Physics I – Exam 3 – Spring 2003
Answer Key
Part A –
1. E, 2. C, 3. E, 4. F
There are some partial credits allowed for the following wrong answers:
2. D = 2 points for knowing the field is perpendicular to equipotential lines.
3. A = 2 points for correct magnitude of KE change but wrong sign.
3. D = 2 points for correct sign of KE change but wrong magnitude.
4. E = 2 points for knowing F is not in the plane of v and B.
B-1
g
G M (6.67  10 11 )( 2.0  10 30 )

1.33  10 12 m / s 2
R2
(10.0  10 3 ) 2
B-2
D, E, C, A, B
B-3
r
m v (9.1  10 31 )(1.4  10 7 )

 1.4 m
q B (1.6  10 19 )(5.7  10 5 )
B-4
V
V
0V


y
y
d
V  d E y  (0.05)(300)  15 Volts
Ey  
C-1
Analysis of the directions of E from each charge reveals that E will be zero somewhere in
Region III, to the right of -4q or x > d.
1 9q
1
 4q

0
2
40 x
40 ( x  d ) 2
9
4

x  3d
2
x
( x  d) 2
C-2
The + charge makes +V and the – charge makes –V. Analysis of the contributions from
each charge reveals that V will be zero somewhere in Region II (between charges) and
also in Region III. Only one is required for the question; either is OK.
Region II:
Region III:
1 9q
1
 4q
1 9q
1
 4q

0

0
40 x 40 (d  x )
40 x 40 ( x  d )
9
4
9
4
x  139 d  0.6923 d
x  95 d  1.8 d


x (d  x )
x ( x  d)
C-3
Calculate the X & Y components of E from each charge, then add. (9q is charge 1.)

1 9q  1
1 
E1 
î 
ĵ  2.86  10 5 î  2.86  10 5 ĵ V / m

2
40 2 d  2
2 

1  4q
E2 
0 î  1 ĵ  0 î  3.60  10 5 ĵ V / m
2
40 d



E total  E1  E 2  2.86 105 î  0.74 105 ĵ V / m


 




Calculate and add V from each charge; V is a scalar.
V
1
40
qi
r
i

1  9q  4q 
1 q 9


 4   2.1  10 3 Volts



40  2d
d  40 d  2
