Physics I – Exam 3 – Spring 2007
Answer Key
Part A – 1: C, 2: A, 3: C
4: F, 5: B, 6: D
Partial credit: 1: D, 4: E
4 pts each.
2 pts partial
Ex (V/m)
200
Part B – 28 Points
100
From the Ex graph to Fx using Fx = (–e) Ex
 
From the Ex graph to V using V    E  dx
-100
0
x (cm)
20
40
60
80
100
20
40
60
80
100
20
40
60
80
100
20
40
60
80
100
20
40
60
80
100
-200
Fx ( N )
From the V graph to U using U = (–e) V
From the U graph to K using K = –U
Looking For
Fx graph:
using Fx = (–e) Ex
shape is mirror image of Ex
correct max value = +3.2 × 10–17 N
3.2e-17
1.6e-17
0
x (cm)
-1.6e-17
-3.2e-17
V(V)
40
30
V graph:
 
+4 for using V    E  dx or other correct idea
20
10
0
correct shape as shown to right
final value = max value ÷ 2
correct max value = +40 V
U graph:
using U = (–e) V or other correct idea
shape is mirror image of V
correct min value = –6.4 × 10–18 J
x (cm)
U(J)
0
x (cm)
-1.6e-18
-3.2e-18
K graph:
 
using K = –U or K  W   F  dx
or other correct idea
shape is mirror image of U
max value = +3.2 × 10–18 – U J
-4.8e-18
-6.4e-18
K (J)
9.6e-18
8.0e-18
–1 for each incorrect or missing unit on axis
6.4e-18
4.8e-18
3.2e-18
1.6e-18
0
x (cm)
Part C
C-1-A (12 points)
T
2r
mv
2m
and r 
→ T
v
qB
qB
B = 7.80 × 10–4 × I = 7.80 × 10–4 × 1.20 = 9.36 × 10–4 Tesla
T = 2  × (9.11 × 10–31) ÷ [(1.60 × 10–19) × (9.36 × 10–4)] = 3.82 × 10–8 sec
Looking For
2r
v
mv
using r 
qB
correct value of B
correct value of T
using T 
C-1-B (12 points)
rqB
mv
→ v
m
qB
v = 0.0557 × (1.60 × 10–19) × (9.36 × 10–4) ÷ (9.11 × 10–31) = 9.16 × 106 m/sec
½ m v2 = K = –U = –(–e) V → V = ½ m v2 ÷ e
V = ½ (9.11 × 10–31) × (9.16 × 106)2 ÷ (1.60 × 10–19) = 239 volts
r
Looking For
mv
or other correct idea to get v
qB
using ½ m v2 = K = –U = –(–e) V or other correct idea to get V
correct value of V
2 V
e
 2 2 must show where it came from.
If student uses
m r B
using r 
C-2 (24 points)
Note: Some judgment is required to grade this because student errors are unpredictable.
E Field (16 Points)
1 q
Let E 1 
be the magnitude of the E field from charge #1.
4  0 r 2
E1 = 9.0 × 10+9 × 1.0 × 10–9 ÷ (√2 × 0.3)2 = 50 N/C or V/m
E2 = 2 E1. E3 = 2 E1. E4 = E1.
By symmetry, the X components cancel.
The Y components add as follows:
Ey = (–E1 + E2 + E3 – E4) × sin(45°)
= (–50 + 100 + 100 – 50) × (0.707)
= 71 N/C or V/m
(-0.3,+0.3)
(+0.3,+0.3)
#1: +q
#2: -2q
Y
E3
Looking For
treated E as a vector
used the correct formula (r–2)
X component is 0
Y component is positive
magnitude of E = 71 V/m (any dir.)
–1 missing or incorrect units
E2
X
E4
E1
#4: -q
Potential (8 Points)
V
1 qi
1

4   0 ri
4  0
(-0.3,-0.3)
1 1
 q1 q 2 q 3 q 4 
q1  q 2  q 3  q 4   0

 
 
r
r
r  4  0 r
 r
Looking For
treated V as a scalar
used the correct formula (r–1) or used symmetry arguments
V = 0 volts (not getting zero is not just a simple numerical error)
–1 missing or incorrect units
#3: +2q
(+0.3,-0.3)