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Test 1 with Answer Key - Electricity, Magnetism and Light | PHYS 241, Exams of Physics

Community College of PhiladelphiaPhysics
Prof. David F. Cattell

Material Type: Exam; Professor: Cattell; Class: Elec Magnetism & Light; Subject: Physics; University: Community College of Philadelphia; Term: Fall 2018;

Typology: Exams

2017/2018

Uploaded on 12/17/2018

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koofers-user-tvb 🇺🇸

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PHYS 241-1 Test 1 Key 9/20/18 p. 1
1. First make a sketch showing the electric fields at the origin that are due to the source charges.
(See Figure 1.)
a.
2
2
2
2
1
12
01
6
95
N m N
1
2C
C
2
22
02
6
95
N m N
2
2C
C
1
4
3.00 10 C
9 10 ; 1.08 10
0.500 m
1
4
3.00 10 C
9 10 ; 1.08 10
0.500 m
q
Er
E
q
Er
E


Rx 2
5
4N
C
0 cos 210
3
1.08 10 2
9.35 10
EE
Ry 1 2
55
4N
C
sin 210
1
1.08 10 1.08 10 2
5.40 10
E E E

44
N
C
ˆˆ
9.35 10 5.40 10 (20 points)E ij
b.
6 4 4 N
C
22
22
22
2
ˆˆ
0.200 10 C 9.35 10 5.40 10
ˆˆ
1.87 10 1.08 10 N
1.87 10 1.08 10 N
2.16 10 N or 21.6 mN
F qE
F
FF


ij
ij
2
2
F
1.08 10
arctan 1.87 10
30.0 below the + axis 330x or


 

c.
44
N
C
44
44
ˆ ˆ ˆ ˆ
: 1.50 3.00 C m 9.35 10 5.40 10
ˆˆ
ˆ
ˆ
1.50 3.00 0 m N; 1.50 5.40 10 3.00 9.35 10 m N
9.35 10 5.40 10 0
ˆ
0.200 m N (20 points)
pE


i j i j
i j k
k
k
Continued on the other side
0.500 m
x
y
Figure 1
q1
q2
E
30
30
2
E
1
E
Figure 2
x
y
q1
q2
F = 330
q
E
F
pf3

Partial preview of the text

Download Test 1 with Answer Key - Electricity, Magnetism and Light | PHYS 241 and more Exams Physics in PDF only on Docsity!

  1. First make a sketch showing the electric fields at the origin that are due to the source charges.

(See Figure 1.)

a.

2 2

2 2

1 (^1 ) 0 1

6 (^9) N m 5 N C^2 1 C

2 (^2 ) 0 2

6 (^9) N m 5 N

C^2 2 C

3.00 10 C

0.500 m

3.00 10 C

0.500 m

q E r E q E r E

 

 

Rx 2

5

(^4) N C

0 cos 210

E   E 

Ry 1 2

5 5

(^4) N C

sin 210

E  E  E 

(^4 4) N C

E  9.35  10 ˆ i^  5.40  10 ˆ j (20 points)

b.

(^6 4 4) N C

2 2

2 2 2 2

2

0.200 10 C 9.35 10 ˆ^ 5.40 10 ˆ
1.87 10 ˆ^ 1.08 10 ˆN
1.87 10 1.08 10 N

2.16 10 N or 21.6 mN

F qE

F
F F

 

 

i j

i j

2

2

F

arctan 1.87 10

30.0 below the + x axis or 330

c.

(^4 4) N C

4 4

4 4

: 1.50 3.00 C m 9.35 10 5.40 10

1.50 3.00 0 m N; 1.50 5.40 10 3.00 9.35 10 m N

0.200 m N (20 points)

  pE           

       ^        

i j i j

i j k

k

k

Continued on the other side

(20 points)

0.500 m

x

y

Figure 1

q 1

E q 2

30 

30 

E 2
E 1

Figure 2

x

y

q 1

q 2

F = 330

q

E
F

x

x

y

P

x 1 - x

x x 1

Charge  q =  x

lies in this interval.

Uniform charge

per unit length .

- L

Figure 4

Flux through S:

S 1 2 3

d   d   d   d

E A E A E A E A

From symmetry E is directed radially away from the line of charge and has a constant magnitude over the

lateral side of S (side 3).

Over the sides 1 and 2 of S E and d A are perpendicular:

1 2

d  0;  d  0

E A E A.

Over side 3 of S E and d A are parallel and E has constant magnitude E :

3 3

3 S

A

A 2 Thus 0 0 2

d Ed

E d E rL d E rL

E A
E A

Gauss' Law:  

S^0 0

so 2 = and. 2

q L d q L E rL E r

E A (20 points)

  1. a.

   

 

 

 

 

2 1 0

0

(^2) L 2 (^0 1 )

0

0 1 0 1 1 L

1 1

0 1 1

0 1 1

4 4 L

L

4 L

1 L

4 L

dq

dE dq dx r x x

r

dx dx

dE E

x x x x

E E

x x x x

x x

E

x x

E

x x

 ^ ^  ^ 

 ^ ^  

Continued on the next page

(12 points)

2 1

1

1 2 2 2 2 1

2

1 1

: For let then and

(We may leave out the " " (why?))

dx Note u x x du dx

x x

dx du du u u du

x x u^ u^ u

dx C

x x x^ x

 

Infinite line of charge

with uniform linear

charge density .

Gaussian surface S in the form of a

right circular cylinder of radius r and

length L coaxial with the line of charge.

L

r

E From symmetry the electric field is directed radially away from

the line of charge.

2 1 3

d A

d A d A

Figure 3