 # Engineering Electromagnetic Chapter 4 problems Anonymous

### Question Description

chapter 4 problem (10 problems) can you solve them?

attached the book if you wanna review the equations

Please don't do nothing similar from any website or solution manual attached. questions.jpeg

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Ace_Tutor
School: University of Maryland  I have the completed answer for the first 7 problems (the last 3 problems (#37, #45 and #48) are not founded in chapter 4)

Problem 8:
With given charge Q1 , the electric field E at Q2 due to Q1 by definition is

E=

Q1

4 0 r 2

r

And the differential of length dl for integration when finding the work can be converted from
rectangular coordinates to spherical coordinates as follow

dl = drr + rd + r sin  d
Thus, their product or the integrand is

E  dl =

Q1

4 0 r 2

 E  dl =

(

r  drr + rd + r sin  d

Q1

4 0 r

2

( )

Q1

dr r  r +

4 0 r

2

)

( )

rd r   +

Q1

4 0 r 2

( )

r sin  d r  

 Q1

 Q1

 Q1

 E  dl = 
dr  (1) + 
rd  ( 0 ) + 
r sin  d  ( 0 )
2
2
2
 4 0 r

 4 0 r

 4 0 r

Q1
 E  dl =
dr
4 0 r 2
a) Hence, the work done WB →C in carrying the charge Q2 from B to C while the 2 last variables
are held constant is calculated by
rA

rA

rB

rB

WB →C = −Q2  E  dl = −Q2 

Q1

4 0 r 2

dr

rA

 WB →C

Q 
= −Q2   − 1 
 4 0 r  rB

 WB →C =

Q1Q2
QQ
− 1 2
4 0 rA 4 0 rB

And the unit for this answer is joule ( J ) .
b) If we write the electric field E in terms of spherical coordinates, it follows that

E=

Q1

4 0 r 2

r = Er r + E  + E 

Q1

Hence, according to the notation, we have Er =

4 0 r 2

, E = 0 and E = 0 .

Similarly, the work done WC → D in carrying the charge Q2 from C ( rA , B , B ) to D ( rA , A , B )
while r and  are held constant is calculated by
A

WC → D = −Q2  E d
B

A

A

B

B

 WC → D = −Q2  0  d = −Q2  0
 WC → D = 0
And the unit for this answer is joule ( J ) .
c) By the result in part b, we have E = 0 . Finally, the work done in carrying the charge Q2 from

D to A while the 2 first variables are held constant is calculated by
A

WD → A = −Q2  E d
B

A

A

B

B

 WD → A = −Q2  0  d = −Q2  0
 WD → A = 0
And the unit for this answer is joule ( J ) .
Problem 12:
a) The coor...

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