PHY 132 Final Exam
May 13, 2004
Version
Put FULL NAME, ID#, and EXAM VERSION on the front cover of the BLUE BOOKLET!
To avoid
problems in grading: do all problems in order
, write legibly
, and
show all work
for partial credit
!
Note: make sketches where appropriate and show your axis conventions. In the problems,
directions
must be
given as “
up
”, “
down
”, “
left
”, “
right
”, “
out of
“ and “
into the paper
”, “
CW
” (clockwise), or “
CCW
”. Include
units, watch the number of significant digits, and check results for consistency. This exam lasts 2.5 hr. Success!
1.
The two metal plates of a parallelplate capacitor have been charged with charges +
Q
and
–Q
respectively,
where
Q=
6.00 μC. The plates have area
A=
0.600 m
2
and separation distance
d=
0.885 mm. The volume be
tween the plates is filled with air (
K=
1.00).
(
50 points
)
a.
Use Gauss’ Law
to prove that the electric field
E
between the plates equals
σ
/
ε
0
, where
σ
≡
Q
/
A
.
Solution:
(10 points)
draw a “pill box” through one of the plates: the surface charge density inside the box is
σ
≡
Q
/
A
. The box
has area
A’
and the field between the plates is
E
. Because of the symmetry, the electric field OUTSIDE
the plates must be zero, and in between the field must be uniform
and pointing from +
Q
to –
Q
.
Gauss’ Law then gives:
EA’
=
σ
A’/
ε
0
Î
E
=
σ
/
ε
0
b.
Calculate the voltage difference between the two plates.
Solution:
∆
V
=
–
0
∫
d
E
·
d
x
=
Ed
=
σ
d/
ε
0
=
Qd/
(
A
ε
0
) = 1000 V
(10 points)
c.
Calculate the capacitance of the parallelplate system.
Solution:
C = Q
/
∆
V
= 6.00 nF (B
: 5.00 nF)
(10 points)
d.
An electron (
m
e
= 9.11×10
–31
kg) is let go from rest near the plate with charge
–Q
; calculate the speed of
the electron when it arrives at the second plate.
Solution:
a = F/m
=
eE/m
=
eV/
(
md
);
v
2
= 2
ad
= 2
eV/m
Î
v
=1.87×10
7
m/s
(10 points)
e.
Calculate the energy stored in the capacitor.
Solution:
U
= ½
QV
= 3.00×10
–3
J (B
: 2.50×10
–3
J)
(10 points)
2.
Consider the circuit shown in the Figure.
(
40 points
)
a.
Calculate the currents through
R
1
,
R
2
, and
R
3
.
Solution:
I
2
=
I
1
+
I
3
(30 points)
8 – 5
I
1
– 10
I
2
= 0
Î
8 – 15
I
1
– 10
I
3
= 0
Î
16 – 30
I
1
– 20
I
3
= 0
12 – 10
I
3
– 10
I
2
= 0
Î
12 – 20
I
3
– 10
I
1
= 0
Î
4 – 30
I
1
+ 10
I
1
= 0
Î
I
1
= 4/20 = 0.20 A
Î
I
3
= 10/20 = 0.50 A
Î
I
2
= 0.70 A
(B:
8 – 10
I
1
– 10
I
2
= 0
Î
8 – 20
I
1
– 10
I
3
= 0
Î
4 – 10
I
1
– 5
I
3
= 0
20
Î
16 – 17
I
– 7
I
3
– 10
I
2
= 0
Î
20 – 17
I
3
– 10
I
1
= 0
A
b.
tween voltages at the points
A
and
B
; beware of the sign!
3
+ 5
I
3
= 0
Î
I
3
= 12/16 = 0.75
Î
I
1
= 1/4/10 = 0.025 A
Î
I
2
= 0.775 A)
Calculate the voltage difference
V
A
–
V
B
be
Solution:
V
A
–
V
B
= 3
I
1
– 6
I
3
= –2.4 V
(B:
6
I
1
– 5
I
3
=–3.6 V)
(10 points)
R
1
=2
Ω
+
−
8 V
3
Ω
R
2
=10
Ω
A
Í
I
2
I
1
Î
B
+ −
12 V
I
3
Î
6
Ω
R
3
=4
Ω
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documenta
B
ext
3.
Consider the rectangular current loop with dimensions
a
=4.00 cm,
b
=15.0 cm as shown,
and lying in the plane of the paper. The current
I=
4.00 A in the loop is in the clockwise
direction. The uniform external magnetic field
B
ext
= 2.00 T, directed as shown.
(
40
points
)
b
a.
Calculate the magnetic forces (magnitude and direction!) on each of the four sides
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '04
 Rijssenbeek
 Physics, Work, Magnetic Field

Click to edit the document details