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305 homewoek solution, Assignments of Electromagnetism and Electromagnetic Fields Theory

electromagnetic theory homework 1 solution

Typology: Assignments

2024/2025

Uploaded on 03/23/2025

yulieth-larobardiere
yulieth-larobardiere 🇺🇸

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Prob. 4.1
2
1
1
12
12
12 3
39 3
()
20(10 )[(3,2,1) ( 4,0,6)] (7,2, 5)
180 10
10 688.88
44 (3,2,1) ( 4,0,6)
36
1.8291 0.5226 1.3065 mN
Q
Q
Q
QQ
xyz
QQ





rr
Frr
aaa
Prob. 4.2 (a)
12
33
00
12
3/2
3
00
12
3/2 3/2
00
3/2
2
1
[(5,0,6) (4,0, 3)] [(5,0,6) (2,0,1)]
(5,0,6) 4 | (5, 0,6) (4, 0, 3) | 4 | (5, 0,6) (2, 0,1) |
(1, 0, 9) (3,0, 5)
4 4 (34)
(82)
If 0, then
95
11
0
4 (82) 4 (34)
582 5
()
934 9
z
QQ
QQ
E
QQ
QQ
 
 
 






E
3/2
82
4( ) nC
34
8.3232 nC
(b)
12
3/2 3/2
00
3/2 3/2
12
1
(5,0,6) (5,0,6)
If 0, then
30
4 (82) 4 (34)
82 82
3 ( ) 12( )
34 34
44.945
x
q
F
qQ qQ
QQ nC
QnC
 

 

FE
pf3
pf4

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Prob. 4.

1 2 1 1 2

12 1 2 3 3 9 3

( ) 20(10 )[(3, 2,1) ( 4, 0, 6)] (7, 2, 5)

1.8291 0.5226 1.3065 mN

Q^ Q Q Q Q

x y z

Q Q

  

r r F r r

a a a

Prob. 4.2 (a)

1 2 3 3 0 0

1 2 3 3/ 2 0 0

1 2 3/ 2 3/ 2 0 0

3/ 2 1 2

[(5, 0, 6) (4, 0, 3)] [(5, 0, 6) (2, 0,1)]

If 0, then

9 1 5 1 0 4 (82) 4 (34)

5 82 5 ( ) 9 34 9

z

Q Q

Q Q

E

Q Q

Q Q

E

4 ( ) nC 34

 8.3232 nC

(b)

1 2 3/ 2 3/ 2 0 0

3/ 2 3/ 2 1 2

1

If 0, then

3 0 4 (82) 4 (34)

82 82 3 ( ) 12( ) 34 34

x

q

F

qQ qQ

Q Q nC

Q nC

F E

Prob. 4.

y

y=3x/2, z=4 plane

3

0 2 x

2 3 / 2 2 2

0 0 2 2 2 5 2 2 2

0 0

10 mC 10(4) 4

40 20 20(9 / 4) 9(32) 288 mC 2 0 4 5 0

x

s S x y

x x

Q dS x yzdxdy x ydydx z

y x x x dx x x dx

 

 

Prob. 4.

2 v^4 cos^ nC v

Q    dv    z  d   d dz

2 1 / 4 (^2) 3 4

0 0 0

4 cos (sin ) 0 2 0 0

z d zdz d

 (16)(0.5)(sin / 4) 5.657 nC

Prob. 4.

(a)

1 3 2

0

12 4 nC 3 0

L L

x

Q   dl  x dx  

(b) Method 1:

3

2 2

2

2 2 3/ 2

2

3 2

(0, 0, h) - (x, 0, 0) = (-x, 0, h), | |

Since x<<h, we can ignore the x-component.

L

z z

dl

r

r h x

x x h dx

h x

h x dx Q

h h

E r

r = r

E

E a a

Method 2:

Prob. 4.

(a) From eq. (4.26), 9

9

12 10 678.58^ V/m, z> 18 (12) 2 10 678.58 V/m, z< 2 36

s^ z n n n o z

a E a a a a

(b) Similarly, 9

9

12 10 678.58^ V/m, z> 18 (12) 2 10 678.58 V/m, z< 2 36

s^ z n n n o z

a E a a a a

For z > 4 and z < 0, the fields cancel out. For 0 < z < 4, they add up. Thus

E  2 678.58 a (^) z V/m= 1.357 a z kV/m

Prob. 4.

_

9

9 0

Let ( , ) 2 5; 2

Since point ( 1, 0,1) is below the plane,

151.7 303.5 V/m

x y

x y n

x y n

s x^ y n

x y

f x y x y f

f

f

a a

a a a

a a a

a a E a

a a

x

y

a n