50999882vakx_electromagnetic field theory_solution manual

8/16/2019 50999882VAKX_Electromagnetic Field Theory_Solution Manual

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Solutions of Examples for Practice

Example 1.6.3

Solution : The origin O (0, 0, 0) while P (3, – 3, – 2) hence the distance vector OP is,

OP = ( ) ( ) ( )3 0 3 0 2 0 3 3 2 a a a a a ax y z x y z

OP = 3 3 22 2 2 = 4.6904Hence the unit vector along the direction OP is,

a OP = OP

| OP|

a a ax y z

3 3 24 6904.

= 0.6396 a x – 0.6396 a y – 0.4264 a z

Example 1.6.4

Solution : The starting point is A and terminating point is B.

Now A = 2 2a a ax y z and B a a ax y z 3 4 2

AB = B A 3 2 4 2 2 1a a ax y z AB = a 6 a ax y z This is the vector directed from A to B.

Now AB = 1 6 12 2 2 = 6.1644Thus unit vector directed from A to B is,

aAB = AB

AB

a a ax y z

6

6.1644

= 0.1622 a x – 0.9733 a y + 0.1622 a z

It can be cross checked that magnitude of this unit vector is unity i.e.

0.1622 0.9733 0.16222 2 2 = 1.

Example 1.7.3 Kept this unsolved example for student's practice.

Example 1.7.4

Solution : Consider the upper surface area, the normal to which is a z . So the differential

surface area normal to z direction is r d dr . Consider the Fig. 1.7.8.

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1 Vector Analysis


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= cos

1 A B

|A| |B|= cos

1 27

45 38= 130.762º

Example 1.10.7

Solution : A = 5 a x and B = 4 a ax y By , AB 45

Now A B = A B A B A Bx x y y z z = 5 4 0 0 20 But A B = |A| |B|cosAB

20 = 5 4 B cos2 2 y 2 45 i.e. 16 By2 = 5.6568 By

2 = 16 i.e. By = 4

Now B = 4 a a ax y z B By zStill A B = 20

20 =

5 452 4 B B2 y

2z

2 cos

16 B By2

z2 = 5.6568 i.e. B By

2z2 = 16

This is the required relation between By and Bz .

Example 1.11.4

Solution : Note that the unit vector normal to the plane containing the vectors A and B is

the unit vector in the direction of cross product of A and B.

Now A B =a a ax y z

3 4 5

6 2 4

= a a ax y z4 5

2 4

3 5

6 4

3 4

6 2

= 26 18 30a a ax y z

a N =

A B

A B

a a ax y z

26 18 30

26 18 302 2 2= 0.5964 a x + 0.4129 a y + 0.6882 a z

This is the unit vector normal to the plane containing A and B.

Example 1.11.5

Solution : The perpendicular vector to the plane containing A and B is given by their

cross product.

A B =a a ar z

A A A

B Br z

r z=

a a ar z

2 1

1 2

= 72

3 4 a a ar z

a n = Unit vector in the direction A B

Electromagnetic Field Theory 1 - 3 Vector Analysis

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=

72

3 4

72

3 42

2 2

a a ar z

( ) ( )

= 3.5

16.9651

a a ar z3 4= 0.648 a 0.1768 a 0.74 ar z


Example 1.12.3

Solution : The scalar triple product is,

A B C =2 0 1

2 1 2

2 3 1

= 14

The vector triple product is,

A B C = B A C C A B

A C = 2 2 0 3 1 1 = 3A B = 2 2 0 1 1 2 = 2

A B C = 3 B 2 C = 3 2 2 2 2 3a a a a a ax y z x y z = 2a 3a 4 ax y z Example 1.13.9

Solution : A = y x za a ax y z A r = A a r a x a r a a r a z a ry y x z

= y cos x sin (Refer Table 1.13.1) … (1)

A = A a a a a a a ax y z

y x z

= y sin x cos … (2)

A z = A a a a a a a az x z z z zy y x z z … (3)Now x = r cos y r sin z z1

Using in equations (1), (2), (3) we get,

A = [r sin cos r sin cos ] [ r sin r c2 a r os ] z2 a a z

= 0 r (sin cos z2 2 a a r az z a z

Example 1.13.10

Solution : B = 10r r cosa a ar

Br = 10

r , B r cos , B 1 … In spherical

B

B

B

x

y

z

=

sin cos cos cos sin

sin sin cos sin

cos

cos sin 0

10r

r cos

1




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B = – sin sin sin cos cos cosr r r 2 2

B = r rsin cos cos cos 2 … cos – sin cos2 2 2

B = B B Br a a ar

At Q, r = 7, 64.623 , = 108.435 hence B at Q is,B = – 0.8571 a – 0.4064 a ar – 6 … At point Q

Example 1.13.12

Solution : Refer example 1.13.11 for P in cylindrical system.

P (6.3245, 108.43 , 3) … Cylindrical

To convert A to cylindrical,

A = A a a a a ax y y( ) (x z) ( )

= y x zcos ( ) sin and x = cos , sin ,y z z A = 2 2 sin cos sin sin sin z z

A = A a a a a ax y y ( x z) ( ) ( )

= y x z(– sin ) ( ) (cos )

= cos2 z cos … cos – sin cos2 2 2

A = [ sin sin ] [ cos 2 2 0 z z cosa a a z

At P, A = – 0.9485a – 6a Example 1.15.3

Solution : From given A, A 2 xy, A z, A yzx y z2

A = div A =

A

x

A

y

A

zx y z

=

x

2 xyy

zz

yz2 = 2y 0 2zy 2y 2yz

At P (2, – 1, 3), x = 2, y = – 1, z = 3

A = (2) 1 2 1 3 8Example 1.15.4

Solution : Given A in cylindrical system,

div A = 1r r

r A 1

r

A A

zrz

where A r = r z sin , A 3 r z2

cos , A z = 0

div A = 1r r r z 1

r r z2 2

sin cos

3 0




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= 1r z 2r 1

r 3r z 2 sin sin = 2z sin 3 z 2 sin

At point P, r = 5,

, z = 1

div A = 2 1 3 1 sin sin

= – 1 at P.

Example 1.16.3

Solution : The outward flux is given by,

= F S dS

over a closed surface S

The cylindrical surface is shown in the Fig. 1.1.The total surface is made up of,

1. Top surface S1 for which z = 1, r varies from0 to 4 and varies from 0 to 2 .

2. Lateral surface for which z varies from 0 to 1, from 0 to 2 and r = 4.

3. Bottom surface S3 for which z = 0, r varies from

0 to 4 and varies from 0 to 2 .For S1, dS = r dr d azFor S2, dS = r dz d arFor S3, dS = r dr d (– az)

F S

dS1

= ( cos sin ) ( )r z r dr d

S

2 2

1

a a ar z

= 0

F S dS 3

= ( cos sin ) [r zS

2 2

3

a a ar z r dr d ( )] = 0F S d

S2

= ( cos sin ) ( )r z r d dS

2 2

2

a a ar r z= r r dz d

z

2 2

0

2

0

1

cos

… a ar z = 1, a a r = 0 r = 4

= (4)3

dz dz

cos2

0

2

0

1

= 64 dz0

1

0

21 2

2

cos

d

= 64 × [ ] [ ] sin

z01

02

0

212

22

= 64

F S dS

= 0 + 64 + 0 = 64



ur aceS1

SurfaceS3

z = 1

SurfaceS2

r = 4

z = 0

x

y

Fig. 1.1


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Let us verify divergence theorem which states that,

F S dS

= ( F)v

dv where dv = r dr d dz

F =

1

r r

(r Fr) + 1

r

F F

z

z

= 1 12 2

r r r r

r z

( cos ) ( sin ) + 0

= cos

( cos )2

23

r

r zr

= 3 r cos2

z

rcos

( F)v

dv = 3 2

0

4

0

2

0

1

r z

rrz

cos

cosr dr d dz

= 33

32

0

2

0

4

0

1

r z rz

cos cos

d dz

= 4 1 2

2 43

0

2

0

1

cos

cos

z

z

d dz = 32 2

2 40

2

02

0

1

sin[sin ]z

z

dz

= { [ ] [ ]}32 2 0 4 0

0

1

zz

dz = 64

0

1

z dz = 64

Thus F S

d

S

= (

F)

v

dv and divergence theorem is verified.

Example 1.16.4

Solution : Using divergence theorem

S

d A S =v

( ) dv A

To evaulate A S d it is necessary to consider all six faces of the cube. Let us find dS foreach surface, for a cube shown in the Fig. 1.2.



x = Constant planes(back and front)

y = Constant planes(sides)

z = Constant planes(top and bottom)

–ax

ax –ay ay

az

–az

Back

FrontLeft Right

Top

Bottom

(a) Cube (b) Directions of dS

Fig. 1.2


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1. Front surface (x = 1), dS = dy dz, direction = a x , dS = dy dz a x

2. Back surface (x = 0), dS = dy dz, direction = a x , dS = dy dz a x3. Right side (y = 1), dS = dx dz, direction = a y , dS = dx dz a y

4. Left side (y = 0), dS = dx dz, direction = a y , dS = dx dz a y

5. Top side (z = 1), dS = dx dy, direction = a z , dS = dx dy a z

6. Bottom side (z = 0), dS = dx dy, direction = a z , dS = dx dy a zA = xy y y z2 3 2a a ax y z

For front, A S d = xy dy dz (x 1) y dy dz2 2

For back, A S d = xy dy dz (x 0) 02

For right, A S d = y dx dz (y 1) dx dz3

For left, A S d = y dx dz (y 0) 03

For top, A S d = y z dx dy (z 1) y dx dy2 2

For bottom, A S d = y z dx dy (z 0) 02

S

d A S =z 0

1

y 0

12

z 0

1

x 0

1

y 0

1

x

y dy dz dx dz

0

12y dxdy

= y

3 [z] [x] [z]

y

3 [x]

13

0

1

01

01

01

3

0

1

01

3

1 1

3

53

A =

A

x

A

y

A

z y 3y y 5yx

y z 2 2 2 2

v A) dv =

z 0

1

y 0

1

x 0

12

3

0

5y dx dy dz 5 y

3

1

01

01[x] [z] = 5

13

1 1 53

Thus divergence theorem is verified.

Example 1.16.5

Solution : A = 2xy y 4 yz2a a ax y z

Using Divergence theorem,S

d A S =v

dv ( A)

A =

A

x

A

y

A

z 2y 2y 4y 8yx

y z

S

d A S =v

(8y)dv ... dv = dx dy dz




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=

z 0

1

y 0

1

x 0

1 2

8y dx dy dz 8 y

2

0

1

01

01[x] [z] = 8

12

4

Example 1.16.6

Solution : The divergence theorem states that

S

d A S = v A dv

Now

S

d A S =side top bottom

d

A S

Consider dS normal to a r direction which is for the side surface.

dS = r d dz a r

A Sd = 30 2e z r d dzr

a a ar z r

= 30 r e d dz r e d dzr r a ar r

side

d A S =

z 0

5r30 r e d dz with r = 2

= 30 2 e z2 05

02

= 255.1

The dS on top has direction a z hence for top surface, dS = r dr d a z

A Sd = 30 e 2 z r dr dr a a ar z z = – 2 z r dr d ... a az z 1

top

d A S =

r 0

2

2 z r dr d with z = 5

=

2 5

r2

2

0

2

02

40

While dS for bottom has direction a z hence for

bottom surface, dS = r dr d a z A Sd = 30 e 2 z r dr dr a a ar z z = 2 z r dr d ... a az z 1

But z = 0 for the bottom surface, as shown in the

Fig. 1.3.

S

d A S = 255.1 – 40 = 129.4363



ar

–az

z = 0

az

dS

z = 5

dS

Fig. 1.3


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This is the left hand side of divergence theorem.

Now evaluate v A dv

A = 1

r r

r A 1

r

A A

zr

z

and A r = 30 e

r , A = 0, A 2 zz

A = 1r r 30 r e z 2 zr 0= 1r 30 r e 30 e 1 2r r = 30 e 30r e 2r r

v

A dv =

z 0

5

r 0

2r r30 e 30

r e 2

r dr d dz

= z 0

5

r 0

2r r30 r e 30 e 2r

dr d dz

=

30 r e

1 30

e1

dr 30 e

1

r r r

2

r2

z2

05

02

Obtained using integration by parts.

= 30 r e 30 e 30 e rr r r 2

0

2

5 2 = 60 e 22 2 10 = 129.437

This is same as obtained from the left hand side.

Example 1.16.7

Solution : The given D is in spherical

co-ordinates. The volume enclosed is shown in the

Fig. 1.4.

According to divergence theorem,

S

d

D S =

v

D dv

The given D has only radial component as given.

Hence D 5 r

4r2

while D D 0.

Hence D has a value only on the surface r = 4 m.

Consider dS normal to the a r direction i.e.

r d d2 sin



z

x

y

45º

a

D

r = 4 m

Dr

ar

Fig. 1.4


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dS = r d d2 sin a r

D Sd = r d d 5r4 r d d22

4sin sin

54

... a ar r 1

S

d D S =

054 r d d4 sin ... r = 4m

= 54 r4 cos

0 02

= 54 4 0 24

cos cos

= 588.896 C

To evaluate right hand side, find D.D = 1

r r

r Dr

Dr

D

22

r

1 1sin

sinsin

= 1r r

r 5

4 r

r r r

22 2

24

0 0

5

4= 5

4 r4 r

23 = 5 r

In spherical co-ordinates, dv = r dr d d2 sin

v D dv =

0r 0

425 r r dr d dsin

= 5 0 02r44

0

4

cos = 5 44 0 2

4

cos cos

= 588.896 C

Example 1.16.8

Solution : The volume bounded by the given planes is a cube. To evaluate total charge

use Gauss's law.

Q =S

d D SBut to evaluate D S

d , it is necessary to consider all six faces of the cube. Let us find dSfor each surface.

1) Front surface (x = 2), dS = dy dz, direction = a x , dS = dy dz a x

2) Back surface ( x = 1), dS = dy dz, direction = a x , dS = – dy dz a x3) Right side (y = 3), dS = dx dz, direction = a y , dS = dx dz a y

4) Left side (y = 2), dS = dx dz, direction = a y , dS = – dx dz a y5) Top side (z = 4), dS = dx dy, direction = a z , dS = dx dy a z

6) Bottom side (z = 3), dS = dx dy, direction = a z , dS = – dx dy a z




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Key Point Remember that though the co-ordinates of x, y and z are positive, the directions of

unit vectors are with respect to region bounded by the planes, as shown in the Fig. 1.5 (b).

For front D Sd = 4x dy dz, x = 2 ... a ax x 1For back D Sd = – 4x dy dz, x = 1 ... a ax x 1For right D Sd = 3 y 2 dx dz, y = 3 ... a ay y 1For left D Sd = 3 y 2 dx dz, y = 2 ... a ay y 1For top D S

d = 2 z 3 dx dy, z = 4 ... a a

z z

1

For bottom D Sd = 2 z 3 dx dy, z = 3 ... a az z 1

S

d D S =z 3

4

y 2

3

z 3

4

y 2

3

4x dy dz + 4x dy dz

Front, x = 2 Back, x = 1

z 3

4

x 1

22

z 3

4

x 1

223y dx dz 3y dx dz

Right, y = 3 Left, y = 2

y 2

3

x 1

23

y 2

3

x 1

2

2z dx dz + z dx d2 3 z

Top, z = 4 Bottom, z = 3

= 4 2 y z 4 1 y z 3 3 x z23

34

23

34 2

12

34



x = constant planes(back and front)

y = constant planes(sides)

z = constant planes(top and bottom)

–ax

ax –ay ay

az

–az

Back

FrontLeft Right

Top

Bottom

(a) Cube (b) Directions of dS

Fig. 1.5


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= 3 2 x z 2 4 x y 2 3 x y2 12 34 3 12 23 3

12

23

= 8 4 27 12 128 54 = 93 CThis is the total charge enclosed.

Let us verify by divergence theorem.

D =

D

x

D

y

D

z 4 6 y 6 zx

y z 2

v D dv =

z 3

4

y 2

3

x 1

224 6 y 6 z

dx dy dz ... Integrate w.r.t. x

= z 3

4

y 2

32

124 6y 6z x

dy dz ... Integrate w.r.t. y

=z 3

4 22

2

3

4y 6 y

2 6 z y

dz

= z 3

42 2 24 3 2

62

3 2 6 z 3 2

dz

= z 3

42

3

3

4

4 15 6 z dz 19 z 6 z

3

= 19 4 3 2 4 33 3 = 93 CThus divergence theorem is verified.

Example 1.17.6

Solution : t = x y e2 z and P(1, 5, – 2)

t = Gradient of t =

t t

y

t

zx a a ax y z = 2xy x e

2 za a ax y z

At P, x = 1, y = 5, z = – 2

t = 10 a a e ax y

2

z

Example 1.17.7

Solution : 1) V = e– z sin 2x cosh y

V =

V

x

V

y

V

zx y za a a

V

x =

x

[e– z sin 2x cosh y] = 2 e– z cos 2x cosh y




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V

y =

y

[e– z sin 2x cosh y] = e– z sin 2x sinh y

V

z =

z

[e– z sin 2x cosh y] = – e– z sin 2x cosh y

V = 2e cos 2x cosh yz

a e sin 2x sinh y a e sin 2x cosh y axx

y

z

z

2) U = 2 2zcos

U =

U U

z za a a 1 U

= 2

z z zcos2 1

2 2 22 2a a a ( sin ) cos

= 2 z z zcos a a a2 2 2 22 sin cos

3) W = 10 r sin2 cos W =

W

r rW

rW

a a ar 1 1

sin

= 10 sin2 cos cos sin cosa ar 1

10 2r

r 1

10 2r

rsin

sin ( sin )

a

= 10 sin2 cos cos sin sin sina a ar 10 2 10

Example 1.17.8

Solution : Gradient of f in spherical system is,

f = f

r1r

f 1r sin

f a a ar

Hence verify the answer as,

f = 100 r sin cos 5 sin a 25r cos cos 2 sinr3 r 3

–

a 25r sin 5 cos

sin a3

–

Example 1.18.7

Solution : A = x y y2 2 2a a ax y z

A =

A

x

A

y

A

z

(x

x +

(y

y +

(y

z

x y z2 2 2

) ) )

= 2x + 2y + 0 = 2 (x + y) ... Divergence

A =

a a ax y z

x y z

x y2 2 y 2

= a a a 2y ax y z x[2y 0] [0 0] [0 0] … Curl




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Example 1.18.8

Solution :

1) P = x2yz a x + xz a z

Px = x2yz, Py = 0, Pz = xz

P =

Py

Pz

Pz

Px

Px

z yx

x zy

y

a a Pyx

z

a

= [0 – 0] a x + [x2y – z] a y + [0 – x

2z] a z = (x2y – z) a y – x

2z a z

2) Q = sin a + 2 z a + z cos a z

Q = sin , Q = 2 z, Qz = z cos

Q = 1 1

Q Q

z

Q

z

Qz z

a aQ )

1 Q

a z

= 1

0 0 1

32

( sin ) [ ] cosz a a 2 z 1

a z

=

z

z zsin

( cos )

2 3a a

3) T = 1

2rrcos sin cos cos a a ar

Tr

=

r 2 cos, T

= r sin cos , T

= cos

T = 1 1r

T T rT

sin

sin

sin

( )

a r

r1r

T

r

a +

1r

T

( )r

r

Tr

a

= 1 1

0r

rsin

(sin cos )sin ( sin )

sin (

a r

1r

) ( cos )

r

r a

+ 1 12

2r

r

r r

( sin cos )

( sin )

a

(sin cos )

= 1

2

2

( sin cos )=

12

2

(sin )=

12

2 2 cos = cos2

T = 1 2 1 1 2r

rr r

rsin

[cos sin sin ] [ cos ] sin cos

a ar

sin

r 2 a

= cos

sin sin

cossin cos

sin22

3

r r rr

a a

a




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Example 1.18.9

Solution : E a a ax y z yz xz xy

E =

E

x

E

y

E

zx y z 0 0 0 0

As E = 0, field E is solenoidal in nature.

E =a a ax y z

x y z

yz xz xy

= a a ax y zx x y y z z = 0

As E = 0, field E is irrotational in nature.Example 1.18.10

Solution : 1) A = yz 4xy ya a ax y z

A =a a ax y z

x y z

yz 4xy y

= a ya [4y – z] ax y z

2) B = z sin cosa a 3 2z … Cylindrical

B = 1

2

a a a z

zz sin 3 z cos 02

= – ( – )6 z cos a sin a 6z cos z cos a2 z

Example 1.19.2

Solution : A L dL

= ab bc cd da

DL = d d a a + dz a z … Cylindrical system

A = cos , A = sin , Az = 0 … From given A

For path ab, the direction is a hence A Ld = (sin ) ( d) A L

dab

= sinº

º

d 60

30

with = 2

= [ c o s ]ºº

6030

(2) = 2 [– 0.866 + 0.5] = – 0.732

For path bc, the direction is a hence A Ld = cos d A L d

bc

=

cos d

2

5

with = 30º




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A = 1 A Az

z a +

A A zz

a

+ 1 A 1 A

a z

Given : A = cos , A = 0, A z = 2

A = 0 0 0 2 0 1 sin

a a a z = 2 sin a a z

As the surface is in x - y plane, dS = d d a z

A SS

d = sin d d a az z … a a z 0

= sin d d cos 202 2

0

1

0

1

0

2

= 0 0 1 12

1

2

Example 1.19.4

Solution : The path L is shown in the Fig. 1.7.

F dL =AB BC CD DA F dL

F dL AB

= x + y xy2 2AB

i j dx i– 2

= (x + y )dx2 2

x = – a

a

= x

3 y x

32

x = – a

a

= 2a

3

3… y = 0 for AB

F dL BC

= x + y xy2 2BC

i j dy j– 2 = –2xy dyy = 0

b

= – 2xy2

2

y = 0

b

= –ab2 … x = + a for BC

F dL CD

= x + y xy2 2CD

i j dx i– 2 = (x + y ) dx2 2x = a

– a

= x

3 y x

32

x = +a

– a

= – – – –

a3

ab a

3 ab

32

32



x

y

z

A D

B C

x = – a

y = b

x = + a

y = 0

Fig. 1.7

z

y

x

1

1

az

Fig. 1.6


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= – –2a

3 ab

322 … y = + b for CD

F dL DA

= x + y xy2 2DA

i j dy j– 2 = –2xy dyy = b

0

= – 2xy

2

2

y = b

0

= – ab2 … x = – a for DA

F dL = 2a

3 ab

2a3

ab ab3

23

2 2– – – –2 = – 4ab2 … L.H.S.

F =

i j k

x y zx + y xy 02 2 2

= (– )2y – 2y k = – 4yk

( ) F dSS

= – 4y dxdy

S

k k ( ) =y = 0

b

x = – a

+a

– 4y dxdy

=

y = 0

b

x = – a+a– 4x ydy = –4 x

y

2–a+a

2

0

b

= – 4ab2

… Stoke's theorem is verified

Example 1.19.5

Solution : According to Stoke's

theorem,

L

d H L = S

d H S

Let us evaluate left hand side. The

integral to be evaluated on a

perimeter of a closed path shown inthe Fig. 1.8. The direction is a-b-c-d-a

such that normal to it is positive a zaccording to right hand rule.

H Ld =ab bc cd da

d H L

ab

d H L = x 2

526xy 3 y dx

a a ax y x



a

bc

d x = 2

x = 5

x

y = –1y = 1

z

Fig. 1.8


24/212

=x 2

5 2

2

5

6 xy dx 6y x

2

= 6y2

25 4 = 63 y

Now y = – 1 for path ab, ab

d H L 63 1 63

Similarly

bc

d H L = y 1

12

33

1

13y dy

3 y

3 y 1 1

= – 2

cd

d H L = x 5

2 2

5

2

6 xy dx 6 x

2 y

6y

2 4 25 63

y

But y = 1 for path cd hencecd

d H L = – 63

da

d H L = y 1

12 3

1

1 3 33y dy y 1 1 1 1 2

H Ld = 63 2 63 2 126 A

Now evaluate right hand side.

H =

a a ax y z

x y z

xy 3y 26 0

= a a a ax y z z0 0 0 0 0 6x 6x

S

d H S = S

6 x dx dy a az z

dS = dx dy a z normal to direction a z

S

d H S = y 1

1

x 2

5 2

2

5

11

6x dx dy 6 x

2 y

= 62 25 4 1 1 3 21 2 126 AThus both the sides are same, hence Stoke's theorem is verified.

Example 1.19.6

Solution : According to Stoke's theorem,

L

d H L = S

d H S




25/212

In spherical system,

dL = dr r d r da a ar sinThe closed path forming its perimeter is composed of three circular arcs. The first path 1 is

r = 3 , 0, 0 as shown in the Fig. 1.9. The second path 2 is r = 3, 90 ,

0 while the path 3 is r = 3, 90 , 0 . For all the three arcs r = 3 m.

Let us evaluate H Ld over these three paths.

H Ld =

Path1 Path2 Path3

H r d H r d H r d

sin

Now, H 0, H 0, H 0 sinr 1 ... Given HThus only second line integtal exists.

H Ld =

= 0

22r sin d 10 r sin 10 0 2sin / ... Path 2

= 10 3 90 22 sin = 47.1238 A ... r = 3 m, 90 for path 2

Now evaluate second side of Stoke's theorem.

H in spherical co-ordinates is,

= 1r sin

H sin H 1r

Hr

a r1

sin

(rH

r

)a

1r

(rH

r

Hr

)a

As H 0, H 0, H 10 sinr

H = 1r sin

sin 1r

(r10 sin

r

2

10 0 0

a r)

a a

1r

0 0



y

x

O r = 3

r = 3

r = 3

Path 1

Path 3

Path 2

rd a

rd a

rsin d a

d =Ld =L

d =L

= 0º = 90º

= 90º

Fig. 1.9


26/212

= 1r sin1r

10 2 10 sin cos sina ar

= 10r sin

10r

sin sin2 a ar

while dS = r sin d d2 a r ... As given in a r direction

H Sd = 10r sin r sin d d2

sin 2 ... a ar r 1

H SdS

=

= 0

2

= 0

2

r sin 2 d d/ /

10

= 10 22 10 20

20

2

02

r r

cos cos ( cos )

//

2

... r = 3 m

= 10 3 1

2

1

2 2

= 47.1238 A ... Thus Stoke's theorem is verified.





27/212


28/212

R1Q = C A a ay z

and R2Q = C B a ay z 4 3

R1Q = 1 1 22 2 And

R2Q = 4 32 2

= 5

F1 = Force on Q due to Q Q Q

R1

1

1Q2

4

a 1Q

and F2 = Force on Q due to Q QQ

R2

2

2Q2

4

a 2Q

Ft = F F a a1 2 1Q 2Q

Q Q

R

Q

R

1

1Q2

2

2Q24

=

Q Q

5

224

2 10

2 2

4 3

5

9

2

a a a ay z y z

= Q 7.071 10 Q12510 24 4 3

a a a ay z y z

Total z component of Ft is,

= Q

7.071 10 3 Q

125

10 2

4

a z

To have this component zero,

7.071 10 3 Q

12510 2 = 0 as Q is test charge and cannot be zero.

Q 2 = 7.071 10 125

3

10= – 29.462 nC

Example 2.2.8

Solution : Let the side of equilateral

triangle is d and is placed in x-y plane asshown in the Fig. 2.3.

l(AB) = l(BC) = l(AC) = d

l(CD) = d d

22

2

= 3

2

d

A (0, 0, 0), B (d, 0, 0),

C d

2 ,

3d2

, 0

Electromagnetic Field Theory 2 - 2 Coulomb's Law and Electric Field Intensity


y

x

z

AB

C

d d

d

P

Dd2

d2

Fig. 2.3


29/212

Key Point For equilateral triangle, the centroid is at a distance of 1

3 rd of height of

perpendicular drawn from any one corner to opposite side, from the side on which

perpendicular is drawn.

l(DP) = 1

3 l(CD) i.e. l(DP) =

1

3

3

2

d=

d

2 3= 0.2886d

Co-ordinates of centroid P d d2

0 2886 0, . ,

The charge at each corner is +Q. Let charge at P is QP. Then net force Ft on charge at A

due to all other charges is,Ft = F F FB C P

= QQ

4 R

QQ

4 R

Q Q

4 RBA2

CA2

P

PA2 0 0 0

a a aBA CA PA

a BA = R

|R |BA

BA

= d

d

a x = – a x , a CA = R

|R |CA

CA

= d d

d2

32

a ax y

a PA =

dd

d2

0 2886

0 5773

a ax y.

.

Ft = Q

d d

d d

2

0 242

32

a a ax x y

d2 +

Q Q P4 0

d2

a 0 . 2886 d a

(0.5773 d) (0.5773 d)

x y

2

= Qd

2

024

0 5 0 866

a a ax x y. . + Q Qd

P

42 5987 1 5

02

. .a ax y

= Qd

Q Q QP p4

1 5 1 50

2[ . [ . ] Q 2.5987 0.866] a ax y

For keeping all charges in equilibrium, Ft = 0

– 1.5 Q – 2.5987 QP = 0

QP = – 0.5773 Q

Thus charge at centroid P must be negative and 0.5773 times the charge Q.

Example 2.2.9

Solution : The square is kept in x-y plane with origin as one of its corners, as shown in

the Fig. 2.4.

The diagonals AC = BD = 8 m

Let AD = DC = BC = AB = l m




30/212

l l2 2 = 8 2

l = 5.656 mz

Hence the co-ordinates of various points are,

A (0, 0, 0), B (0, 5.656, 0), C (5.656, 5.656, 0),

D (5.656, 0, 0)The point E is centroid hence E (2.828, 2.828, 0).

The point P is 3 m above the centre E hence theco-ordinates at P are (2.828, 2.828, 3).

To find force on charge at P which is

Q 150 C2 due to charges at A, B, C and D of Q 30 C1 each. FP = F F F FA B C D

FA = Q Q

4 R

Q Q

4 R |

1 2

0 AP2

1 2

0 AP2

a R

R

APAP

A

P|

, FB = Q Q

4 R

Q Q

4 R |

1 2

0 BP2

1 2

0 BP2

a R

R

BPBP

BP

|

FC = Q Q

4 R

Q Q

4 R |1 2

0 CP2

1 2

0 CP2

a R

RCP

CP

C

P|

, FD = Q Q

4 R

Q Q

4 R |1 2

0 DP2

1 2

0 DP2

a R

RDP

DP

D

P|

RAP = (2.828 0) (2.828 0) (3 0) 2.828 a a a ax y z x y za a 2.828 3

RBP = (2.828 0) (2.828 ) (3 0) 2.8 a a ax y z5656. 28 2.828 3a a ax y z

RCP = (2.828 ) (2.828 ) (3 0) 5 656 5 656. .a a ax y z 2.828 2.828 3 a a ax y z

RDP = (2.828 ) (2.828 ) (3 0) 2. 5 656 0. a a ax y z 828 2.828 3a a ax y z

| |RAP =| |R R RBP CP DP| | | | (2.828) (2.828) 32 2 2 5

F F F FA B C D = Q Q

4 (5)[ ]1 2

03

R R R RAP BP CP DP

FP = 30 10 150 10

4 8.854 10

6 6

12 35[12 ]

a 3.8827 a Nz z

Example 2.2.10

Solution. : The arrangement of charges is shown

in the Fig. 2.5. The charge at P is test charge i.e.

QP = 1C.

F1P = Q Q

R

P1

0 124

a R1 ... Force due to Q1

R1 = 1 1 1 2 0 0a a ax y z= – 2 a x – a y

R1 = R1 2 1 52 2



z

y

x

0

P (–1, 1, 0)

Q (1, 2, 0)1Q (2, 0, 0)2 R2

R1

Fig. 2.5

z

x

y A B

P

C

D E At A, B, C, D

Q = 30 C1

Q = 150 C2

l

l

(AC) = 8 m(BD) = 8 m

Fig. 2.4


31/212


32/212


33/212

F3 = 1 10 1 10

4 8.854 10 12

3 3

12

a x 3 2 2

12

a ay z

= 216.211 374.489 611.538 Na a ax y z

The total force on charge at P4

is,

F = F F F a a1 2 3 z z3 611.446 1834.338 N i.e. |F| = 1834.338 NThe magnitude of force on each charge remains same as above due to symmetrical

distribution of charges.

Example 2.2.13

Solution : The arrangement is shown in the

Fig. 2.8.

The co-ordinates of the vertices of triangle are,

Point O (0, 0, 0)Point Q (0.1, 0, 0)

Point P (0.05, 0.0866, 0).

Let us find the force on P due to the charges

at O and Q.

F1 = Q Q

4 R

Q Q

4 R |1 2

0 OP2

1 2

0 OP2

a R

ROP

OP

OP

|

ROP = 0.05 0.0866a ax y , | | (0.05) (0.0866) 0.12 2ROP

F1 = 0.25 10 0.25 10

4 8.854 10

6 6

12

0.1

[0.05 0.0866 ]

0.12

a ax y= 0.02808 0.0486 Na ax y

F2 = Q Q

4 R

Q Q

4 R | |1 2

0 QP2

1 2

0 QP2

a R

RQP

QP

QP

RQP = (0.05 0.1) 0.0866 0.05 0.0866 a a ax y x a y

|RQP| = (0.05) (0.0866)2 2 = 0.1

F2 = 0.25 10 0.25 10

4 8.854 10 0

6 6

12

.1

[ 0.05 0.0866

0.12

a a ]x y= 0.02808 0.0486 Na ax y

F = F F 0.09729 a N1 2 y … Direction a y

|F| = 0.09729 N … Magnitude



y

xO

d

10 cm 10 – 52 2

= 8.66 cm

Q

5 cm

10 cm

5 cm 0.25 C0.25 C

0.25 CP

R

Fig. 2.8


34/212

Example 2.3.6

Solution : E1 = Q

4 R

1

0 12

a R1

a R1

= R

R

a a

R

x z1

1

0 1 1 0

1

= a ax z

2

E1 =

8

4 2 20

02

a ax z = 12

a ax z

E2 = Q

4 R

2

0 22

QR2

a R2 = R

R

a a

R

a a2

2

x

2

x z0 1 1 0 z

2

E2 =

44 2 2

0

02

a aa ax z x z

1

2

E = E E 0.3535 a 1.0606 a1 2 x z V m

Example 2.3.7

Solution : The various points and charges are

shown in the Fig. 2.10.

The position vectors of points A, B and P are,A = 2 a x , B a x 2

P = a a ax y z 2 2EA is field at P due to Q 1, and will act along aAP .

EB is field at P due to Q 2 and will act along a BP .

EA = Q

R

Q

R

1

AP2

1

AP24 4

a P A

P AAP

EB = Q

R

Q

R

2

BP2

2

BP2

4 4

a P B

P B

BP

E at P = E EA B

= 14

Q

R

Q

R

1

AP2

2

BP2

P A

P A

P B

P B

=

1

4

1 2 2

9 1 2 2

3 2 2

17 3 22 2 2 2

2

2 2 2

a a a a a ax y z x y zQ

2 2



Q1

O

Q2

R AP

RBPa AP

aBP

A(2, 0, 0)

B(–2, 0, 0)

P(1, 2, 2)

y

x

z

Fig. 2.10

P (0, 0, 1)

(–1, 0, 0)Q1

Q2

R2

R1

B

A

(1, 0, 0)

y

z

x

Fig. 2.9


35/212


36/212


37/212

y component of EA i.e. E EA y A cos 45

Similarly l (OB) = 2 m, l (OP) = 2 mhence PBO = 45

y component of EB i.e. E EBy B cos 45

But EAy is in a y direction while EBy is ina y direction. From symmetry of thearrangement E EAy By . Hence they cancel

each other.

While z components of EA and EB help

each other as both are in a z direction.

EAz = E E E aBz A B z or sin 45Similarly there are 4 more pairs of charges

which will behave identically and their y

components are going to cancel while z components are going to add.

Thus total z component of E at P is,

Etotal = (E due to any charge) 10 45sin a z = Q

R2410 45

sin a z

where R = 2 2 82 2

Etotal =

500 10

4 810 45

6

2

sin a z = 3.972 10

6 a z V/m

FP = Q 10 3.972 10P6 6E atotal z

20 = – 79.44 ( )a z N

This is the force on the charge at P. In general, force acts normal to the plane in which

circle is kept, i.e. – 79.44 a n where a n is unit vector normal to the plane containing the

circle.

Example 2.3.11

Solution : a) A (2, –1, 3) and P (0, 0, 0)

E at P = Q

RAP24 aAP

Now aAP =

0 2 0 1 0 3

2 1 32 2 2

a a ax y z

E =

5 10

4

2 3

14

9

8.854 10 1412 2

a a ax y z

= – 1.715 a x + 0.857 a y – 2.573 a z V/m



Oy

x

z

45º 45º

2 2

2

ABQ

Q

QQ Q

Q

Q Q Q

Q

R R

P(0,0,2)EAy

EA

EBy

EB

45º45º

Fig. 2.13


38/212

b) Let point P is now (x, 0, 0).

aAP =

r

r

a a aAP

AP

x y z

x 2

x 2 1 32 2 2

3

E =

Q

x 2 1 9

x 2

x 2 1 92 24

3

a a ax y z

=

5 10

x 2 10

x 29

2 3/ 2

4

3

a a ax y z

=

44.938

x 2 10

x 22 3/ 2

a a ax y z3

| |E =

44.938

x 2 10

x 2 1 32

3/ 22 2 2

=

44.938

x 2 102 V/m

To find x at which| |E is maximum,

d

dx

| |E= 0

44.938

2 x 2

x 2 102 2 = 0

(x – 2) = 0

x = 2 where| |E is maximum.The graph of | |E against x is shown in the Fig. 2.14.

c) Hence| |maxE is at x = 2,

| |maxE = 44.938

10 = 4.4938 V/m

Example 2.3.12

Solution : E = Q

R024

a R

a R = R

R

QP

QP

= P Q

P Q

P Q = 0 2 0 2 0 01 2 3 2 5. . . . .a a ax y z= 0 4 01 0 2. . .a a ax y z



4.49| |E max

| |E in V/m

10 –10 20x

Fig. 2.14

y

z

xaR

P

Q5 nC

Fig. 2.15


39/212

a R =

0 4 01 0 2

0 4 0 1 0 22 2 2

. . .

. . .

a a ax y z

= 0 4 01 0 2

0 45825

. . .

.

a a ax y z

= 0 8728 0 2182 0 4364. . .a a ax y z R = P Q = 0.45825

E =

5 10

4 8 854 10 0 45825

9

12 2

. . a R = 214 a R

Substituting value of a R ,

E = 186.779 a 46.694 a 93.389 ax y z V/m … E at P

Example 2.3.13

Solution : The arrangement is shown in the Fig. 2.16. Letthe charges are placed at A, B and C while E is to be

obtained at fourth corner O.

E at 'O' = E E EA B C

= Q

4 R

Q

4 R

Q

4 R0 A2

0 B2

0 C2

a a aRA RB RC

RA = – 0.05 a x , RA = 0.05, a RA = – a x

RB = – 0.05 – 0.05a ax y , RB = 0.0707,

a RB = – 0.707 – 0.707a ax y

RC = – 0.05 a y , RC = 0.05, a RC = – a y

E = Q4

–

(0.05)

– 0.707 – 0.707

0.07070 2( ) ( )

( )

a a ax x y

2

( )–

(0.05) 2a y

= 100 10

4 8 854 10400 141 4 141 4 40

9

12

–

–.– – . – . –

a a ax x y 0 a y

= – . – .486 6 486 6a ax y kV/m, |E| at 'O' = 688.156 kV/mExample 2.3.14

Solution : The arrangement is shown in the Fig. 2.17.

F3 = F F13 23 = Q Q

4 R

Q Q

4 R

1 3

0 132

2 3

0 232

a aR13 R23

R13 = – 3 2a a ax y z , R13 = 14, a R13 = R

R13

13



x

z

y

AB

CO

RA

RC

RB

(0,0.05,0)(0,0,0)

(0.05,0,0) (0.05,0.05,0)

Fig. 2.16

x

z

y

R13

(–1, ,4) –1

(0,3,1)

R23

Q = – 2 mC2

Q = 1 mC1

Q = 10 nC3

(3,2, ) –1

Fig. 2.17


40/212

R23 = a a ax y z 4 3– , R23 = 26, a R23 = R

R23

23

F

a a a3

x y z14

1 10 10 10

14 14

3 9

2

– –

( )

– 3 + + 2

–

( )

– –2 10 10 10

26 26

3 9

2

a a ax y z+ 4 – 3

= – 6.503 a 3.707 a 7.4985 ax y z– mN

E3 = F

Q3

3

= – 650.3 a 370.7 a 749.85 ax y z– KV/m

Example 2.4.4

Solution : Given : v = 10 z e y C m2 0.1 x 3 sin .

Consider differential volume in cartesian system as, dv = dx dy dz

dQ = v dv = 10 z e y2 0.1 x

sin dx dy dz Q =

vol

v dv

But now it becomes triple integration

Q =z 3

4

y 0

1

x 2

22 0.1 x10 z e y

sin dx dy dz

=

z 3

4

y 0

12

0.1 x

2

2

10 z y e

0.1 dy dz

sin =

z 3

42

0.2 0.210 z

y e0.1

e0.1

cos

0

1

dz

= 10 0z

3

3

3

4

cos cos

4.0267 = 10 4 3

31 13 3

4.0267 = 316.162 C

Example 2.4.5

Solution : i) 0 < x < 5 m, L = 12x mC m2

Q = L dL = 12 x dx mC20

5

= 12 x

3

3

0

5

= 500 mC = 0.5 C

ii) S = z nC m2 2 , = 3, 0 < z < 4 m

Q = SS

dS = SS

d dz =

z

2 –9z 10 d dz

0

4

0

2

... = 3

= 3 10 z3

2 –902

3

0

4

= 1.206 C

iii) v = 10rsin

C m 3

, r = 4 m




41/212

Q =vol

v dv =vol

v2r sin dr d d =

0

2

0 0r

210r sin

r sin dr d d

= 102

2

0

4

0 02r

= 1579.136 C.

Example 2.4.6

Solution : n e = 1000

r cos

4

electrons/m 3

1 electron = 1.6 10 19 C charge

v = n e charge on 1 electron = 1.6 10

r cos

4 C / m

163

The volume is defined as sphere of r = 2 m.

dv = r sin dr d d2 ... spherical system

Q =vol

vr 0

2 162dv

1.6 10r

r

0

2

04

cos sin dr d d

=

1.6 10 r

2 cos

sin4

14

162

0

2

0

0

2

= 1.6 10 2 2 4 1 2.56 1016 15 C

Example 2.6.6Solution : i) For origin let r = r1

E =

L

r2 0 1a r1

Point on the line is (x, 3, 5). Origin is (0, 0, 0)

Do not consider x co-ordinate as the charge is parallel

to x-axis.

r1 = (0 – 3) a y + (0 – 5) a z

= – 3 a y – 5 a z , |r 1| = 34

E = 30 10

2 34

3 5

34

9

8.854 10-12a ay z

= – 47.582 a y – 79.303 a z V/m

ii) P(5, 6, 1)

r2 = (6 – 3) a y + (1 – 5) a z = 3 a y – 4 a z , |r2 | = 5



z = 5

z

O y = 3 y

x

Parallelto x-axis

Fig. 2.18


42/212

E = 30 10

2 8.854 10 5

3 4

5

9

12

a ay z= 64.711 a y – 86.2823 a z V/m

Example 2.6.7

Solutino : a) The line charge is shown in the

Fig. 2.19.It is parallel to the x axis as y = 1 constant and

z = 2 constant. The line charge is infinite hence

using the standard result,

E =

L

2 r a r

To find a r , consider a point on the line charge

(x, 1, 2) while P (6, –1, 3). As the line charge is

parallel to x axis, do not consider x coordinate while finding a r .

r = 1 1 3 2 2a a a ay z y z , r = 2 1 52 2

a r = r|r|

a ay z

2

5

E =

L

122 8.854 105

2

5

24 10 2

2

9

a a a ay z y z

5

= – 172.564 a y + 86.282 a z V/m

b) Consider a point charge QA at A (–3, 4, 1).

The electric field due to QA at P (6, –1, 3) is, EA = Q

R

A

AP24

aAP

RAP = 6 3 1 4 3 1 9 5 2 a a a a a ax y z x y z , RAP = 10.488

aAP = R

R

a a aAP

AP

x y z

9 5 210.4888

EA

=

Q

10.4888 10.4888

A

24

9 5 2

a a ax y z

The total field at P is now, Et = E+ EA

The y component of total Et is to be made zero.

172.564 Q

10.4888

A3

5

4 a y = 0 i.e.

5

4

Q

10.4888

A3

= – 172.564

QA = 172.564 8.854 10 10.488812 34

5

= – 4.4311 C



r P(6,–1,3)

1

2 –

O y

x

z

Fig. 2.19


43/212

Example 2.6.8

Solution : The charge is shown in the Fig. 2.20.

Key Point Charge is not infinite hence basic method of

differential charge dQ must be used.

dQ = L dL = L dz

dE = dQ

4 R2 0a R =

L dz

4 R20

R

|R |

i) To find E at origin

R = – z a z , |R| = z, a R = – a z

dE =

L

o

dz( )

4 z 2a z i.e. E =

L

o4dz

z 2z 1

3

a z

E =

20 10

4

19

1

3

8.854 10 12 z z

z

a z = – 119.824 a z V/m

ii) To find E at P(4, 0, 0)

R = (4 – 0) a x + (0 – z) a z = 4 a x – z a z , |R| = 16 2 z

Example 2.6.9

Solution : The line is shown in the Fig. 2.21.

The line with x = – 3 constant and y = 4 constant

is a line parallel to z axis as z can take any value.The E at P (2, 3, 15) is to be calculated.

The charge is infinite line charge hence E can be

obtained by standard result,

E =

L

2 r a r

To find r, consider two points, one on the line

which is (–3, 4, z) while P (2, 3, 15). But as line is

parallel to z axis, E cannot have component in

a z direction hence z need not be consideredwhile calculating r.

r = 2 3 3 4 5 a a a ax y x y ... z not considered

| |r = 5 1 262 2

a r = r

r

a ax y

| |

5

26



z

z = 3

dQ (0,0,z)

z = 1

y

x

Fig. 2.20

–34

O

P(2,3,15)

–

z

y

x

ar

r

Fig. 2.21


44/212

E =

L

2 8.854 10

1

26

5 25 10 5

2

9a a

26

a ax y x y

12 26

= 86.42 a x 17.284 a y V/m

Example 2.6.10


The charge is parallel to x-axis hence E cannot have anycomponent in x direction hence do not consider x whilecalculating E.

i) E at P(0, 0, 0)

r = (0 3) (0 5) a ay z = 3 5a ay z

r = | r | 3 5 342 2

E = L0r L

02 r 2 ra r| r|

= 30 10

2 8.854 10 34

3 5

34

9

12

a ay z

47.58 a 79.3 a V my z

ii) E at P(0, 6, 1)

r = (6 3) (1 5) 3 4 , 3 a a a a |r|y z y z2 24 5

E = 30 10

2 8.854 10 5

3 4

5

9

12

a ay z

64.71 a 86.28 a V my z

iii) E at P(5, 6, 1)As E does not have any component in x direction and y, z, co-ordinates are same as in (ii)hence E also remains same as obtained in (ii).

E = 64.71 a 86.28 a V my z

Example 2.6.11


Key Point As charge is along z-axis, E can not have any

component in a z direction.

Do not consider z co-ordinate while calculating r . r = ( 2 0) (2 0) a ax y

= 2 2 , 4 4 8a a rx y

E =

L

0

L

02 r 2 ra

r

rr

= 40 10 2 a 2 a

2 8.854 10 8 8

9x y

12

179.754 a x 179.754 a V my



z

x

y

(x, 3, 5)5

L = 30 nC/m

3

Parallel tox-axis

Fig. 2.22

zL = 40 nC/m

P(–2,2,8)

x

(0,0,z)

Fig. 2.23


45/212

Example 2.6.12

Solution : Consider the charge along z-axis as shown in

the Fig. 2.24. Consider the differential charge at a

distance z.

dQ = L dl = L dz

dE =

L

02

dz

4 Ra R

R = 0 ( h 0) (0 z)a a ax y z

= h za ay z

| |R = h z| |

2 2R , a

R

R

dE =

L

0 2 2 2 2

dz

4 (h z )

h z

h z

a ay z

E =

L

0 z1

z2

2 2 3/ 2z1

z2

2 24

h dz

(h z )

z dz

(h z

a ay z

) 3/ 2

I1

I2

…(1)

I 1 =z1

z2

2 2 3/ 2

h dz

(h z )

, z = h tan , dz = h sec2 d

I 1 =z1

z2 2 2

3

h sec d

h sec 3

= 1h

z1

z2

cos d = 1h z1

z2[sin ]

= 1

hz

h z2 2 z1

z2

= 1h

z

h z

z

h z

2

222

1

212

I2 =

z1

z2

2 2 3/ 2

z dz

(h z )

, h2 + z2 = u2, 2z dz = 2u du

I2 =z1

z2

3

u du

u =

1u z1

z2=

1

h z2 2 z1

z2

=

1

h z

1

h z

1

h z

1

h z222 2

12 2

22 2

12

–

Using I1 and I2in equation (1),



P(0, –h,0)

(0, 0, z )2

(0, 0, z )1

R

y

A

B

z

dl

x

z

(0, 0, z)

Fig. 2.24

h

zh +z2 2

Fig. 2.24 (a)


46/212

E =

L

0

2

222

1

212

L

0 24

z

h h z

z

h h z 4

1

ha y

z

1

h z22 2

12

a V / mz

Example 2.6.13

Solution : Q = 1 C and placed betweenA(0, 0, 1) and B(0, 0, 2) m. L = 2 – 1 = 1 m

L

= Q

L =

11

= 1 C/m

Consider an elementary charge dQ at a distance

z as shown in the Fig. 2.25.

dQ = L

dz

i) For point P1(0, 0, 0),

R = za z , a R = a z| |R = z

dE = dQ

4 R02

a R =

L

02

dz

4 z( ) a z

E =z 1

z 2L

02

dz

4 z( )

a z

=

L

0z 1

2

24dz

z

a z

=

1 10

4 10

6

12 8.854

1z 1

2

a z = 8987.7424 1

2

1 a z

= – 4493.8712 a z V/m

ii) For point P2(0, 1, 1)

R = 0 (1 0) (1 z)a a ax y z , | |R = 1 (1 z)2

dE = dQ

4 R02

a R = dQ

4 R02

R

|R |

=

L

0 2

dz

4

[ + (1 z) ]

1 + (1 z)[ ( ) ]1 1 2

z

a ay z

dE = 1 10

4

6

8.854 10

dz

[1 (1 z) ]

(1 z) dz

[112 2 3/ 2

a ay z

(1 z) ]2 3 / 2

E = dE = 8987.7424z 1

2

2 3/ 2 2 3/

dz

[1 (1 z) ]

(1 z) dz

[1 (1 z) ]

a ay z2



(0, 0, z)

(0, 0, 2)

(0, 0, 1)

P (0, 1, 1)2

P (0, 0, 0)1

y

z

x

z

B

A

dz

Fig. 2.25


47/212

I 1 =z 1

2

2 3 / 2

dz

[1 (1 z) ]

put 1 – z = tan , – dz = sec2 d

For z = 1, = 0 and z = 2, 2 = – 45

I1 =

1

2 2

3/ 2sec d[1 ] tan 2 =

1

2

1 dsec =

1

2

dcos

I1 =

[sin ]0

45= [sin ( )]45 = + 0.7071

I2 =

z 1

2

(1 z) dz

[1 (1 z) ]2 3 / 2

put [1 + (1 – z)2] = u2

2(1 – z) (– dz) = 2u du i.e. (1 – z)dz = –u du

for z = 1, u1 = 1 and z = 2, u2 = 2

I2 =u 1

u 2

3

u du

u

=

1u 1

2

= 1

21

= – 0.2928

E = 8987.7424 [0.7071 a y – 0.2928 a z ] = 6355.2326 a y – 2631.6109 a z V/m

Example 2.6.14

Solution : The charge is shown as in the

Fig. 2.26.

Key Point If L is not distributed all along thelength then standard result can not be used. The

basic procedure is to be used.

As charge is not infinite, let us use basic

procedure of considering differential charge.

Consider the differential element dl in the z

direction hence,

dl = dz

dQ = L

dl = L

dz

dE = dQ

R

dz

4 R2L

24

a aR R

Any point on z axis is (0, 0, z) while point P at which E to be calculated is 2 0 0, , .R = 2 0 0 z z a a a ax z x z2

|R| = 2 z 4 z2 2 2 , a R = R

|R |

a ax z

2 z

4 z 2



P(2,0,0)

xdE

– 5

5

y0R

aR

dQ

L

L

z

Fig. 2.26


48/212

dE =

L

22 2

dz

4 z

z

4 z4

2

a ax z =

L

2 3/ 2

dz

4 z

z

4

2

a ax z

Now there is no charge between – 5 to 5 hence to find E, dE to be integrated in two zones

to – 5 and 5 to in z direction.

E =

5

5

d dE+ E

Looking at the symmetry it can be observed that z component of E produced by charge

between 5 to will cancel the z component of E produced by charge between – 5 to .Hence for integration a z component from dE can be neglected.

E =

5

5

2

4

2

4

L

2 3/ 2L

2 3/ 2

dz

4 z

dz

4 z

a ax x

Solving, E = 13 V ma xTo find cylindrical co-ordinates find the dot product of E with a ar , and a z , at point P,

referring table of dot products of unit vectors.

Er = E a a ar x r 13 13 cos E = E a a ax 13 13 sin Ez = E a a az x z 13 0At point P, x = 2, y = 0, z = 0

r = x y2 2

= 2 and tan tan

1 1

0 0

y

x

cos = 1 and sin = 0

Er = 13, E = 0, Ez = 0

Hence the cylindrical co-ordinate systems E is,

E = E E Er za a ar z = 13 a r V/m

Example 2.7.2 Kept this example for student's practice.

Example 2.8.5

Solution : Case 1 : Point charge Q 1 = 6 C at A (0, 0, 1) and P (1, 5, 2)

E1 = Q

R

Q

R

1

AP2

1

AP24 4

a R

| RAP

AP

AP

|

RAP = 1 0 5 0 2 1 a a ax y z = a a ax y z 5

| |RAP = 1 5 1 272 2 2




49/212

E1 =

6 10

4

56

8.854 10 2712 2

a a a

27

x y z

E1 = 384.375 a x + 1921.879 a y + 384.375 a z V/mCase 2 : Line charge L along x-axis.

It is infinite hence using standard result,

E2 =

L

2 r a r =

L

2 rr

r

Consider any point on line charge i.e. (x, 0, 0) while P (1, 5, 2). But as line is along x-axis,

no component of E will be along a x direction. Hence while calculating r and a r , do not

consider x co-ordinates of the points.

r = 5 0 2 0 5 2 a a a ay z y z

| r| = 5 2 292 2

E2 =

L

2 8.854

29

5 2 180 10 5 2

1

9a a

29

a a

2

y z y z

0 2912

= 557.859 a y + 223.144 a z V/m

Case 3 : Surface charge S over theplane z = – 1. The plane is parallel to xyplane and normal direction to the planeis a an z , as point P is above theplane. At all the points above z = – 1plane the E is constant along a zdirection.

E3 =

S

2 a n

= 25 10

2 8.854 10

9

12

a z

= 1411.7913 a z V/m

Hence the net E at point P is,E = E E E1 2 3

= 384.375 a x + 1921.879 a y + 384.375 a z + 557.859 a y + 223.144 a z + 1411.7913 a z

= 384.375 a x + 2479.738 a y + 2019.3103 a z V/m

Example 2.8.6

Solution : The sheet of charge is shown in the Fig. 2.28.

Consider the differential area dS carrying the charge dQ. The normal direction to dS is a zhence dS r dr dz .



z

x

yaz

S

P(1, 5, 2)

Fig. 2.27


50/212

dQ = S dS = S r dr d = 10 4

r

r dr d

dQ = 10 4 dr d

dE = 10 4 dr d

4 R2

a R

Consider R as shown in the Fig. 2.29, which has

two components in cylindrical system,

1. The component along a r having radiusr i.e. r a r .

2. The component z = 3 along a z i.e. 3 a z .

R = r a ar z3

|R| = r 3 r 92 2 2

a R = R

|R |

a ar z

r

r 923

dE = 10 34

dr d

4 r 9

r

r 922 2

a ar z

It can be seen that due to symmetry about

z-axis, all radial components will cancel each

other. Hence there will not be any component

of E along a r . So in integration a r need not beconsidered.

E =

0 0

4 4103

r 2 3/ 2

dr d

4 r 9

a z

As there is no r dr in the numerator, use

r = 3 tan , dr = 3 sec2 d

For r = 0, = 0For r = 4, = tan– /1 4 3

...Change of limits

E =

0 0

4 2

2 3 2

10 3

4 9

3sec

tan/

d da z

E =

0 0

2 2

2 3 21

299.5914 10 d d3 sec

tan/

a z



R

aRr = 4 O

P(0, 0, 3)

S

dS

y

x

Fig. 2.28

R

O

P(0, 0, 3)3

y

x –ar

az

r

Fig. 2.29


51/212

=

0 0

299.5914 10d d

3

sec a z =

0 1 0

2

299.5914 10 d d3 cos a z

= 299.5914 10 3 02

0

sin a z ... Separating variables

= 1.8823 10 6 sin a z ... sin 0º = 0

Now = tan1 4

3 i.e. tan

43

sin = 45

= 0.8

E = 1.8823 10 0.86 a z

= 1.5059 10 6 a z V/m

= 1.5059 a z MV/mExample 2.8.7

Solution : The sheet is shown in the Fig. 2.31.

The point P is on the back side of the plane.

The normal to the plane in the direction of P is a x .

a n = a x

EP =

S

0

12

1225 10

2 8.854 10

( )a an x

EP = – 0.2823 a x V/m

Example 2.8.8

Solution : Q = 100 C, r = 10 cm = 0.1 m, area = r 2 = 0.03141 m2

S = Qarea

= 100 10

0.03141

– 6= 3.1831 10 m– 3 2 C

The disc is shown in the Fig. 2.32.

Consider differential surface area dS.

Using cylindrical system,dS = r dr d, R = – r zr za a

a R = – r a za

r z

r z

2 2

Key Point All radial components of E

at P will cancel each other due to

symmetry.



4

3

2

4 +3= 5

2 2

Fig. 2.30

– ax

x

O y

z

P (–5, 0, 0)

S = 5 pC/m2

Sheetat x = 0

Fig. 2.31

z

y

x

R

P

z

r

ds

P

Rzaz

– r a r

0

(a) (b)

Fig. 2.32


52/212

E =S

dQ

4 R02

a R

=

0 0 02 2 2 2

[r dr d

4 [r + z ]

z

r z

2 0 1

r

S. ] a z =

S

r

z

4r dr d

[r + z ]0 0 02 2 3/ 2

2 0 1.

a z

Use r z u2 2 2 i.e. r dr = u du

Limits : r = 0, u z1 and r = 0.1, u 2 012 2. z

E =

S

u

uz

4u du d

u0 03

2

1

2

a z =

S

z

u4 0 u 1

u 2[ ] –

02 1

a z

=

S

z

u u4 0

2 1 1

1 2– a z

Using z = 20 cm = 0.2 m, u 1 = 0.2 and u 2 = 0.05 = 0.2236

E = 3.1831 10 0.2

4 8.854 10

10.2

– 1

0.2236

– 3

–12

2 a z = 18.9723 a z MV/m

Example 2.8.9

Solution : The plane is shown in the Fig. 2.33

Consider the differential surface area dS carrying

charge dQ.

dQ = S dS where dS = dxdy

dQ = 2 x y 92 2 3/2 dx dy nC

dE = dQ

4 Ro2

a R

R = 0 – x 0 – y 0 – –3a a ax y z

R = –x – y 3a a ax y z ,

| | = x y 92 2R , a R = RR| |

dE =

2 x y 9 dx dy

4 x y 9

–x – y 3

x y

2 2 3/ 2

o2 2 2 2

a a ax y z

910 9–

Due to symmetrical distribution, x and y components of dE will cancel each other and

only z component will exist.



z

y

x

O

P Q (2,2,–3)

S(–2, 2, –3)

R

z = –3 plane

(2,–2,–3)

dS

Fig. 2.33


53/212

dE = 6 10

4 dx dy

–9

o

a z

E = 6 104

dx dy 6 10

4 [x]

–9

ox – 2

2

y = –2

2 –9

o –

a z 2

2–22[y] a z = 862.82 a z V/m.

Example 2.8.10

Solution : The sheets are shown in the

Fig. 2.34.

E =

S

2 0a N

i) PA = (2, 5, – 5)

It is below the plane z = – 4.

Hence a N for this point due to all the

sheets is –a z .

Et =

S1 S2 S3

2 2 20 0 0–a –a –az z z = –56.47 a z V/m

ii) PB = (4, 2, –3)

It is above z = – 4 and below other two plane. Hence a aN z for S1 and –a z for S2and S3 .

Et

=

3 10

2

6 10

2

8 10

2

9

0

9

0

9

0

– – ––

a –a –a

z z z = 282.358 a

z V/m

iii) PC = (–1, –5, 2)

It is above z = 1 and below z = 4. Hence a N = a z for S1 and S2 while –a z for S3 .

Et =

3 102

6 102

8 10

50999882vakx_electromagnetic field theory_solution manual

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