4

1E

(2) Gausss law : An appropriate Gaussian surface S is chosen. The charge enclosed is

determined. Then

=

=

l

S

enc

voltdlEVAlso

determinedareEhenceandDThen

QdsnD

(3) Laplace equation : The Laplace equation 0V2 = is solved subjecting to differentboundary conditions to get V. Then, V-E =

23


24/37

Solutions to Problems on Electrostatics :-

1. Data : Q1 = 12 C , Q2 = 2 C , Q3 = 3 C at the corners of equilateral triangle d m.To find : 3QonF

Solution :

y

23

2323

y

13

1313

232

2132

1

0

33

231333

zy3

y2

1

3

2

1

321

21321

a0.866a0.5-r-r

r-ra

d

0.866ad0.5

r-r

r-ra

ad

Qa

d

Q

4

QF

FFFisFforceThea

0.866a

d0.5r

adr

0r

md)0.866d,0.5(0,P

m0)d,(0,P

m(0,0,0)P

norigintheat

withplane,YZinliePandP,PIf

andP,PatlieQandQ,QLet

+==

+==

+

=

+=

+=

=

=

=

=

=

Substituting,

)a0 . 9 2 4a( 0 . 3 8aw h e r eNa0 . 3 5 4F

1 3 . 1 11 2 . 1 25

a1 2 . 1 2a5

d

1 0x2 7

0 . 8a0 .5-(d

1 0x2)a0 . 8 6 6a0 .5(

d

1 0x1 21 0x9)1 0x( 3F

zyFF3

22

zy

2

3-

y2

- 6

zy2

- 696-

3

+==

+

+=

+++=

2. Data : At the point P, the potential is V)zy(xV222

p++=

To find :

(1)

VforexpressiongeneralusingbyV(3)(1,1,2)QandP(1,0.2)givenV(2)EPQPQp

Solution :

24


25/37

/mV]a3zay2ax2[-

az

Va

y

Va

x

V-V-E)1(

z

2

yx

z

p

y

p

x

p

pp

++=

+

+

==

]V1-V-VV(3)

V1-0y0

dz3zdy2ydx2xdl.E-V)2(

PQPQ

02

1

1

0

1

2

2

2

P

Q

pPQ

==

=++=

++==

3. Data : Q = 64.4 nC at A (-4, 2, -3) m A

To find : m(0,0,0)0atE 0E Solution : 0

20a29

9x64.4E

)AO(29

1

AO

AOa

2)-(0a4)(0AO

[

(AO)36

10x4

10x64.4

/Na(AO)4

QE

AO0

AO

x

29-

9-

AO2

0

0

=

==

+=

=

=

4. Q1 = 100 C at P1 (0.03 , 0.08 , - 0.02) mQ2 = 0.12 C at P2 (- 0.03 , 0.01 , 0.04) mF12 = Force on Q2 due to Q1 = ?

Solution :

Na9F

0.11

0.121x10x100F

0.636-a0.545-(a

0.06-(

(-0.03R-RR

aR4

QQF

1212

2

6-

12

x12

1212

122

120

2112

25


26/37

5. Q1 = 2 x 10-9 C , Q2 = - 0.5 x 10

-9 C C

(1) R12 = 4 x 10-2 m , ?F12 =

(2) Q1 & Q2 are brought in contact and separated by R12 = 4 x 10-2 m ?F`12 =

Solution :

(1)

(repulsiveN12.66F

x4(x36

10x4

)10x1.5(F

contactintobroughtWhen(2)

10x4(x36

10x4

10x0.5-x10x2F

`

12

9-

29-`

12

-9-

--9

12

=

=

=

6. Y

P3x x

P1 P2x x

0 X

Q1 = Q2 = Q3 = Q4 = 20 CQP = 200 C at P(0,0,3) m

P1 = (0, 0 , 0) m P2 = (4, 0, 0) m

P3 = (4, 4, 0) m P4 = (0, 4, 0) m

FP = ?

Solution :

R

Q

36104

QF

a3a4-R

a4-a4-R

a3a4-R

Ra3R

FFF

2

1p

1

9-

p

p

zy4p

yx3p

zx2p

1pz1p

2p1pp

=

=

=

=

=

=

26


27/37

6-

zy2

zyx2zx2z296- 12

)a

0 . 6a

0 . 8-(5

1

)a0 .a0 . 6-a( - 0 . 66 . 4

1

)a0 . 6a0 . 8-(5

1

a3

1

1 0x9 x1 0x0 02

++

++++

=

)a0 . 6a0 . 8-(

2 5

1 0 0

)a0 .a0 .-a( - 0 .4 0 . 9

1 0 0)a0 . 6a0 . 8( -

2 5

1 0 0a

9

1 0 0

1 0xx 1 0x 1 0x 9 x 1 00 0 x 1 02

zy

zyxzxz2-6-996-

++

++++

=

Na17.23N)a17a1.7-a1.7(-

(11.11a3.2)-(-1.5266.4

1a)526.12.3(36.0

pzyx

y2x

=+=

++=

7. Data : Q1 , Q2 & Q3 at the corners of equilateral triangle of side 1 m.

Q1 = - 1 C, Q2 = -2 C , Q3 = - 3 C

To find : E at the bisecting point between Q2 & Q3 .Solution :

27


28/37

Z

P1 Q1

Q2 P E1P Q3

YP2 E2P E3P P3

[

[

a410x9

1.3310x9

36

104

1E

a0.5-R

a0.5R

a0.866-R

R

4

1

EEE

3

3

9-P

y3P

y2P

z1P

10

2P1PP

=

=

=

=

+=

=

=

+=

Z

E1P EP ( EP ) = 37.9 k V / m

Y

E2P (E3P E2P) E3P

8. Data Pl = 25 n C /m on (-3, y, 4) line in free space and P : (2,15,3) m

To find : EPSolution :

Z l = 25 n C / m

A

R (2, 15, 3) m

P

Y

28

P1 : (0, 0.5, 0.866) m

P2 : (0, 0, 0) m

P3 : (0, 1, 0) m

P : (0, 0.5, 0) m


29/37

X

The line charge is parallel to Y axis. Therefore EPY = 0

m/Va88.23E

36

102

25a2E

0.167-a(0.834R

Ra

(3a(-3))-(2APR

RP

R

0

lP

xR

x

=

=

==

==

R

9. Data : P1 (2, 2, 0) m ; P2 (0, 1, 2) m ; P3 (1, 0, 2) m

Q2 = 10 C ; Q3 = - 10 CTo find : E1 , V1

Solution :

R

Q

4

1V

14.14]aa[10

(0.679

1010x9E

a2a2aR

a2-aa(2R

4

1EEE

21

2

0

1

yx

3

6-9

1

zyx31

zyx21

0

21211

= += = ++=+=

=+=

V3000VmV/14.14E11

==

29


30/37

10. Data : Q1 = 10 C at P1 (0, 1, 2) m ; Q2 = - 5 C at P2 (-1, 1, 3) mP3 (0, 2, 0) m

To find : (1) 0Efor0)0,(0,atQ(2)E 3x3 =Solution :

[[ a1.23-

16-a(8

1010x9E

0.3RRa

(R

Ra

a1)(0R

a1)-(2R

R

4

1E(1)

x

y

9

3

23

23

23

13

13

13

x23

y13

10

3

=

=

=

=

=

+=

=

=

zerobecannotE

a1.23-E

a

R

Q

a

R

Q

a

R

Q

10x9E(2)

3x

x3x

03203

23223

2

13213

19

3

=

++=

11. Data : Q2 = 121 x 10-9 C at P2 (-0.02, 0.01, 0.04) m

Q1 = 110 x 10-9 C at P1 (0.03, 0.08, 0.02) m

P3 (0, 2, 0) m

To find : F12Solution :

R]a[

10x7.8x36

104

10x110x10x121F

-a0.05-R;NaR4

QQF

1212

3-9-

9-9-

12

x12122

120

21

12

=

=

=

Na0.015F1212

=

30


31/37

12. Given V = (50 x2yz + 20y2) volt in free space

Find VP , m3)-2,(1,PataandE npP

Solution :

[

/Va6.562a150a600 a(-3)(2)100-E -azyx100-E Vx-V-E

(-3)(2)(1)50V

P

x

P

x

2

P

31


32/37

Additional Problems

A1. Find the electric field intensity E at P (0, -h, 0) due to an infinite line charge of density

l C / m along Z axis.

+Z

A dz

APR

z

dEPy P Y

dEPz h 0

d PE

Pa X

-

Solution :

Source : Line charge l C / m. Field point : P (0, -h, 0)

[

aR

h

R4

dz-dE

aR

h-

R4

dzdE

-ah-R

1

R

Ra

aR4

dQdE

2

0

lPy

y2

0

lP

yR

R2

0

P

=

=

==

=

Expressing all distances in terms of fixed distance h,

h = R Cos or R = h Sec ; z = h tan , dz = h sec2 d

0]Cos[h4

E

dSinh4

-Sech

tanhxSech4

dSech-dE

ah2

-2x

h4

-]Sin[

h4

-E

dCosh4

-Cosx

Sech4

dSech-dE

2/

2/-

0

lPz

0

l

22

0

2l

Pz

y

0

l

0

l2/

2/-

0

lPy

0

l

22

0

2

lPy

=

=

=

/

/

=

=

=+

=

=

=

mV /ah2

-E y

0

l

=

An alternate approach uses cylindrical co-ordinate system since this yields a more general

insight into the problem.

Z +

A dz

32


33/37

z R

P ( , / 2, 0)0 Y

P

/ 2

APX

-

4E dE(ii)

4E

4

dE(i)

dz,tanz

Taking

4dE(i)

4

dzdE

dQ

Rwhere

4

dQdE

intensityfieldThe

isdzdQ

P P

P

lP

P

0

lP

0

P

l

z z

m/Va2

E

0

lP

=

Thus, directioninradialisE

A2. Find the electric field intensity E at (0, -h, 0) due to a line charge of finite length along

Z axis between A (0, 0, z1) and B(0, 0, z2)

Z

B (0, 0, z2)

dz

P 2 A(0, 0, z1) 1

Y

X

Solution :

[

2-,

2

t-fromextendingislinetheIf

)Sin-(Sinh4

E

)Sin(-h4

dCosh4

-EdE

aR

z-a

R

h

R4

dzdE

12

21

0

lP

0

l

0

l

z

z

PP

z2

0

lP

2

1

2

1

==

+=

+=

==

=

y

33


34/37

mV /ah2

-E y

lP

0=

A3. Two wires AB and CD each 1 m length carry a total charge of 0.2 C and are disposedas shown. Given BC = 1 m, find BC.ofmidpointP,atE

P

A B . C

1 m

1 m

D

Solution :

(1)

1 = 1800 2 = 1800

A B P

1 m

[ ] [ ]{ } n a t( I n d e t e r0

0aC o s-C o sa)S i n-( S i n-

h4

E z12y12

0

lPA B

=+

=

az(2) Pay

C

1 1 = - tan-15.0

1= - 63.430

2 = 0

D

[ ]

[ ]

[ ] )a1989.75a(-3218a0.447)-(1a0.894-10x3.6E

a63.43)Cos-0(Cosa(-63.43))(Sin-

0.536

104

10x0.2

a)Cos-(Cosa)Sin-(Sin-h4

E

zyzy

3

P

zy9-

6-

z12y12

0

lP

CD

CD

+=+=

+=

+

=

Since EAB

is indeterminate, an alternate method is to be used as under :

Z

34


35/37

dEPz

d

dy

y B Y P dEPyA

L R

t4

E

dtt4

-dE

;t-y-dLLet

(L4

adE

;ay)-d(LR

aR4

dydE

d

1

L0

l

P

2

0

lP

0

yl

Py

R

2

0

lP

mV /dL

1-

d

1

4

E lP

+=

0

mV/a2400a0.67]-2[1800E

a1.5

1-

0.5

1

36

104

10x0.2E

yyP

y9-

-6

P

AB

AB

==

=

a0.381(-awhere

EEE

yP

PPP CDAB

A4. Develop an expression for E due to a charge uniformly distributed over an infinite

plane with a surface charge density of S C / m2.

Solution :

Let the plane be perpendicular to Z axis and we shall use Cylindrical Co-ordinates. The

source charge is an infinite plane charge with S C / m2 .

dEP Z

RAP =z P

0 Y

d

X A

)aza-(R1a

)aza-(R

OPOA-OPAOAP

zR

z

+=

+=

+=+=

35


36/37

The field intensityPdE due to dQ = S ds = S (d d) is along AP and given by

dd)aza-(R4

a

R4

dddE z3

0

R2

0

P

+

=

= SS

Since radial components cancel because of symmetry, only z components exist

dR

z2x

4

R

dzd

4

dEE

ddR4

zdE

0

3

00

3

2

00S

PP

3

0

P

=

==

=

SS

S

z is fixed height of above plane and let APO = be integration variable. All distances

are expressed in terms of z and

= z tan , d = z Sec2 d ; R = z Sec ; = 0, = 0 ; = , = / 2

(na2

[-2

dSin

2

dSecz

Secz

tanzz

2

E

z

0

S

0

S

2/

00

S2

0

33

0

SP

=

=

=

=

A5. Find the force on a point charge of 50 C at P (0, 0, 5) m due to a charge of 500 Cthat is uniformly distributed over the circular disc of radius 5 m.

Z

P

h =5 m

0 Y

X

Solution :

Given : = 5 m, h = 5 m and Q = 500 CTo find : fp & qp = 50 C

36


37/37

Na56.55f

x50xa10x1131f

EwhereqxEf

zP

z

3

P

PPPP

=

=

=

Documents

Field Theory Compiled Basvaraj