pan (zp695) – midterm 03 review – chiu – (56565)
1
This printout should have 26 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001 (part 1 of 2) 0.0 points
The electric feld is measured all over a cu
bical surFace, and the pattern oF feld detected
is shown in the above fgure. On the right side
oF the cube and the bottom oF the cube, the
electric feld has the value
±
336
,

336
,
0
²
N/C
.
On the top oF the cube and the leFt side oF the
cube, the electric feld is zero. On halF oF the
Front and back Faces, the electric feld has the
value
±
336
,

336
,
0
²
;o
nt
h
eo
t
h
e
rh
a
l
F
oF the Front and back Faces, the electric feld
is zero. One edge oF the cube is
d
=0
.
55
m
long. (Hint: Notice that he electric feld vec
tors are perpendicular to the diagonal area oF
the cube)
Select the correct choices For the unit vector
oF
7
E
Ia.
ˆ
E
=
ˆ
i

ˆ
j
√
2
Ib.
ˆ
E
=
ˆ
i
+
ˆ
j
√
2
Ic.
ˆ
E
=
ˆ
i
+
ˆ
j
Outgoing unit normal For the right side
shaded area denoted by ˆ
n
right
is:
IIa. ˆ
n
right
=

ˆ
i
IIb. ˆ
n
right
=
ˆ
i
IIc. ˆ
n
right
=
ˆ
i

ˆ
j
√
2
1.
Ib, IIb
2.
Ib, IIc
3.
Ib, IIa
4.
Ia, IIc
5.
Ic, IIc
6.
Ia, IIb
correct
7.
Ic, IIb
8.
Ic, IIa
9.
Ia, IIa
Explanation:
By inspection, the unit vector For
7
E
is
ˆ
E
=
ˆ
i

ˆ
j
√
2
Hence, the answer is Ia.
The outgoing normal vector For the right
side shaded area is ˆ
n
right
=
ˆ
i
.H
e
n
c
e
,t
h
e
correct choice is IIb.
002 (part 2 of 2) 0.0 points
±ind the electric charge within the cube.
Correct answer: 1
.
799
×
10

9
C.
Explanation:
Let :
5
0
=8
.
85
×
10

12
J
/
V
2
·
m
,
E
x
=336
N/C ,
and
d
.
55
m.
The total outgoing ²ux by symmetry equals
twice the outgoing ²ux through the shaded
area at the right side.
φ
cube
=2
φ
right
φ
cube
E
±
ˆ
i

ˆ
j
√
2
²
·
(
d
2
ˆ
i
)
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documentpan (zp695) – midterm 03 review – chiu – (56565)
2
φ
cube
=
√
2
Ed
2
Here,
E
is given by
E
=
±
E
2
x
+
E
2
y
=
√
2
E
x
=475
.
176 N
/
C
hence, the total fux enclosed is given by
φ
cube
=2
E
x
d
2
Based on Gauss law, the chage enclosed is
Q
=
5
0
φ
cube
=
5
0
(2
E
x
d
2
)=1
.
799
×
10

9
C
003
0.0 points
An alpha particle (consisting oF two pro
tons and two neutrons) is moving in a
circle at constant speed, perpendicular to
au
n
i
F
o
rmm
a
g
n
e
t
i
c±
e
l
da
p
p
l
i
e
db
ys
om
e
currentcarrying coils.
The alpha particle
makes on clockwise revolution every 82 ns.
IF the speed is small compared to the speed
oF light, what is the magnitude
B
oF the mag
netic ±eld made by the coils?
Correct answer: 1
.
63 T.
Explanation:
let :
e
=1
.
6
×
10

19
C
,
m
P
≈
m
n
.
7
×
10

27
kg
,
and
T
=82ns=8
.
2
×
10

8
s
.
Given the period,
ω
=
2
π
T
.T
h
e
c
y
c
l
o
t
r
o
n
Frequency is
ω
=
qB
m
.So
lv
ing
,
B
=
2
πm
qT
=
2
π
(2
m
N
+2
m
P
)
2
eT
=
4
P
=
4
π
(1
.
7
×
10

27
kg)
(1
.
6
×
10

19
C)(8
.
2
×
10

8
s)
.
63 T
.
004
0.0 points
Alongwirecarriesacurrent
I
1
=1Aupward
,
and a rectangular loop oF height
h
=0
.
2m
and width
w
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Magnetic Field, IIA

Click to edit the document details