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Electromagnetics 1st Edition By Branislav Notaros (Solution Manual)
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ELECTROMAGNETICSPROBLEMS (in Chapters 1-14)
Solutions Manual
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
NOTE: (For Complete File, Download link at the end of this File) P1SOLUTIONS TO PROBLEMS
ELECTROSTATIC FIELD IN FREE SPACE
Section 1.1 Coulomb’s Law
PROBLEM 1.1 Three unequal charges in a triangle. (a)The new situation
is shown in Fig.P1.1(a), where charge 2 is now assumed to be 3 Q. From Coulomb’s
law, Eq.(1.1), magnitudes of the individual partial electric forces that charges 1, 2,
and 3 exert on one another are
Fe12=Fe21=Fe23=Fe32=3Q2
4πε0a2, Fe13=Fe31=Q2
4πε0a2,(P1.1)
and all forces are repulsive. In the adopted xy-coordinate system, the resultant
force on charge 1 is expressed as
Fe1=Fe21+Fe31=Fe21cos60◦(−ˆ x) +Fe21sin 60◦(−ˆ y) +Fe31(−ˆ x)
=−Q2
8πε0a2(5ˆ x+ 3√
3ˆ y). (P1.2)
Its magnitude, as well as the magnitude of the resultant force on charge 3, and the
angle αin Fig.P1.1(a), determining the direction of both vectors Fe1andFe3, come
out to be
O
Fe3Fe2
Fe31
Fe23/c973Q
Q
aa a
Q
Fe160
xy
Fe21Fe13Fe12Fe32
30
12
3OFe3
Fe2Fe31Fe23
/c98−3Q
Qaa a
QFe1
xy
Fe21
Fe13
12
3
(a) (b)
Figure P1.1 The same as in Fig.1.3(a) but with one of the point charges amou nting
to (a) 3 Qand (b) −3Q.
1
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 2
Fe1=Fe3=Q2
8πε0a2/radicalBig
52+ (3√
3)2=√
13Q2
4πε0a2, α= arctan3√
3
5= 46.1◦(P1.3)
(arctan ≡tan−1).
The resultant electric force on charge 2 is, as in Fig.1.3(b) and Eq.(1.10), given
by
Fe2= 2Fe12cos30◦=Fe12√
3 =3√
3Q2
4πε0a2, (P1.4)
andFe2is directed upward (perpendicularly to the opposite side of the equilateral
triangle), that is, Fe2=Fe2ˆ y, in Fig.P1.1(a).
(b)For charge 2 amounting to −3Q, Fig.P1.1(b), we have
Fe1=Fe21cos60◦ˆ x+Fe21sin 60◦ˆ y+Fe31(−ˆ x) =Q2
8πε0a2(ˆ x+ 3√
3ˆ y),(P1.5)
and hence
Fe1=Fe3=Q2
8πε0a2/radicalBig
1 + (3√
3)2=√
7Q2
4πε0a2, β= arctan3√
3 = 79 .1◦.(P1.6)
The force Fe2(magnitude) is the same as in Eq.(P1.4), and the direction of Fe2
is downward, Fe2=Fe2(−ˆ y).
PROBLEM 1.2 Three charges in equilibrium. The resultant Coulomb force
on charge 3 in Fig.1.50 being zero, we have that
Fe3=Fe13+Fe23= 0 −→Fe13=Fe23 −→Q1Q3
4πε0d2=Q2Q3
4πε0(D−d)2
−→(D−d)2=Q2
Q1d2−→D−d=±/radicalBigg
Q2
Q1d−→d= 2 cm ,(P1.7)
where we eliminate the other solution, d= 6 cm, because d<D.
From the condition that the total force on charge 1 must also be zero,
Fe21+Fe31= 0 −→Q1Q2
4πε0D2=−Q1Q3
4πε0d2−→Q3=−d2
D2Q2=−4 pC.
The condition Fe2= 0 gives the same result.
PROBLEM 1.3 Four charges at rectangle vertices. With reference to
Fig.P1.2,
Fe14=Fe14ˆ x, Fe14=Q2
4πε0a2, (P1.8)
Fe24=Fe24cosαˆ x+Fe24sinαˆ y, Fe24=Q2
4πε0c2,
c=/radicalbig
a2+b2,cosα=a
c,sinα=b
c, (P1.9)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. P1. Solutions to Problems: Electrostatic Field in Free Space 3
Fe34=Fe34ˆ y, F e34=Q2
4πε0b2, (P1.10)
so that the total electric force on charge 4 amounts to
Fe4=Fe14+Fe24+Fe34=Q2
4πε0/bracketleftbigg/parenleftbigg1
a2+a
c3/parenrightbigg
ˆ x+/parenleftbiggb
c3+1
b2/parenrightbigg
ˆ y/bracketrightbigg
= (9.637ˆ x+ 24.48ˆ y)µN. (P1.11)
The magnitude of this vector comes out to be |Fe4|= 26.31µN;Fe4makes an angle
ofβ= 68.5◦with thex-axis (Fig.P1.2).
O/c97Q
aFe34
xy
1
2 3b4
Q QQ Fe14Fe24
c
Figure P1.2 Computing the electric forces on charges placed at vertices o f a rect-
angle.
Similarly, the resultant forces on other charges are
Fe1= (−9.637ˆ x+ 24.48ˆ y)µN,Fe2=−(9.637ˆ x+ 24.48ˆ y)µN =−Fe4,
Fe3= (9.637ˆ x−24.48ˆ y)µN =−Fe1. (P1.12)
These vectors are also shown in Fig.P1.2.
PROBLEM 1.4 Five charges in equilibrium. Because of symmetry, the
r
esultant electric force on the charge (small ball) Q2at the square center is zero,
as indicated in Fig.P1.3, so it is always in the electrostatic equilibrium, regardless
of the actual value of Q2(which is unknown). Because of symmetry as well, if we
findQ2such that one of the four charges (balls) at the square vertices is in the
equilibrium, then this condition automatically applies to the remaining three balls
with charge Q1.
From Fig.P1.3, Q2that makes the resultant force on the lower left charge (charge
1) be zero obviously must be negative and is obtained as follows:
Fe1=Fe21+Fe31+Fe41+Fe51= 0−→ 2Fe21cos45◦+Fe31=Fe51
−→ 2Q2
1
4πε0a2√
2
2+Q2
1
4πε0(√
2a)2=−Q1Q2
4πε0(√
2a/2)2(Q2<0)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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