Type:
Solution Manual
Fundamentals of Resource: Electromagnetics with Engineering Applications Edition:
1st Edition
Author(s):
Stuart M. Wentworth
2-1 Solutions for Chapter 2 Problems 1. Vectors in the Cartesian Coordinate System P2.1: Given P(4,2,1) and APQ=2ax +4ay +6az, find the point Q. APQ = 2 ax + 4 ay + 6 az = (Qx-Px)ax + (Qy-Py)ay+(Qz-Pz)az Qx-Px=Qx-4=2; Qx=6 Qy-Py=Qy-2=4; Qy=6 Qz-Pz=Qz-1=6; Qz=7 Ans: Q(6,6,7)
P2.2: Given the points P(4,1,0)m and Q(1,3,0)m, fill in the table and make a sketch of the vectors found in (a) through (f). Vector Mag Unit Vector a. Find the vector A AOP = 4 ax + 1 ay 4.12 AOP = 0.97 ax + 0.24 ay from the origin to P b. Find the vector B BOQ = 1 ax + 3 ay 3.16 aOQ = 0.32 ax + 0.95 ay from the origin to Q c. Find the vector C CPQ = -3 ax + 2 ay 3.61 aPQ = -0.83 ax + 0.55 ay from P to Q d. Find A + B A + B = 5 ax + 4 ay 6.4 a = 0.78 ax + 0.62 ay e. Find C – A C - A = -7 ax + 1 ay 7.07 a = -0.99 ax + 0.14 ay f. Find B - A B - A = -3 ax + 2 ay 3.6 a = -0.83 ax + 0.55 ay a. AOP = (4-0)ax + (1-0)ay + (0-0)az = 4 ax + 1 ay.
AOP = 42 + 12 = 17 = 4.12 4 1 ax + a y = 0.97a x + 0.24a y 17 17 (see Figure P2.2ab) aOP =
b. BOQ =(1-0)ax + (3-0)ay + (0-0)az = 1 ax + 3 ay.
BOQ = 12 + 32 = 10 = 3.16
Fig. P2.2ab
1 3 ax + a y = 0.32ax + 0.95a y 10 10 (see Figure P2.2ab) aOQ =
c. CPQ = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + 2 ay.
CPQ = 32 + 22 = 13 = 3.61 −3 2 ax + a y = −0.83ax + 0.55a y 13 13 (see Figure P2.2cd) a PQ =
Fig. P2.2cd
2-2 d. A + B = (4+1)ax + (1+3)ay + (0-0)az = 5 ax + 4 ay.
A + B = 52 + 42 = 41 = 6.4 5 4 ax + a y = 0.78ax + 0.62a y 41 41 (see Figure P2.2cd) a=
e. C - A = (-3-4)ax + (2-1)ay + (0-0)az = -7 ax + 1 ay.
C − A = 72 + 12 = 50 = 7.07 −7 1 ax + a y = −0.99ax + 0.14a y 50 50 (see Figure P2.2ef) a=
FigP2.2ef
f. B - A = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + 2 ay.
B − A = 32 + 22 = 13 = 3.6 −3 2 ax + a y = −0.83ax + 0.55a y 13 13 (see Figure P2.2ef) a=
P2.3: MATLAB: Write a program that will find the vector between a pair of arbitrary points in the Cartesian Coordinate System. A program or function for this task is really overkill, as it is so easy to perform the task. Enter points P and Q (for example, P=[1 2 3]; Q=[6 5 4]). Then, the vector from P toQ is simply given by Q-P. As a function we could have: function PQ=vector(P,Q) % Given a pair of Cartesian points % P and Q, the program determines the % vector from P to Q. PQ=Q-P; Running this function we have: >> P=[1 2 3]; >> Q=[6 5 4]; >> PQ=vector(P,Q) PQ = 5
3
1
Alternatively, we could simply perform the math in the command line window:
2-3 >> PQ=Q-P PQ = 5 3 1 >> 2. Coulomb’s Law, Electric Field Intensity, and Field Lines P2.4: Suppose Q1(0.0, -3.0m, 0.0) = 4.0nC, Q2(0.0, 3.0m, 0.0) = 4.0nC, and Q3(4.0m, 0.0, 0.0) = 1.0nC. (a) Find the total force acting on the charge Q3. (b) Repeat the problem after changing the charge of Q2 to –4.0nC. (c) Find the electric field intensity for parts (a) and (b). (a) F13 =
Q1Q2 a13 , where R13 = 4 ax + 3 ay =, R13 = 5m, a13 = 0.8 ax + 0.6 ay. 4 o R132
so
( 4a + 3a ) 4 x10 C )(1x10 C ) ( 5 FV NM F = C VC 4 (10 F 36 m ) ( 5m ) −9
13
−9
x
−9
y
2
= 1.15 x10−9 ax + 0.86 x10−9 a y N . Similarly, F23 = 1.15x10−9 ax − 0.86 x10−9 a y N , so FTOT = 2.3a x nN (b) with Q2 = -4 nC, F13 is unchanged but F23 = −1.15x10−9 ax + 0.86 x10−9 a y N , so
FTOT = 1.7a y nN . −9 FTOT ( 2.3x10 ax N ) VC V (c) Ea = = = 2.3ax . -9 Q3 m (1x10 C ) Nm
Likewise, Eb = 1.7a y
V . m
Fig. P2.4 P2.5: Find the force exerted by Q1(3.0m, 3.0m, 3.0m) = 1.0 C on Q2(6.0m, 9.0m, 3.0m) = 10. nC. Q1Q2 a12 , where 4 o R122 R12 = (6-3)ax + (9-3)ay + (3-3)az = 3 ax + 6 ay m 3a + 6ay R12 = 32 + 62 = 45m, a12 = x ,and 45 F12 =
2-4
(1x10 C )(10 x10 C ) 3a + 6a FV NM , so F = 0.89a + 1.8a N . F = C VC 45 4 (10 F 45m ) 36 m ) ( −6
−9
x
−9
12
y
12
2
x
y
Fig. P2.5
P2.6: Suppose 10.0 nC point charges are located on the corners of a square of side 10.0 cm. Locating the square in the x-y plane (at z = 0.00) with one corner at the origin and one corner at P(10.0, 10.0, 0.00) cm, find the total force acting at point P. We arbitrarily label the charges as shown in Figure P2.6. Then ROP = 0.1 ax + 0.1 ay ROP = 0.141 m aOP = 0.707 ax + 0.707 ay.
FOP =
(10nC )(10nC )( 0.707 ) ( ax + ay ) −9 2 ( 4 ) 10 F 36 m ( 0.141m )
(
)
= 32 ( ax + a y ) N
FTP =
FSP =
(10nC )(10nC ) ay = 90a y N −9 2 10 F 0.1m ) ( 4 ) 36 m (
(
)
(
)
Fig. P2.6
(10nC )(10nC ) ax = 90ax N −9 2 ( 4 ) 10 F 36 m ( 0.1m )
and then the total (adjusting to 2 significant digits) is: FTOT = 120 ( ax + ay ) N .