Physics I – Review for Exam 3 – Spring 2004
Answer Key
Part A Multiple Choice
1.
ABCD. (All can be concluded from Newton’s Theory with some math.)
2.
ABCD.
The magnetic force on a moving charged particle can never change its speed.
B2 curves the particle clockwise, B1 curves the particle counter-clockwise.
The radius in B1 is smaller than in B2. r = m v / (q B)
If v and B are doubled, r in the above expression stays the same.
3.
III
4.
II, III
5.
A. Electrons gain KE going to higher potential.
6.
D. Electric field points in the “downhill” direction of electric potential.
B Graphing
Features looking for:
1. Using K + U = 0 in some form.
2. Using U = q V in some form.
3. Proton and electron curves are mirrored.
4. Electron has the max. KE & proton has min. K.E.
5. All graphs have straight lines.
6. Electron has final K.E. = 3.2 x 10-17 J.
7. Proton has final K.E. = 3.2 x 10-17 J.
8. Max. K.E. = 4.8 x 10-17 J at 40 cm (either particle).
9. Min. K.E. = 1.6 x 10-17 J at 40 cm (either particle).
Alternative to 1. and 2. above, constant acceleration:
1'. Finding E = –V/x.
2'. Using a = F/m = ±e E/m and eqns. of motion.
Potential (V)
100
0
d (cm)
20
40
60
80
100
Proton
KE (J)x 10^-17
3.2
1.6
0
d (cm)
20
40
60
80
100
40
60
80
100
Electron
KE (J)x 10^-17
4.8
3.2
0
d (cm)
20
C-1 Electric Field and Potential

1 qi
E
r̂i Let #1 be the + charge and #2 be the – charge.
4   0 ri 2

1.6  10 19
E1  9.0  10 9
ĵ  4.0  1011 ĵ
11 2
(6.0  10 )

1.6  10 19
1 
 1
E 2  9.0  10 9
î 
ĵ   1.4  1011 î  1.4  1011 ĵ

11 2
( 2  6.0  10 )  2
2 
 

11
11
E  E1  E 2  1.4  10 î  2.6  10 ĵ N / C
V
 1.6  10 19
1 qi
 1.6  10 19 
  7.0 V
 9.0  10 9 

11
4   0 ri
2  6.0  10 11 
 6.0  10
C-2 e/m for an Unknown Particle
mv
m2 v2
2
r
r  2 2
qB
q B
K  U  q(V)  q (1.000  10 5 )  12 m v 2
2 q (1.000  10 5 )
m
2 q (1.000  10 5 )
m2
m (2.000  10 5 )
2
m
r 

q 2 B2
q B2
v2 
q 2.000  10 5
2.000  10 5


 8.51  10 8 C / kg
2
2
2
2
m
r B
(0.03066) (0.5)