ELECTROMAGNETICS - Solutions Manual© 2011 Pearson Education, ...

ELECTROMAGNETICS

PROBLEMS (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) 1 / 4

P1SOLUTIONS TO PROBLEMS

ELECTROSTATIC FIELD IN FREE SPACE

Section 1.1Coulomb’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 3Q. 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= 3Q 2 4πε0a 2 , Fe13=Fe31= Q 2 4πε0a 2

, (P1.1)

and all forces are repulsive. In the adoptedxy-coordinate system, the resultant force on charge 1 is expressed as Fe1=Fe21+Fe31=Fe21cos 60 ◦ (−ˆx) +Fe21sin 60 ◦ (−ˆy) +Fe31(−ˆx) =− Q 2 8πε0a 2 (5ˆx+ 3 √ 3ˆy).(P1.2) Its magnitude, as well as the magnitude of the resultant force on charge 3, and the angleαin Fig.P1.1(a), determining the direction of both vectorsFe1andFe3, come out to be O F e3 F e2 F e31 F e23 a 3Q Q a aa Q F e1 60 x y F e21 F e13 F e12 F e32 30 1 2 3 O F e3 F e2 F e31 F e23 b −3Q Q a aa Q F e1 x y F e21 F e13 1 2 3 (a)(b) Figure P1.1The same as in Fig.1.3(a) but with one of the point charges amounting to (a) 3Qand (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 / 4

2 Fe1=Fe3= Q 2 8πε0a 2 q 5 2

• (3

√ 3) 2 = √ 13Q 2 4πε0a 2 , α= arctan 3 √ 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= 2Fe12cos 30 ◦ =Fe12 √

• =

3 √ 3Q 2 4πε0a 2

, (P1.4)

andFe2is directed upward (perpendicularly to the opposite side of the equilateral triangle), that is,Fe2=Fe2ˆy, in Fig.P1.1(a).(b)For charge 2 amounting to−3Q, Fig.P1.1(b), we have Fe1=Fe21cos 60 ◦ ˆx+Fe21sin 60 ◦ ˆy+Fe31(−ˆx) = Q 2 8πε0a 2 (ˆx+ 3 √ 3ˆy),(P1.5) and hence Fe1=Fe3= Q 2 8πε0a 2 q

1 + (3

√ 3) 2 = √ 7Q 2 4πε0a 2 , β= arctan 3 √

3 = 79.1

◦

.(P1.6)

The forceFe2(magnitude) is the same as in Eq.(P1.4), and the direction ofFe2 is downward,Fe2=Fe2(−ˆy).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 4πε0d 2 = Q2Q3 4πε0(D−d) 2 −→(D−d) 2 = Q2 Q1 d 2 −→D−d=± s Q2 Q1 d−→d= 2 cm,(P1.7) where we eliminate the other solution,d= 6 cm, becaused < D.From the condition that the total force on charge 1 must also be zero, Fe21+Fe31= 0−→ Q1Q2

4πε0D

2 =− Q1Q3 4πε0d 2

−→Q3=−

d 2 D 2 Q2=−4 pC.The conditionFe2= 0 gives the same result.PROBLEM 1.3 Four charges at rectangle vertices.With reference to Fig.P1.2, Fe14=Fe14ˆx, Fe14= Q 2 4πε0a 2

,(P1.8)

Fe24=Fe24cosαˆx+Fe24sinαˆy, Fe24= Q 2 4πε0c 2 , c= p a 2 +b 2 ,cosα= a c ,sinα= b c , (P1.9)©ψ2011ψPearsonψEducation,ψInc.,ψUpperψSaddleψRiver,ψNJ.ψψAllψrightsψreserved.ψThisψpublicationψisψ protectedψbyψCopyrightψandψwrittenψpermissionψshouldψbeψ obtainedψfromψtheψpublisherψpriorψtoψanyψprohibitedψreproduction,ψstorageψinψ aψ retrievalψsystem,ψorψtransmissionψinψ anyψformψorψbyψanyψmeans,ψelectronic,ψ mechanical,ψphotocopying,ψrecording,ψorψlikewise.ψForψinformationψψregardingψpermission(s),ψwriteψto.ψRightsψandψPermissionsψDepartment,ψ PearsonψEducation,ψInc.,ψUpperψSaddleψRiver,ψNJψ07458. 3 / 4

P1. Solutions to Problems: Electrostatic Field in Free Space3

Fe34=Fe34ˆy, Fe34= Q 2 4πε0b 2

,(P1.10)

so that the total electric force on charge 4 amounts to Fe4=Fe14+Fe24+Fe34= Q 2

4πε0

1 a 2 + a c 3 ´ ˆx+ ` b c 3 + 1 b 2 ´ ˆy λ = (9.637ˆx+ 24.48ˆy)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 a Q a F e34 x y 1 23 b 4 QQ Q F e14 F e24 c Figure P1.2Computing the electric forces on charges placed at vertices o f a rect- angle.Similarly, the resultant forces on other charges are Fe1= (−9.637ˆx+ 24.48ˆy)μN,Fe2=−(9.637ˆx+ 24.48ˆy)μN =−Fe4, Fe3= (9.637ˆx−24.48ˆy)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 ofQ2(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 chargeQ1.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−→2Fe21cos 45 ◦ +Fe31=Fe51

−→2

Q 2 1 4πε0a 2 √ 2 2 + Q 21

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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