Field Theory

Field Theory 10EE44

Department of EEE, SJBIT Page 1

Field theory(10EE44)

Subject Code : 10EE44 IA Marks : 25

No. of Lecture Hrs./ Week : 04 Exam Hours : 03

Total No. of Lecture Hrs. : 52 Exam Marks : 100

Part-A

Unit-1:a. Coulombs Law and electric field intensity: Experimental law of Coulomb, Electric

field intensity, Field due to continuous volume charge distribution, Field of a line charge

b. Electric flux density, Gauss law and divergence: Electric flux density, Gauss law,

Divergence, Maxwells First equation(Electrostatics), vector operator and divergence theorem

Unit-2: a. Energy and potential : Energy expended in moving a point charge in an electric

field, The line integral, Definition of potential difference and Potential, The potential field of a

point charge and system of charges, Potential gradient, Energy density in an electrostatic field

b. Conductors, dielectrics and capacitance: Current and current density, Continuity of current,

metallic conductors, Conductor properties and boundary conditions, boundary conditions for

perfect Dielectrics, capacitance and examples.

Unit-3: Poissons and Laplaces equations:Derivations of Poissons and Laplaces Equations,

Uniqueness theorem, Examples of the solutions of Laplaces and Poissons equations

Unit-4: the steady magnetic field: Biot-Savarts law, Ampere circuital law, stokes theorem, magnetic flux and flux density , scalar and vector magnetic potential

Part-B

Unit-5: Magnetic forces: forces on moving charges, differential current element, force between

differential current element, force and torque on closed circuit.

Magnetic material and inductance: magnetization and permeability, magnetic boundary

condition, magnetic circuit, inductance and mutual inductance

Unit-6: Time varing field and maxwells equation: Faradays law, displacement current, maxwells equation in point and integral form, retarded potentials.

Unit-7: uniform plane wave: wavce propagation in free space and dielectrics, Poyenting

theorem and wave power, propagationin good conductors, skin effect.

Unit-8: Plane waves at boundaries and in dispersive media: reflection of uniform plane wave

at normal incidence, SWR, Plane wave propagation in general direction

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Field Theory 10EE44


Table of contents

Sl.no Contents Page no

1 Vectors 4 to 18

2

Unit-1:a. Coulombs Law and electric field

intensity: Experimental law of Coulomb

19 to 38

Electric field intensity

Field due to continuous volume charge

distribution

Field of a line charge

b. Electric flux density, Gauss law and

divergence: Electric flux density,

Gauss law

Divergence

Maxwells First equation(Electrostatics)

vector operator and divergence theorem

3

Unit-2: a. Energy and potential : Energy expended

in moving a point charge in an electric field,

39 to 61

The line integral,

Definition of potential difference and

Potential

The potential field of a point charge and

system of charges

Potential gradient

Energy density in an electrostatic field

b. Conductors, dielectrics and capacitance:

Current and current density,

Continuity of current

metallic conductor

Conductor properties and boundary

conditions,

boundary conditions for perfect Dielectrics,

capacitance and examples.

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Field Theory 10EE44


4

Unit-3: Poissons and Laplaces

equations:Derivations of Poissons and Laplaces

Equations,

62 to 71 Uniqueness theorem

Examples of the solutions of Laplaces and

Poissons equations

5

Unit-4: the steady magnetic field:

Biot-Savarts law

72 to 89 Ampere circuital law

stokes theorem

magnetic flux and flux density

scalar and vector magnetic potential

6

Unit-5: Magnetic forces: forces on moving charges

90 to 101

differential current element

force between differential current element

force and torque on closed circuit.

Magnetic material and inductance: magnetization

and permeability

magnetic boundary condition

magnetic circuit

inductance and mutual inductance

7

Unit-6: Time varing field and maxwells equation:

102 to 113

Faradays law

displacement current

maxwells equation in point and integral form

retarded potentials.

8

Unit-7: uniform plane wave: wave propagation in

free space and dielectrics

118 to 171 Poyenting theorem and wave power

propagationin good conductors

skin effect.

9

Unit-8: Plane waves at boundaries and in

dispersive media: reflection of uniform plane wave

168 to 189 at normal incidence

SWR

Plane wave propagation in general direction

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Field Theory 10EE44


Introduction to vectors

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 and

strength 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

law eAssociativ C )B A( )C B ( A

law eCommutativ A B B A

(2) Subtraction

)B (- A B - A

(3) Multiplication by a scalar

law veDistributi B m A m )B A( m

law veDistributi An A m A n) (m

law eAssociativ )A (mn )A(n m

law eCommutativ m A A m

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Field Theory 10EE44


A vector is represented graphically by a directed line segment.

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

A a Aor A / A a Also

actor Unit ve AVector

A

AA

A

Product of two or more vectors :

(1) Dot Product ( . )

0 , B } COS A { OR COS B ( A B . A

B

B

Cos A

A

Cos B

A

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

(2) CROSS PRODUCT (X)

C = A x B = n SIN B A

C x A B x A )C B ( x A

A x B - B x A

vectorsof system handedright a form C B Asuch that directed

B and A of plane lar toperpendicur unit vecto is n and

) 0 ( B and Abetween angle is ' ' where

Ex.,

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Field Theory 10EE44


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 , az

2. Cylindrical , , z a , a , az

3. 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 , zy x i.e.,

RECTANGULAR CO-ORDINATE SYSTEM :

Z

x=0 plane

az p

y=0 Y

plane ay

ax z=0 plane

X

yxz

xzy

zyx

xzzyyx

a a x a

a a x a

a a x a

0 a . a a . a a . a

az is in direction of advance of a right circular screw as it is turned from ax 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

a r a z ay a x r rzyx

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Field Theory 10EE44


Equation of Vector AB :

Zand Y X, along B of components are B & B , B and

Zand Y X, alongA of components are A & A , A where

A - B ABor B AB A

thena B a B a B B OB and

a A a A a A A OA If

zys

zys

zzyyxx

zzyyxx

Dot and Cross Products :

get wegrouping, and by term termproducts' Cross' Taking

)a B a B a (B x )a A a A a (A B x A

C A B A B A ) a B a B a (B . )a A a A a (A B . A

zzyyxxzzyyxx

zzyyxxzzyyxxzzyyxx

zyx

zyx

zyx

BBB

AAA

aaa

B x A

)AB . (AB AB ABlength Vector

where

AB

AB a

AB alongVector Unit

C and B , A sides of oidparallelop a of volume therepresents )C x B ( . A (ii)

parallel are B andA 0 0 Sin then 0 B x A

larperpendicu are B andA 90 i.e., 0 Cos then 0 B . A (i)

vectors,zeronon are C and B ,A If

CCC

BBB

AAA

) C x B( . A

AB

0

zyx

zyx

zyx

B

B

AB

0 A

A

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Field Theory 10EE44


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

zyx

zyx

a dz ady adx rd

dz z

r dy

r dx

x

r rd

a z ay a x r

y

Differential length 1 ----- ]dz dy dx [ rd 1/2222

Differential surface element, sd

1. zadxdy : z tor

2. zadxdy : z tor ------ 2

3. zadxdy : z tor

Differential Volume element

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

z

dx p p dz

dy

r

r d r

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

y

z z

y / x tan

y x

1-

22

x

0 ww

w.city

stude

ntsgro

up.co

m

Field Theory 10EE44


1 z

r h a

r

r

h ; a a

r a Cos a Sin -

r

1

r h ; a h a

r a Sin a Cos

r

1 ------ dz z

r d

r d

r rd

a z a Sin a Cos r

a z ay a x r

zz

y x

y x

zyx

zyx

z

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

22 22

z

zz z

y

xyx

(dz) )d ( d rd and

2 ------ a dz a d a d rd ,Further

orthogonal are a and a , a ; a a

a Cos a Sin a a Cos a Sin - a

a Sin a Cos a a Sin a Cos a

yx

Differential areas :

zz

a dz) (d a ds

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

a . )d ( ) (d a ds

Differential volume :

4 ----- dz d d dor

(dz) )d ( ) (d d

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

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Field Theory 10EE44


d ddr Sin r vd

ddr r S d

ddr Sin r S d

d d Sin r S d

a d Sin r a dr adr Rd

d R

d R

dr r

R Rd

a Cos a Sin - R

/R

a

a Sin a Sin Cos a Cos Cos R

/R

a

a Cos a Sin Sin a Cos Sin r

R /

r

Ra

a Cosr a Sin Sin r a Cos Sin r R

2

2

2

r

r

yx

zyx

zyxr

zyx

General Orthogonal Curvilinear Co-ordinates :-

z u1 a3 u3

a1 u2 a2 y

x

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

Here

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



3

3

2

2

1

1

321

333222111

3

3

2

2

1

1

321zyx

133221

321321

u

R h ,

u

R h ,

u

R h

; factors scale are h , h , h where

a du h a du h a du h

du u

R du

u

R du

u

R R dthen

u & u , u of functions are z y, x,& a z ay a x R If

0 a . a & 0 a . a , 0 a . a if Orthogonal is System

u & u , u tol tangentiaorsubnit vect are a , a , a

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Field Theory 10EE44


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

) u , u , (u A A and ) u , u , (u A A

) u , u , (u A A wherefieldVector a is a A a A a A A &

fieldScalar a )u , u , u ( V V where

A h A h A h

u u u

a h a h a h

h h h

1 A x

) A h (h u

) A h (h u

) A h (h u

h h h

1 A .

a u

v

h

1 a

u

v

h

1 a

u

v

h

1 V

3213332122

32111332211

3 2 1

332211

32 1

33221 1

321

321

3

231

2

132

1321

3

33

2

2 2

1

11



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Field Theory 10EE44


Vector Transformation from Rectangular to Spherical :

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 a 2 - a 4 - a PQ

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

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) m Z

PQ P (1,2,3)

Q(2,-2,-1)

0

Y

X

)a 2 - a 4 - a(

a 3)- (-1 a 2)- (-2 a 1) - (2

a )z - (z a )y - (y a ) x- (x PQ vector The

zyx

zyx

zpqypqx pq

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

G

G

0

z

y

x

zyx

zyx

rzryrxr

zyxr

rr

RrrRS

zzyyxxR

A

A

A

a . a a . a a . a

a . a a . a a . a

a . a a . a a . a

A

A

A

as A , A , A torelated are A , A , A where

a A a A a A

a ) a . A( a )a A( a ) a A ( A : Spherical

a A a A a A A :r Rectangula

R



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Field Theory 10EE44


)a 0.333 - a 0.667 - a (0.667 a

3 (-1) (-2) 2 G

G

G a ,or unit vect The

)a - a 2- a (2

a 0) - (z a )0 - (y a )0 - (x G vector The

zyxg

222

g

zyx

zgygxg

3. Given

zyx

zyx

a5 a 2 - a 4 - B

a a 3 - a 2 A

B x A (2) and B . A (1) find

Solution :

)a 5 a 2 - a (-4 . )a a 3 - a (2 B . 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

B x A

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 between two points, the co-ordinates are to be in Cartesian (rectangular). The corresponding

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

2.64 2 1.73 AB)(

a 2 a 1.73 -

a 0) - (2 a 1)- (1 a 1.73-

a )A - (B a )A - (B a ) A - (B AB

a 2 a a 2 a 2

Sin a 6

Cos B &

a a 1.73 a 6

Sin 2 a 6

Cos 2 A

22

zx

zyx

zzzyyyxxx

zyzyx

yxyx

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,



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Field Theory 10EE44


1.732 0.5) 0.5 (2

) AB . AB ( AB

a (-0.707) a 0.707) (- a 1.414 -

a )A - (B a )A - (B a )A - (B AB

) a 0.707 a 0.707 (

)a 4

3 Cos a Sin

4

3Sin a Cos

4

3Sin ( B

) a 0.707 a 0.707 (

) a 4

Cos 1 a 0Sin 4

Sin 1 a 0 Cos 4

Sin (1 A

1/2

1/2

zyx

zzzyyyxxx

yx

zyx

yx

zyx

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

AB

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

1.732

1

= ) a 408.0 a 0.408 - a 0.816 - ( zyx

7. Transform ordinates.-Co lCylindricain F into )a 6 a 8 - a (10 F zyx

Soln :

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

01-

22

z

xxzyx

yxzyx

yxzyx

38.66 - x

y tan Sin y

12.81 y x Cos x

a 6 a ) Cos 8 - Sin (-10 a ) Sin 8 - Cos (10

a )]a( . )a 6 a 8 - a (10 [

a )]a Cos a Sin (- . )a 6 a 8 - a 10( [

a )]a Sin a (Cos . )a 6 a 8 - a [(10

)a 6 a (12.8

a 6 a 38.66)] (- Cos 8 - 38.66) (-Sin 10 [- a ] 38.66) - (Sin 8 - 38.66) (- Cos 10 [ F

z

zpCyl

8. Transform a z a x - ay B zyx

into Cylindrical Co-ordinates.

x

y



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Field Theory 10EE44


z

z

22

zyxzyx

yxzyx

zz Cyl

zyx

a z a -

a z a ] Cos - Sin - [ a ]Sin Sin - Cos Sin [

a z a )]a Cos a Sin (- ). a z a Cos - a Sin ( [

a )]a Sin a (Cos ). a z a Cos - a Sin ( [

a )a . (B a )a (B. a )a (B. B

a z a Cos - a Sin B

Sin y , Cos x

9. Transform xa 5 into Spherical Co-ordinates.

a Sin 5 a Sin Cos 5 a Cos Sin 5

a )]a Cos a Sin (- . a 5 [

a ]a Sin - a Sin Cos a Cos (Cos . a 5 [

a )]a Cos a Sin Sin a Cos (Sin . a 5 [

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

r

yxx

zyxx

rzyxx

rrSph

10. Transform to Cylindrical Co-ordinates z) , ,( Qat a 4x) -(y - a y) x (2 G yx

Soln :

Sin y , Cos x

a ] Cos )4x -(y - Sin y) x (2 - [

a ]Sin 4x) -(y - Cos y) 2x ( [

0 a ] a Cos a Sin - [ . ]a 4x) -(y - a y) x (2 [

a ] a Sin a Cos [ . ]a 4x) -(y - a y) x (2 [ G

a ) a (G. a ) a (G. a ) a (G. G

yxyx

xyxCyl

Cyl

zz

a ) Cos Sin 3 - Sin Cos 4 (

a ) Sin - Cos Sin 5 Cos 2 (

a ] Cos 4 Cos Sin - Sin - Cos Sin 2 - [

a ] Cos Sin 4 Sin - Cos Sin Cos 2 [

a ] Cos ) Cos 4 - Sin ( - Sin ) Sin Cos 2 ( - [

a ]Sin ) Cos 4 - Sin ( - Cos ) Sin Cos 2 ( [ 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



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Field Theory 10EE44


)a 0.72 a 0.24 - a 0.65 (- AB

BA a

14.77 10.6 3.53 9.65 AB

a 10.6 a 3.53 - a 9.65- A - B AB

a 3.53 a 3.53 - B a 7.07 - a 3.53 a 6.12 A

a 4

Cos 5 a Sin 4

Sin 5 a Cos 4

Sin 5 B

a 4

3 Cos 10 a

6Sin

4

3Sin 10 a

6 Cos

4

3Sin 10 A

a Cosr a Sin Sin r a Cos Sin r OA

zyxAB

222

zyx

zxzyx

zyx

zyx

zyx

12. Transform a 6 a 8 - a 10 F zyx

into F

in Spherical Co-orindates.

)a 0.783 a 5.38 a (11.529 F

a 0.781) x 8 - 0.625 - x (-10

a (0.625)) x 0.42 x 8 - 0.781 x 0.42 x 10 (

a (-0.625)) x 0.9 x 8 - 0.781 x 0.9 x (10 F

0.781 (-38.66) Cos Cos 0.42 64.69 Cos Cos

0.625- (-38.66)Sin Sin 0.9 64.69Sin Sin

38.66- 10

8- tan

64.89 200

6 Cos

r

z Cos ; 200 6 8 10 r

a ) Cos 8 - Sin 10 (-

a ) Sin 6 - Sin Cos 8 - Cos Cos (10

a ) Cos 6 Sin Sin 8 - Cos Sin (10

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

a Cos a Sin - a

a Sin - a Sin Cos a Cos Cos a

a Cos a Sin Sin a Cos Sin a

r

r

01-

01-1-222

r

rrSph

yx

zyx

zyxr

Line Integrals

In general orthogonal Curvilinear Co-ordinate system

C

333

C

222

C C

111

332211

333222111

du F h du F h du F h dl . F

a F a F a F F

a du h a du h a du h dl



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Field Theory 10EE44


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

contour. closed a around taken isit if zero is and a and bbetween potential therepresent dl .

andintensity field electric therepresent - E

then flux, ticelectrosta is If ).!path on the dependnot (does (a) - b)( d dl .

b

a

b

a

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

EXAMPLES :

13. Evaluate line integral dl . a I

where yx a x)-(y a y) (x a

(4,2) B to(1,1)A from x y along

2

Soln : ady adx dl yx

2

1

2

1

22

2

dy )y -(y dy 2y y) (y dl . a

dx dy dy 2or x y

dy ) x -(y dx y) (x dl . a

2

1

223 dy ) y -y y 2 y (2

2

1

23 dy y) y y (2

3

1 11

3

1 1 -

3

2 12

3

4 - 2

3

8 8

2

1

3

1

2

1 -

2

2

3

2

2

2

2

y

3

y

2 4

y 2

234

2

1

234

14. Evaluate the Integral S

ds . E I

where a radius of hunisphere is S and a x E x

Soln:

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



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Field Theory 10EE44


3

a 2 x

3

2 x a d Cos d Sin a ds . E

2 0 , 2 / 0

d d Cos Sin a ds . E

d d Sin a . a ) Cos Sin ( a ds . E ds . E

Cos Sin a x ; a Cos Sin x E

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

a d d Sin a ds

a d )Sin (a )d (a ds

; 0 z , a z y x

3/2

0

2

0

3233

23

2

r

2

r

rr

rr

r

2

r

2222

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.



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Field Theory 10EE44


Unit-1

a. Coulombs Law and electric field intensity: Experimental law of Coulomb

Electric field intensity

Field due to continuous volume charge distribution

Field of a line charge

b. Electric flux density, Gauss law and divergence:

Electric flux density

Gauss law

Divergence, Maxwells First equation(Electrostatics)

vector operator and divergence theorem



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Field Theory 10EE44


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 is coulomb (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

2 surface tonormalor unit vect is n whereC ds n D or c/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 / N a r 4

q

q

F E 12

10

1

C E Dor c / N D

c / N a r 4

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

C a r 4

Q Dor C E D and

m) / Vor ( c / N a r 4

Q E

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

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.



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Field Theory 10EE44


Q1 1r

nE

P1

Q2 2r

P 2E

1E

P2 1r

Qn 0

Pr nr

m / V a r -r

Q

4

1 E 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

Q d and dv Q V

V

V

C / N a )r -(r 4

)(r E

a R 4

V a

R 4

Q E

R21

0

1

r

R2

0

R2

0

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

density of l C / m Ra

P

R

dl l C / m L

C / N a R

dl

4

1 E R

L

2

0

p l



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Field Theory 10EE44


4. E

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

Ra

P (Field point)

ds R

(Source charge)

C / N a R

ds

4

1 E 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

(2) --------- dl . V dV

dz V

dy y

V dx

x

V dV Also,

(1) -------- dl . E - dl q

f - dV

z

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

la

dR R

0

+ Q R

P Ra

At point P, C / N a R 4

Q E R2

0

Therefore, the force f

on a unit charge at P.



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Field Theory 10EE44


N a R 4

Q E x 1 f R2

0

p

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

field)scalar (a Volt R 4

Q V

dR R 4

Q - )a . a(

R 4

dl Q - V

dl . E - dl . f - dV

2

0

P

2

0

lR

R

2

0

p

R

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

voltR

1 -

R

1

4

Q )V - (V V

qp0

qppq

Electric Potential at a point due to different charge configurations.

1. Discrete charges . Q1 . P

Q2 Rm

Qm

V R

Q

4

1 V

n

1 m

m

0

1P

2. Line charge

x P V dl R

l

4

1 V

l

l

0

2P

l C / m

3. Surface charge

x P V R

ds

4

1 V

S

S

0

3P

s C / m2

4. Volume charge

x P

R V R

dv

4

1 V

V

V

0

4P

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

v C/ m3



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Field Theory 10EE44


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

potential surface ialequipotent twopotential of difference is V

volt R 4

Q V

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) C Q ds n . D

where S is called the Gaussian Surface.

By Divergence Theorem,

S V

dv D . ds n . D

----------- (2)

Also, V

V dv Q ---------- (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

In equation 4, E D 0

equation Laplace 0 V 0, If

equationPoisson - V

/ V) (- . or / E .

2

0

2

00

0

+Q



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Field Theory 10EE44


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

2

0

Volt dl E V

MV/ R

Q

4

1 E

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

l

S

enc

voltdl E V Also

determined are E hence and DThen

Q ds n D

(3) Laplace equation : The Laplace equation 0 V 2 is solved subjecting to different

boundary conditions to get V. Then, V - E

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 : 3Qon F

Solution :



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Field Theory 10EE44


zy

23

2323

zy

zy

13

1313

232

2132

1

0

33

231333

zy3

21y2

1

3

2

31

1321

321321

a 0.866 a 0.5 - r - r

r - r a

a 0.866 a 0.5 d

a d 0.866 a d 0.5

r - r

r - r a

a d

Q a

d

Q

4

Q F

X F F F is F force The

a 0.866 a d 0.5 r

P P a d r

Y d 0 r

m d) 0.866 d, 0.5 (0, P

d d m 0) d, (0, P

P m (0,0,0) P

Z n origin theat

P with plane, YZin lie P and P , P If

meter. d side of trianglelequilatera of corners theP and P , Pat lie Q and Q , QLet

Substituting,

) a 0.924 a (0.38 a whereN a 0.354 F

13.11 12.12 5

a 12.12 a 5

d

10 x 27

) a 0.866 a 0.5 - ( d

10 x 2 )a 0.866 a 0.5 (

d

10 x 12 10 x 9 ) 10 x (3 F

zyFF3

22

zy

2

3-

zy2

-6

zy2

-696-

3

2. Data : At the point P, the potential is V )z y (x V 222p

To find :

(1) Vfor expression general usingby V (3) (1,1,2) Q and P(1,0.2)given V (2) E PQPQp

Solution :

/mV ] a 3z ay 2 a x 2 [ -

a z

V a

y

V a

x

V - V - E )1(

z

2

yx

z

p

y

p

x

p

pp

V 1- V - V V (3)

V 1- 0 y 0

dz 3z dy 2y dx 2x dl . E - V )2(

PQPQ

02

1

1

0

1

2

2

2

P

Q

pPQ



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Field Theory 10EE44


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

To find : m (0,0,0) 0at E

0E

Solution : 0

C / N a 20 a 29

9 x 64.4 E

)a 0.56 a 0.37 - a (0.743 )AO( 29

1

AO

AO a

a 3 a 2 - a4 a 3) (0 a 2) - (0 a 4) (0 AO

CN/ ]a [

(AO) 36

10 x 4

10 x 64.4

C / N a (AO) 4

Q E

AOAO0

zyxAO

zyxzyx

AO

29-

9-

AO2

0

0

4. Q1 = 100 C at P1 (0.03 , 0.08 , - 0.02) m

Q2 = 0.12 C at P2 (- 0.03 , 0.01 , 0.04) m F12 = Force on Q2 due to Q1 = ?

Solution :

N a 9 F

a 10 x 9 x 0.11

10 x 0.121 x 10 x 100 F

)a 0.545 a 0.636 - a 0.545- ( a

m 0.11 R ; )a 0.06 a 0.07 - a 0.06 - (

)a0.02 -a 0.08 a (0.03 - )a 0.04 a 0.01 a (-0.03 R -R R

a R 4

Q Q F

1212

12

9

2

6-6-

12

zyx12

12zyx

zyxzyx1212

122

120

2112

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 :



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Field Theory 10EE44


(1)

)(repulsive N 12.66 F

a N 12.66 a 10 x 9 x 16

1.5 a

)10 x 4 ( x 36

10 x 4

)10 x 1.5 ( F

C 10 x 1.5 )Q (Q 2

1 Q Qcontact intobrought When (2)

e)(attractiv N 5.63 a 10 x 16

9- a

)10 x 4 ( x 36

10 x 4

10 x 0.5 - x 10 x 2 F

`

12

1212

13 18-2

12

22-9-

29-`

12

9-

21

`

2

`

1

12

5-

12

22-9-

-9-9

12

6. Y P3

x x

P1 P2

x x

0 X

Q1 = Q2 = Q3 = Q4 = 20 C

QP = 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 :

a R

Q a

R

Q a

R

Q a

R

Q

36

10 4

Q F

a 0.6 a 0.8 - a ; m 5 R ; a 3 a 4 - R

a 0.47 a 0.625 - a 0.625 - a ; m 6.4 R ; a 3 a 4 - a 4- R

a 0.6 a 0.8 - a m 5 R ; a 3 a 4- R

a a m 3 R a 3 R

F F F F F

4p2

4p

43p3

3p

32p2

2p

21p2

1p

1

9-

p

p

zy 4p4pzy 4p

zyx 3p3pzyx3p

zx 2p2pzx2p

z1p1pz1p

4p3p2p1pp

6-

zy2

zyx2zx2z296- 10 x 20

)a 0.6 a 0.8- ( 5

1

)a 0.47 a 0.625 - a (-0.625 6.4

1 )a 0.6 a 0.8 - (

5

1 a

3

1

10 x 9 x 10 x 002



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Field Theory 10EE44


)a 0.6 a 0.8- ( 25

100

)a 0.47 a 0.625 - a (-0.625 40.96

100)a 0.6 a 0.8 (-

25

100a

9

100

10x x10x10x9x1000x102

zy

zyxzxz2- 6-996-

N a 17.23 N ) a 17a 1.7 - a 1.7 (-

)a 2.4)1.152.4 (11.11 a 3.2) - (-1.526 6.4

1 a )526.12.3( 36.0

pzyx

zy2x

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

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

To find : E

at the bisecting point between Q2 & Q3 .

Solution :

Z

P1 Q1

Q 2 P E1P Q3

Y

P2 E2P E3P P3

V/mk 18 37.9 m / V 0 a 12 a 36 a 1.33 a 4 10 x 9

a 12 a 8 - a 1.33 10 x 9

)a - ( 0.5

10 x 3 - )a - (

0.5

10 x 2 - )a - (

0.866

10 x 1 -

36

10 4

1 E

a - a 0.5 R a 0.5 - R

a a 0.5 R a 0.5 R

a - a 0.866 R a 0.866 - R

a R

Q a

R

Q a

R

Q

4

1

E E E E

03

zyzy

3

yyz

3

y2

6-

y2

6-

z2

6-

9-P

y3P 3Py3P

y2P 2Py2P

z1P 1Pz1P

3P2

3P

32P2

2P

21P2

1P

1

0

3P2P1PP

Z

E1P EP ( EP ) = 37.9 k V / m

Y

E2P (E3P E2P) E3P

P1 : (0, 0.5, 0.866) m

P2 : (0, 0, 0) m

P3 : (0, 1, 0) m

P : (0, 0.5, 0) m



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Field Theory 10EE44


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

Solution :

Z l = 25 n C / m

A

R

(2, 15, 3) m P

Y

X

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

m / Va 88.23 E

a

5.1 x 36

10 2

x 25 a

2 E

) a 0.167 - a (0.834 R

R a

m 5.1 R ; ) a - a (5 a 4) - (3 a (-3)) - (2 AP R

RP

R9-R

0

lP

zxR

zxzx

R

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

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

Solution :

V 3000 3

10

3

10 10 x 9

R

Q

R

Q

4

1 V

m / V )a 0.707 a (0.707 14.14 ]a a [ 10

) a 0.67 a 0.67 a (0.33 9

10 )a 0.67 - a 0.33 a (0.67

9

10 10 x 9 E

a 0.67 a 0.67 a 0.33 a 3 R a 2 a 2 a R

a 0.67 - a 0.33 a 0.67 a 3 R )a 2 - a a (2 R

a R

Q a

R

Q

4

1 E E E

6-6-9

31

3

21

2

0

1

yxyx

3

zyx

6-

zyx

6-9

1

zyx31 31zyx31

zyx21 21zyx21

312

31

3212

21

2

0

21211

V 3000 V m V / 14.14 E 11

10. Data : Q1 = 10 C at P1 (0, 1, 2) m ; Q2 = - 5 C at P2 (-1, 1, 3) m P3 (0, 2, 0) m

To find : (1) 0 Efor 0) 0, (0,at Q (2) E 3x3

Solution :



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Field Theory 10EE44


m / V 10 a 12.32 - a 6.77 a 1.23 -

)a 3.68 a 1.23 - a (-1.23 )a 16 - a (8

)a 0.9 - a 0.3 a (0.3 )11(

10 x 5- )a 0.894 - a (0.447

)5(

10 x 10 10 x 9 E

a 0.9 - a 0.3 a 0.3 R

R a

) a 0.894 - a 0.447 ( R

R a

11 R a 3 -a a a 3) - (0 a 1) - (2 a 1) (0 R

5 R a 2 - a a 2) - (0 a 1) - (2 R

a R

Q a

R

Q

4

1 E (1)

3

zyx

zyxzy

zyx2

6-

zy2

6-9

3

zyx

23

2323

zy

13

1313

23zyxzyx23

13zyzy13

232

23

2132

13

1

0

3

zero becannot E

a 1.23 - E

a 2 R ; a R

Q a

R

Q a

R

Q 10 x 9 E (2)

3x

x3x

y03032

03

232

23

2132

13

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 : F12

Solution :

0.088 R ]a[

10 x 7.8 x 36

10 4

10 x 110 x 10 x 121 F

a 0.02 a 0.07 - a 0.05- R ; N a R 4

Q Q F

1212

3-9-

9-9-

12

zyx12122

120

2112

N a 0.015 F 1212

12. Given V = (50 x2yz + 20y

2) volt in free space

Find VP , m 3)- 2, (1, Pat a and E npP

Solution :



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Field Theory 10EE44


a 0.16 - a 0.234 a 0.957 a ; m / V a 6.5 62

a 100 - a 150a 600

a (2) 50 - a (-3) 50 - a (-3) (2) 100 - E

ay x50 - a z x50 - a zy x 100 - E

a V

- a V

- a V x

- V - E

V 220- (2) 20 (-3) (2) (1) 50 V

zyxPP

zyx

zyxP

z

2

y

2

x

zyx

22

P

zy

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)

a R

z

R 4

dz - dE a

R

h

R 4

dz - dE

a E d a E d a R

z - a

R

h -

R 4

dz dE

a z - ah - R

1

R

R a

h z AP R

ah - a z - AP R ; m / V a R 4

dz a

R 4

dQ dE

z2

0

lPzy2

0

lPy

z PzyPyzy2

0

l P

zyR

22

yzR2

0

lR2

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



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Field Theory 10EE44


0 ]Cos [ h 4

E

d Sin h 4

-

Sec h

tan h x

Sec h 4

d Sech - dE

a h 2

- 2 x

h 4

- ]Sin [

h 4

- E

d Cos h 4

- Cos x

Sec h 4

d Sech - dE

2 /

2 / -

0

lPz

0

l

22

0

2

lPz

y

0

l

0

l2 /

2 / -

0

lPy

0

l

22

0

2

lPy

m V / a h 2

- 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

z R

P ( , / 2, 0) 0 Y

P

/ 2

AP X

-



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Field Theory 10EE44


0 ]Cos [ 4

E

d )Sin (- 4

d

Sec 4

Sec x tan x dE (ii)

2

2 x

4

]Sin [

4

E

d Cos 4

d

Sec 4

Sec x x dE (i)

Sec Cos

R and d Sec dz , tan z

and of in terms distances all expressing and ,n variableintegratio asA PO Taking

dz z R 4

- dE (ii) ; dz

R 4

dE (i)

a dE a dE a R

z - a

R

R 4

dz dE

C dz dQ

)a z - a ( R

1 a and a z - a R where

m / V a R 4

dQ dE

is dQ todue dEintensity field The

at Z. change elemental theis dz dQ

2 /

2 / -

0

lP

0

l

33

0

2

lP

0

l

0

l2 /

2 / -

0

lP

0

l

33

0

2

lP

2

2

0

lP2

0

lP

PPz2

0

lP

zRz

R2

0

P

P

l

z

z

z

zz

l

m / V a 2

E

0

lP

Thus, directionin radial is E

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 :



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Field Theory 10EE44


2 - ,

2

, to - from extending is line theIf

m / V a ) Cos - (Cos a ) Sin - (Sin h 4

E

a )Cos ( h 4

a )Sin (-

h 4

a d Sin h 4

- a d Cos

h 4

- E d E

a R

z - a

R

h

R 4

dz dE

12

z21y21

0

lP

z

0

ly

0

l

z

0

ly

0

l

z

z

PP

z2

0

lP

2

1

2

1

2

1

2

1

2

1

y

m V / a h 2

- E y

lP

0

A3. Two wires AB and CD each 1 m length carry a total charge of 0.2 C and are disposed

as shown. Given BC = 1 m, find BC. ofmidpoint P,at E

P

A B . C

1 m

1 m

D

Solution :

(1)

1 = 1800 2 = 180

0

A B P

1 m

nate)(Indetermi 0

0 a Cos - Cos a ) Sin - (Sin -

h 4

E z12y12

0

lPAB

az (2) Pay C

1 1 = - tan-1

5.0

1 = - 63.43

0

2 = 0

D



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Field Theory 10EE44


)a 1989.75 a (-3218 a 0.447) - (1 a 0.894 - 10 x 3.6 E

a 63.43) Cos - 0 (Cos a (-63.43))(Sin -

0.5 36

10 4

10 x 0.2

a ) Cos - (Cos a ) Sin - (Sin - h 4

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

dEPz

d

dy

y B Y P dEPy A

L R

d L

1 -

d

1

4

t 4

E

dt t 4

- dE

d

1 t ; L y

d L

1 t , 0 y ;dt - dy - ; t - y - d LLet

dy y) - d (L 4

a dE

)a(- R

1 a ; a y) - d (L R

m / V a R 4

dy dE

0

ld

1

d L

10

lP

2

0

lP

2

0

yl

Py

yRR

R2

0

lP

m V / d L

1 -

d

1

4

E lP

0

mV/ a 2400 a 0.67] - 2 [ 1800 E

a 1.5

1 -

0.5

1

36

10 4

10 x 0.2 E

yyP

y9-

-6

P

AB

AB



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Field Theory 10EE44


)a 0.925 a 0.381 (- a where

m / V a 2152

) a 1990 a (-820

a 1990 a 3218 - a 2400 E E E

zyP

P

zy

zyyPPP 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

R AP

z P

0 Y

d

X A

)a z a - ( R

1 a

)a z a - ( R

OP OA - OP AO AP

zR

z

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

d d )a z a - ( R 4

a

R 4

d d dE z3

0

R2

0

P

SS

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

d R

z 2 x

4

R

d z d

4

dE E

d d R 4

z dE

0

3

00

3

2

00S

PP

3

0

P

SS

S

z is fixed height of above plane and let A PO be integration variable. All distances are expressed in terms of z and

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



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Field Theory 10EE44


plane) to(normal a 2

a ] Cos [- 2

d Sin

2

d Sec z

Sec z

tan z z

2

E

z

0

S

z

2 /

0

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 C that 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 C

To find : fp & qp = 50 C

N a 56.55 f

10 x 50 x a 10 x 1131 f

C / N a 10 x 1131

a 10 x 36 x 25 x 2

500

a

36

10 x )5 ( 2

10 x 500 a

2

A

Q

a 2

E whereq x E f

zP

6-

z

3

P

z

3

z

3

z9-2

6-

z

0

z

0

SPPPP



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Field Theory 10EE44


Unit-2

a. Energy and potential : Energy expended in moving a point charge in an electric field

The line integral

Definition of potential difference and Potential

The potential field of a point charge and system of charges

Potential gradient, Energy density in an electrostatic field

b. Conductors, dielectrics and capacitance: Current and current density

Continuity of current metallic conductors

Conductor properties and boundary conditions

boundary conditions for perfect Dielectrics, capacitance



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Field Theory 10EE44


An electric field surrounds electrically charged particles and time-varying magnetic fields. The

electric field depicts the force exerted on other electrically charged objects by the electrically

charged particle the field is surrounding. The concept of an electric field was introduced

by Michael Faraday.

The electric field is a vector field with SI units of newtons per coulomb (N C1

) or,

equivalently, volts per metre (V m1

). The SI base units of the electric field are kgms3A1. The strength or magnitude of the field at a given point is defined as the force that would be

exerted on a positive test charge of 1 coulomb placed at that point; the direction of the field is

given by the direction of that force. Electric fields contain electrical energy withenergy

density proportional to the square of the field amplitude. The electric field is to charge as

gravitational acceleration is to mass and force densityis to volume.

An electric field that changes with time, such as due to the motion of charged particles in the

field, influences the local magnetic field. That is, the electric and magnetic fields are not

completely separate phenomena; what one observer perceives as an electric field, another

observer in a differentframe of reference perceives as a mixture of electric and magnetic fields.

For this reason, one speaks of "electromagnetism" or "electromagnetic fields". In quantum

electrodynamics, disturbances in the electromagnetic fields are called photons, and the energy of

photons is quantized.

Consider a point charge q with position (x,y,z). Now suppose the charge is subject to a

force Fon q due to other charges. Since this force varies with the position of the charge and by

Coloumb's Law it is defined at all points in space, Fon q is a continuous function of the charge's

position (x,y,z). This suggests that there is some property of the space that causes the force which

is exerted on the charge q. This property is called the electric field and it is defined by

Notice that the magnitude of the electric field has units of Force/Charge. Mathematically,

the E field can be thought of as a function that associates a vector with every point in space.

Each such vector's magnitude is proportional to how much force a charge at that point would

"feel" if it were present and this force would have the same direction as the electric field

vector at that point. It is also important to note that the electric field defined above is caused

by a configuration of other electric charges. This means that the charge q in the equation

above is not the charge that is creating the electric field, but rather, being acted upon by it.

This definition does not give a means of computing the electric field caused by a group of

charges.

From the definition, the direction of the electric field is the same as the direction of the force

it would exert on a positively charged particle, and opposite the direction of the force on a

negatively charged particle. Since like charges repel and opposites attract, the electric field is

directed away from positive charges and towards negative charges.

Array of discrete point charges

Electric fields satisfy the superposition principle. If more than one charge is present, the total

electric field at any point is equal to the vector sum of the separate electric fields that each point

charge would create in the absence of the others.

The total E-field due to N point charges is simply the superposition of the E-fields due to

each point charge:



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Field Theory 10EE44


where ri is the position of charge Qi, the corresponding unit vector.

Continuum of charges

The superposition principle holds for an infinite number of infinitesimally small elements of

charges i.e. a continuous distribution of charge. The limit of the above sum is the integral:

where is the charge density (the amount of charge per unit volume), and dV is

the differential volume element. This integral is a volume integral over the region of the charge

distribution.

The electric field at a point is equal to the negative gradient of the electric

potentialthere,

Coulomb's law is actually a special case of Gauss's Law, a more fundamental description of the

relationship between the distribution of electric charge in space and the resulting electric field.

While

Columb's law (as given above) is only true for stationary point charges, Gauss's law is true for all

charges

either in static or in motion. Gauss's law is one of Maxwell's

equations governing electromagnetism.

Gauss's law allows the E-field to be calculated in terms of a continuous distribution of charge

density

where is the divergence operator, is the total charge density, including free and bound charge, in other words all the charge present in the system (per unit

volume).



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Field Theory 10EE44


where is the gradient. This is equivalent to the force definition above, since electric potential is defined by the electric potential energy U per unit (test) positive charge:

and force is the negative of potential energy gradient:

If several spatially distributed charges generate such an electric potential, e.g. in a solid,

an electric field gradient may also be defined

where is the potential difference between the plates and d is the distance separating the plates. The negative sign arises as positive charges repel, so a positive charge will experience a

force away from the positively charged plate, in the opposite direction to that in which the

voltage increases. In micro- and nanoapplications, for instance in relation to semiconductors, a

typical magnitude of an electric field is in the order of 1 volt/m achieved by applying a voltage

of the order of 1 volt between conductors spaced 1 m apart.



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Parallels between electrostatic and gravitational fields

is similar to Newton's law of universal gravitation:

.

This suggests similarities between the electric field E and the gravitational field g, so sometimes

mass is called "gravitational charge".

Similarities between electrostatic and gravitational forces:

Both act in a vacuum.

Both are central and conservative.

Both obey an inverse-square law (both are inversely proportional to square of r).

Differences between electrostatic and gravitational forces:

Electrostatic forces are much greater than gravitational forces for natural values of charge and

mass. For instance, the ratio of the electrostatic force to the gravitational force between two

electrons is about 1042

.

Gravitational forces are attractive for like charges, whereas electrostatic forces are repulsive for

like charges.

There are not negative gravitational charges (no negative mass) while there are both positive and

negative electric charges. This difference, combined with the previous two, implies that

gravitational forces are always attractive, while electrostatic forces may be either attractive or

repulsive.

Electrodynamic fields are E-fields which do change with time, when charges are in motion.

An electric field can be produced, not only by a static charge, but also by a changing magnetic

field. The electric field is given by:



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Field Theory 10EE44


in which B satisfies

and denotes the curl. The vector field B is the magnetic flux density and the vector A is the magnetic vector potential. Taking the curl of the electric field equation we obtain,

which is Faraday's law of induction, another one of Maxwell's equations.

Energy in the electric field

The electrostatic field stores energy. The energy density u (energy per unit volume) is given by

where is the permittivity of the medium in which the field exists, and E is the electric field

vector.

The total energy U stored in the electric field in a given volume V is therefore

Line integral

Line integral of a scalar field

Definition

For some scalar field f : U Rn R, the line integral along a piecewise smooth curve C U is

defined as

where r: [a, b] C is an arbitrary bijective parametrization of the curve C such that r(a)

and r(b) give the endpoints of Cand .



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The function f is called the integrand, the curve C is the domain of integration, and the

symbol ds may be intuitively interpreted as an elementary arc length. Line integrals of scalar

fields over a curve C do not depend on the chosen parametrization r of C.

Geometrically, when the scalar field f is defined over a plane (n=2), its graph is a

surface z=f(x,y) in space, and the line integral gives the (signed) cross-sectional area bounded

by the curve C and the graph of f. See the animation to the right.

Derivation

For a line integral over a scalar field, the integral can be constructed from a Riemann

sum using the above definitions off, C and a parametrization r of C. This can be done by

partitioning the interval [a,b] into n sub-intervals [ti-1, ti] of length t = (b a)/n, then r(ti)

denotes some point, call it a sample point, on the curve C. We can use the set of sample

points {r(ti) : 1 i n} to approximate the curve C by a polygonal path by introducing a

straight line piece between each of the sample points r(ti-1) and r(ti). We then label the

distance between each of the sample points on the curve as si. The product of f(r(ti)) and

si can be associated with the signed area of a rectangle with a height and width of f(r(ti))

and si respectively. Taking the limit of the sum of the terms as the length of the partitions

approaches zero gives us

We note that, by the mean value theorem, the distance between subsequent points on the

curve, is

Substituting this in the above Riemann sum yields

which is the Riemann sum for the integral

Line integral of a vector field

For a vector field F : U Rn Rn, the line integral along a piecewise smooth curve C U, in the direction of r, is defined as

where is the dot product and r: [a, b] C is a bijective parametrization of the curve C such

that r(a) and r(b) give the endpoints of C.

A line integral of a scalar field is thus a line integral of a vector field where the vectors are

always tangential to the line.

Line integrals of vector fields are independent of the parametrization r in absolute value, but they

do depend on its orientation. Specifically, a reversal in the orientation of the parametrization

changes the sign of the line integral.



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Field Theory 10EE44


Electric Potential Difference

In the previous section of Lesson 1, the concept of electric potential was introduced. Electric

potential is a location-dependent quantity that expresses the amount of potential energy per unit

of charge at a specified location. When a Coulomb of charge (or any given amount of charge)

possesses a relatively large quantity of potential energy at a given location, then that location is

said to be a location of high electric potential. And similarly, if a Coulomb of charge (or any

given amount of charge) possesses a relatively small quantity of potential energy at a given

location, then that location is said to be a location of low electric potential. As we begin to apply

our concepts of potential energy and electric potential to circuits, we will begin to refer to the

difference in electric potential between two points. This part of Lesson 1 will be devoted to an

understanding of electric potential difference and its application to the movement of charge in

electric circuits.

Consider the task of moving a positive test charge within a uniform electric field

from location A to location B as shown in the diagram at the right. In moving the

charge against the electric field from location A to location B, work will have to

be done on the charge by an external force. The work done on the charge changes

its potential energy to a higher value; and the amount of work that is done is equal to the change

in the potential energy. As a result of this change in potential energy, there is also a difference in

electric potential between locations A and B. This difference in electric potential is represented

by the symbol V and is formally referred to as the electric potential difference. By definition,

the electric potential difference is the difference in electric potential (V) between the final and

the initial location when work is done upon a charge to change its potential energy. In equation

form, the electric potential difference is

The standard metric unit on electric potential difference is the volt, abbreviated V and named in

honor of Alessandro Volta. One Volt is equivalent to one Joule per Coulomb. If the electric

potential difference between two locations is 1 volt, then one Coulomb of charge will gain 1

joule of potential energy when moved between those two locations. If the electric potential

difference between two locations is 3 volts, then one coulomb of charge will gain 3 joules of

potential energy when moved between those two locations. And finally, if the electric potential

difference between two locations is 12 volts, then one coulomb of charge will gain 12 joules of

potential energy when moved between those two locations. Because electric potential difference

is expressed in units of volts, it is sometimes referred to as the voltage.

Electric Potential Difference and Simple Circuits

Electric circuits, as we shall see, are all about the movement of charge between varying locations

and the corresponding loss and gain of energy that accompanies this movement. In the previous

part of Lesson 1, the concept of electric potential was applied to a simple battery-powered

electric circuit. In thatdiscussion, it was explained that work must be done on a positive test

charge to move it through the cells from the negative terminal to the positive terminal. This work

would increase the potential energy of the charge and thus increase its electric potential. As the

positive test charge moves through the external circuit from the positive terminal to the negative

terminal, it decreases its electric potential energy and thus is at low potential by the time it

returns to the negative terminal. If a 12 volt battery is used in the circuit, then every coulomb of

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