fieldtheory compiled notes

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FIELD THEORY Sub Code : EC44 IA Marks : 25 Hrs/Week : 04 Exam Hrs. : 03 Total Hrs. : 52 Exam Marks : 100 1. Electric Fields 18 hours a. Coulomb’s law and Electric field intensity b. Electric flux density, Gauss law and divergence c. Energy and potential d. Conductors, dielectrics and capacitance e. Poisson’s and Laplace’s equations 2. Magnetic fields 14 hours a. The steady magnetic field b. Magnetic forces, materials and inductance 3. Time varying fields and Maxwell’s 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 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

+=+

==+

=

=

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

A a Aor A / A a Also

actor Unit ve AVector

A

AA

A

==

Product of two or more vectors :

(1) Dot Product ( . )

π θ 0 , B } θ COS A { OR θ COS B ( A B . A ≤≤=

B

B

θ θ Cos A

A

θ θ Cos B

A

A . B = B . A (A Scalar quantity) (2) CROSS PRODUCT (X)

C = A x B = n θ 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

) θ 0 ( B and Abetween angle is ' θ ' where

Ex.,

+=+

=

≤≤ π

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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 , az 2. Cylindrical ρ, φ, z aρ , aφ , az 3. Spherical r, θ, φ ar , aθ , aφ

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., φθφρ 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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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

BBBAAA 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 θ 0 θSin then 0 B x A

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

vectors,zeronon are C and B ,A If

CCCBBBAAA

) C x B( . A

AB

0

zyx

zyx

zyx

==

=

===

===

=

B B

AB

0 A

A

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

zyx

zyx

a dz ady adx rd

dz zr dy r dx

xr 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(ρ, φ, z) x = ρ Cos φ y = ρ Sin φ az r ρ z = z

φ ap r y z z

y / x tan φ

y x ρ1-

22

==

+=

ρ x

0

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1 z r h a

r

ρ r h ; a ρ a r a Cos ρ a Sin ρ - r

1 ρr h ; a h a

ρr a Sin a Cos

ρr

1 ------ dz zr d r ρd

ρr rd

a z a Sin ρ a Cos ρ r

a z ay a x r

zz

y x

ρρρy x

zyx

zyx

=∂∂

==∂∂

=∂∂

==∂∂

=+=∂∂

=∂∂

==∂∂

=+=∂∂

∂∂

+∂∂

+∂∂

=

++=

++=

z

φφφφ

φ

φφ

φφ

φφ

φφφ

ρ

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

22 22

z

zz z

y

xyxρ

(dz) )d ρ( ρd rd and

2 ------ a dz a d ρ a ρd 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

++=

++=

=

+=+=

−=+=

φ

φ

φφφφ

φφφφ

φρ

φρ

ρρρ

φρ

yx

Differential areas :

zz

a dz) ρ (d a ds3 ------- a . )d ρ( (dz) a ds

a . )d ρ( )ρ (d a ds

φφ

ρρ φφ

=

==

Differential volume :

4 ----- dz d ρd ρ dor (dz) )d ρ( )ρ (d d

φτφτ

==

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Spherical Co-ordinate Systems :- Z X = r Sin θ Cos φ Y = r Sin θ Sin φ z p Z = r Cos θ R θ r 0 y Y x φ r Sinφ X

φθθ

θφθ

φθθ

φθθ

φφ

θθ

φφφφ

θφθφθθθ

θφθφθ

θφθφθ

φ

θ

φθ

φ

θ

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 rR Rd

a Cos a Sin - R /Ra

a Sin a Sin Cos a Cos Cos R /Ra

a Cos a Sin Sin a Cos Sin rR /

rRa

a Cosr a Sin Sin r a Cos Sin r R

2

2

2r

r

yx

zyx

zyxr

zyx

=

==

=

++=

∂∂

+∂∂

+∂∂

=

+=∂∂

∂∂

=

−+=∂∂

∂∂

=

++=∂∂

∂∂

=

++=

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

33

22

11

321

333222111

33

22

11

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 Systemu & u , u tol tangentiaorsubnit vect are a , 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 ρ φ z a ρ aφ az 1 ρ 1

Spherical r θ φ ar aθ aφ 1 r r sin θ Transformation equations (x,y,z interms of cylindrical and spherical co-ordinate system variables) Cylindrical : x = ρ Cos φ , y = ρ Sin φ , z = z ; ρ ≥ 0, 0 ≤ φ ≤ 2π -∞ < z < ∞ Spherical x = r Sinθ Cosφ , y = r Sinθ Cosφ , z = r Sinθ r≥ 0 , 0 ≤ θ ≤ π , 0 ≤ φ ≤ 2π

) 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

3213

2312

1321321

3 33

22 2

111

===++=

=

∂∂

∂∂

∂∂

=∇

∂∂

+∂∂

+∂∂

=∇

∂∂

+∂∂

+∂∂

=∇

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Vector Transformation from Rectangular to Spherical :

=

++=

++=

++=

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

φφφ

θθθ

φ

θ

φθ

φφθθ

φφθθ

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 (φ) 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)

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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) 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

)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) 524132aaa

B x Azyx

−−−=

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

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

=+=∴

+=

++=

++=∴

+=++=

+=+=∴

ππ

ππ

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

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

=++=

=

++=

++=

−=

++=

+=

++=

πππππ

πππ

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

++= φφ

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01-

22

z

xxzyx

yxzyx

yxzyx

38.66 - xy 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

===

=+==

++=

++

+++

++=

φφρ

ρφρ

φφφφ

φφ

φφ

φρ

φ

ρ

)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

+=

++=

ρ

φ

8. Transform a z a x - ay B zyx +=

into Cylindrical Co-ordinates.

z

z22

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

+=

++=

++++

++=

++=

+=

==

φ

φρ

ρ

ρ

φφρρ

ρ

φρφρφφρφφρ

φφφρφρ

φφφρφρ

φρφρ

φρφρ

9. Transform xa 5 into Spherical Co-ordinates.

φθ

φ

θ

φφθθ

φφθφθ

φφ

θφθφθ

θφθφθ

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 φρ+= 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

φρφρ

φφ

ιφ

φφ

φφ

φ

ρ

φ

ρρ

φφρρ

==

++

+=

+++

+++=

++= zz

ρ

φ

x

y

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φ

ρ

φ

ρ

φ

ρ

φφρφρφρ

φρφφρφρ

φφφρφρφφρ

φφρφρφφρφρ



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, 3π/4, π/6) to (5, π/4, π) Soln : A(r, θ, φ) expressed in rectangular co-ordinates

)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 6

Sin 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

+==

=++=

+==

+=+=

++=

++=∴

++=

πππππ

πππππ

θφθφθ

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 Cos0.625- (-38.66)Sin Sin 0.9 64.69Sin Sin

38.66- 10

8- tan

64.89 2006

Cos rz

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

φθ

φ

θ

φ

θ

φφθθ

φ

θ

φθφθ

φ

θ

φφθφθφθθφθφθ

φφ

θφθφθ

θφθφθ

++=

++=

========

==

====++=

++

+=

++=

+=

+=

++=

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Line Integrals In general orthogonal Curvilinear Co-ordinate system

∫∫∫ ∫ ++=

++=

++=

C333

C222

C C111

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

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

C

=Φ∇∫

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

∫

∫ ∫

Φ∇

Φ∇=

ΦΦ=Φ=Φ∇

φ

∫ =Φ∇ 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 ydy ) 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

31 11

31 1 -

32 12

34 - 2

38 8

21

31

21 -

22

32

22

2y

3y

2 4y 2

234

2

1

234

==

++=

++

++=

++

//

=

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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

3a 2 x

32 x a d Cos d Sin a ds . E

2 0 , 2 / 0d 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

2r

2r

rr

rr

r2

r

2222

ππφφθθ

πφπθφθφθ

φθθφθ

φθφθ

φθθ

φθθ

π π

φφθθ

===

<<<<=

==

==

++=

=

=

≥=++

∫ ∫ ∫

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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 ψ 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 ψor c/m Aψ 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 qF E 12

10

1

∈==

π

C E Dor c / N D c / N a r 4

q 1 00

121

1

0

∈=∈

=

∈

=π

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

r20

π

π

=∈∈=

∈∈=

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

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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 1r

Qn 0 Pr nr

m / V a r -r

Q 4

1 E m2m

m

0r ∑∈

=π

2. E

due to continuous volume charge distribution

Ra R

P

ρv C / m3

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

C / m3 i.e, vd

Q d and dv Q VV

V == ∫ ρρ

C / N a )r -(r 4

)(r E

a R 4V

a R 4

Q E

R210

1

r

R20

R20

1∫ ∈

=

∈∆

=∈∆

=∆

V

V

V

πρ

πρ

π

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 ρl C / m Ra P R

dl ρl C / m L

C / N a R

dl 4

1 E RL

20

p ∫∈= lρ

π

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4. E

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

Ra P (Field point) ds R

(Source charge)

C / N a R

ds 4

1 E RS

20

p ∫∈= Sρ

π

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 yV dx

xV dV Also,

(1) -------- dl . E - dl qf - dV

∇=

∇=

∂∂

+∂∂

+∂∂

=

==

z

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

dR R +

0 θ + Q R

P Ra

At point P, C / N a R 4

Q E R20∈

=π

Therefore, the force f

on a unit charge at P.

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N a R 4

Q E x 1 f R20

p ∈==∴

π

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

20

P

20

lR

R

20

p

∈=

∈=

∈=∴

==

∫∫∞∞

π

ππ

R

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

voltR1 -

R1

4Q )V - (V V

qp0qppq

∈==

π

Electric Potential at a point due to different charge configurations. 1. Discrete charges

. Q1 . P Q2 Rm Qm

V RQ

41 V

n

1 m

m

01P ∑∈

=π

2. Line charge

x P V dl Rl

41 V

l

l

02P ∫∈

=π

ρl C / m

3. Surface charge

x P V R

ds 4

1 VS

S

03P ∫∈

=ρ

π

ρs C / m2

4. Volume charge x P

R V R

dv 4

1 VV

V

04P ∫∈

=ρ

π

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

ρv C/ m3

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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

0R ∈=

π

Gauss’s 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, ∫=

VV dv Q ρ ---------- (3)

From 1, 2 & 3,

ρ D . =∇

----------- (4) is point form (or differential form) of Gauss’s law while equation (1) is Integral form of Gauss law. Poisson’s equation and Laplace equation In equation 4, E D 0

∈=

equation Laplace 0 V 0, If

equationPoisson - V

/ V) (- . or / E .

20

2

00

=∇=

∈∇

∈=∇∇∈=∇∴

ρ

ρρρ

0 +Q

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Till now, we have discussed (1) Colulomb’s 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 Coulomb’s law is fundamental in approach while the other two use the physical concepts involved in the problem. (1) Coulomb’s law : Here E

is found as force f

per unit charge. Thus for the simple case of

point charge of Q C,

∫=

∈=

l

20

Volt dl E V

MV/ RQ

41 E

π

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

determined. Then

∫

∫

=

=

l

Senc

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 ∇=

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Solutions to Problems on Electrostatics :-

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

Solution :

zy 23

2323

zyzy

13

1313

2322

1321

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 dQ a

dQ

4Q F

X F F F is F force The

a 0.866 a d 0.5 r

P P a d rY d 0 r

m d) 0.866 d, 0.5 (0, Pd 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

+==

+=+

==

+

∈=

+=

+=

=====

π

Substituting,

) a 0.924 a (0.38 a whereN a 0.354 F

13.11 12.12 5

a 12.12 a 5

d10 x 27

) a 0.866 a 0.5 - ( d10 x 2 )a 0.866 a 0.5 (

d10 x 12 10 x 9 ) 10 x (3 F

zyFF3

22

zy2

3-

zy2

-6

zy2

-696-

3

+==

+

+=

+++=

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

p ++= 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

yV

a x

V - V - E )1(

z2

yx

zp

yp

xp

pp

++=

∂

∂+

∂

∂+

∂

∂=∇=

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]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

2P

QpPQ

===++=

++== ∫ ∫ ∫∫

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( 291

AO

AO a

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

CN/ ]a [ (AO)

3610 x 4

10 x 64.4

C / N a (AO) 4

Q E

AOAO0

zyxAO

zyxzyx

AO2

9-

9-

AO20

0

==

+===

+=++++=

=

∈=

ππ

π

4. Q1 = 100 µC at P1 (0.03 , 0.08 , - 0.02) m Q2 = 0.12 µC 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

129

2

6-6-

12

zyx12

12zyx

zyxzyx1212

122120

2112

=

=

+=

=+=

+++==

∈=

π

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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)

)(repulsive N 12.66 F

a N 12.66 a 10 x 9 x 16

1.5 a )10 x 4 ( x

3610 x 4

)10 x 1.5 ( F

C 10 x 1.5 )Q (Q 21 Q Qcontact intobrought When (2)

e)(attractiv N 5.63 a 10 x 16

9- a )10 x 4 ( x

3610 x 4

10 x 0.5 - x 10 x 2 F

`12

121213 18-

2

1222-

9-

29-`

12

9-21

`2

`1

125-

1222-

9-

-9-9

12

µ

µ

ππ

µ

ππ

=

===

=+==

+===

+

6. Y P3 x x P1 P2 x x 0 X Q1 = Q2 = Q3 = Q4 = 20 µ C QP = 200 µ 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 RQ a

RQ a

RQ a

RQ

3610 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

4p24p

43p3

3p

32p2

2p

21p2

1p

19-

pp

zy 4p4pzy 4p

zyx 3p3pzyx3p

zx 2p2pzx2p

z1p1pz1p

4p3p2p1pp

+++=

+==+=

+==+=

+==+=

===

+++=

ππ

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6-

zy2

zyx2zx2z296- 10 x 20

)a 0.6 a 0.8- ( 51

)a 0.47 a 0.625 - a (-0.625 6.4

1 )a 0.6 a 0.8 - ( 51 a

31

10 x 9 x 10 x 002

++

++++=

)a 0.6 a 0.8- (

25100

)a 0.47 a 0.625 - a (-0.625 40.96100)a 0.6 a 0.8 (-

25100a

9100

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 = - 1µC, Q2 = -2 µC , Q3 = - 3 µC 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 -

3610 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 RQ a

RQ a

RQ

41

E E E E

03zyzy

3

yyz3

y2

6-

y2

6-

z2

6-

9-P

y3P 3Py3P

y2P 2Py2P

z1P 1Pz1P

3P23P

32P2

2P

21P2

1P

1

0

3P2P1PP

∠=++=+=

+=

++=

===

==+=

===

++

∈=

++=

ππ

π

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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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 ρl = 25 n C / m A R

ρ (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

3610 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-R0

lP

zxR

zxzx

=

=∈

=

==

===+==

πππ

ρR

9. Data : P1 (2, 2, 0) m ; P2 (0, 1, 2) m ; P3 (1, 0, 2) m Q2 = 10 µC ; Q3 = - 10 µC To find : E1 , V1 Solution :

V 3000 3

10 3

10 10 x 9 RQ

RQ

41 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 RQ a

RQ

41 E E E

6-6-9

31

3

21

2

01

yxyx3

zyx

6-

zyx

6-9

1

zyx31 31zyx31

zyx21 21zyx21

31231

3212

21

2

021211

=

+=

+

∈=

+=+=

++++=

++==++=

+==+=

+

∈=+=

π

π

V 3000 V m V / 14.14 E 11 ==

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10. Data : Q1 = 10 µC at P1 (0, 1, 2) m ; Q2 = - 5 µC 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 :

[ ][ ] 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 RQ a

RQ

41 E (1)

3zyx

zyxzy

zyx2

6-

zy2

6-9

3

zyx23

2323

zy13

1313

23zyxzyx23

13zyzy13

23223

2132

13

1

03

+=

++=

++=

+==

==

=+=+++=

==+=

+

∈=

π

zero becannot E

a 1.23 - E

a 2 R ; a R

Q a RQ a

RQ 10 x 9 E (2)

3x

x3x

y0303203

23223

2132

13

193

=

=

++=

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

3610 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

12123-

9-

9-9-

12

zyx12122120

2112

==

+=∈

=

ππ

π

N a 0.015 F 1212 ˆ=

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12. Given V = (50 x2yz + 20y2) volt in free space Find VP , m 3)- 2, (1, Pat a and E npP

Solution :

[ ]

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

z2

y2

x

zyx

22P

+==

+=

=

=

∂∂

∂∂

∂∂

=∇=

=+=

zy

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Additional Problems A1. Find the electric field intensity E

at P (0, -h, 0) due to an infinite line charge of density

ρl C / m along Z axis. +∞ Z A dz APR

z dEPy P θ Y dEPz h 0 d PE

Pa X - ∞ Solution : Source : Line charge ρl C / m. Field point : P (0, -h, 0)

[ ]

a Rz

R 4dz ρ - dE a

Rh

R 4dz ρ - dE

a E d a E d a Rz - a

Rh -

R 4dz ρ dE

a z - ah - R1

R R a

h z AP R

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

dz ρ a R 4

dQ dE

z20

lPzy2

0

lPy

z PzyPyzy20

l P

zyR

22

yzR20

lR2

0 P

∈=

∈=

+=

∈=

==

+==

==∈

=∈

=

ππ

π

ππ

Expressing all distances in terms of fixed distance h, h = R Cos θ or R = h Sec θ ; z = h tan θ , dz = h sec2 θ dθ

0 ]Cos [ h 4

ρ E

d Sin h 4

ρ - Sec h

tan h x Sec h 4

d Sech ρ - dE

a h 2

ρ - 2 x h 4

ρ - ]Sin [ h 4

ρ - E

d Cos h 4

ρ - Cos x Sec h 4

d Sech ρ - dE

2 / 2 / -

0

lPz

0

l22

0

2l

Pz

y0

l

0

l2 / 2 / -

0

lPy

0

l22

0

2l

Py

=∈

=

∈=

//

∈=

∈=

∈=+

∈=

∈=

∈=

ππ

ππ

θπ

θθπθ

θθπθθ

ππθ

π

θθπ

θθπθθ

m V / a h π2

ρ - E y0

l ˆ∈

=

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An alternate approach uses cylindrical co-ordinate system since this yields a more general insight into the problem. Z + ∞ A dz z R

θ P (ρ , π / 2, 0) 0 Y P π / 2 AP X - ∞

0 ]Cos [ ρ 4

ρ E

d )Sin (- ρ 4

ρ d Sec ρ 4

Sec ρ x tan ρ x ρ dE (ii)

ρ 2ρ 2 x

ρ 4ρ ]Sin [

ρ 4ρ E

d Cos ρ 4

ρ d Sec ρ 4 Sec ρ x ρ x ρ dE (i)

Sec ρ Cos

ρ R and d Sec ρ dz , tan ρ z

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

dz z R 4

ρ- dE (ii) ; dz ρ R 4

ρ dE (i)

a dE a dE a Rz - a

Rρ

R 4dz ρ dE

C dz ρ dQ

)a z - a ρ ( R1 a and a z - a ρ R where

m / V a R 4

dQ dE

is dQ todue dEintensity field The

at Z. change elemental theis dz ρ dQ

2 / 2 / -

0

lP

0

l33

0

2l

P

0

l

0

l2 / 2 / -

0

lPρ

0

l33

0

2l

Pρ

2

20

lP2

0

lPρ

PρPρzρ20

lP

zρRzρ

R2

0

P

P

l

=∈

=

∈=

∈−

=

∈=

∈=

∈=

∈=

∈=

====

=

∈=

∈=

+=

∈=

=

==

∈=

=

ππ

ππ

θπ

θθπ

θθπ

θθ

ππθ

π

θθπ

θθπθ

θθ

θθθ

θθ

ππ

π

π

z

z

z

zz

l

m / V a ρ 2

ρ E ρ0

lP ∈=∴

π

Thus, directionin radial is E

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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 θ2 A(0, 0, z1) θ1 Y X Solution :

[ ]

2 - ,

2

, to - from extending is line theIf

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

ρ E

a )Cos ( h 4

ρ a )Sin (- h 4

ρ

a d Sin h 4

ρ - a d Cos h 4

ρ - E d E

a Rz - a

Rh

R 4dz ρ dE

12

z21y210

lP

z0

ly

0

l

z0

ly

0

lz

zPP

z20

lP

2

1

2

1

2

1

2

1

2

1

πθπθ

θθθθπ

θπ

θπ

θθπ

θθπ

π

θθ

θθ

θ

θ

θ

θ

==

∞∞

−∈

+=

∈+

∈+=

∈∈==

−

∈=

∫∫∫

y

m V / a h 2

- E yl

P ˆρ

0∈=

π

A3. Two wires AB and CD each 1 m length carry a total charge of 0.2 µ 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) θ1 = 1800 θ2 = 1800 A B P 1 m

[ ] [ ]{ } nate)(Indetermi 00 a Cos - Cos a ) Sin - (Sin -

h 4ρ E z12y12

0

lPAB

=+∈

= θθθθπ

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az (2) Pay C

θ1 θ1 = - tan-1 5.0

1 = - 63.430

θ2 = 0 D

[ ]

[ ]

[ ] )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

3610 4

10 x 0.2

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

ρ E

zyzy3

P

zy9-

6-

z12y120

lP

CD

CD

+=+=

+=

+∈

=

ππ

θθθθπ

Since E

ABρ

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 - d1

4ρ

t 4ρ E

dt t 4

ρ - dE

d1 t ; L y

d L

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

dy y) - d (L 4

a ρ dE

)a(- R1 a ; a y) - d (L R

m / V a R 4

dy ρ dE

0

ld1

d L10

lP

20

lP

20

ylPy

yRR

R20

lP

+∈=

∈

+=

∈=

==

+====+

+∈

−=

=+=

∈=

+ππ

π

π

π

m V / d L

1 - d1

4ρ E l

P

+∈=∴

0π

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

a 1.51 -

0.51

3610 4

10 x 0.2 E

yyP

y9-

-6

P

AB

AB

==

=

ππ

)a 0.925 a 0.381 (- a wherem / 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 ρS 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 ρS C / m2 . dEP Z θ R AP

=

z P 0 Y dφ φ X A ρ

)a z a ρ - ( R1 a

)a z a ρ - ( R

OP OA - OP AO AP

zρR

zρ

+=

+=

+=+=

The field intensity PdE due to dQ = ρS ds = ρS (dφ dρ) is along AP and given by

dρ ρ d )a z a ρ - ( R 4

ρ a R 4

dρ d ρ ρ dE zρ30

R20

P φππ

φ+

∈=

∈= SS

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

ρd R

ρ z 2 x 4

ρ R

ρd ρ z d 4

ρ dE E

ρd ρ d R 4

z ρ dE

03

003

2

00SPP

30

P

∫∫∫∫∞∞

∈=

∈==

∈=∴

ππ

φπ

φπ

πSS

S

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

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ρ = z tan θ , d ρ = z Sec2 θ dθ ; R = z Sec θ ; ρ = 0, θ = 0 ; ρ = ∞ , θ = π / 2

plane) to(normal a 2ρ

a ] Cos [- 2ρ d Sin

2ρ d Sec z

Sec z tan z z

2ρ E

z0

S

z2 /

00

S2/

00

S2

033

0

SP

∈=

∈=

∈=

∈= ∫∫

∞π

π

θθθθθθθ

A5. Find the force on a point charge of 50 µC at P (0, 0, 5) m due to a charge of 500 πµC that is uniformly distributed over the circular disc of radius 5 m. Z P h =5 m 0 Y φ ρ X Solution : Given : ρ = 5 m, h = 5 m and Q = 500 πµC To find : fp & qp = 50 µC

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

3610 x )5 ( 2

10 x 500 a 2AQ

a 2ρ E whereq x E f

zP

6-z

3P

z3

z3

z9-2

6-

z0

z0

SPPPP

=

=

=

=

/

/=∈

=

∈==

π

ππ

π

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fieldtheory compiled notes

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