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PHY2061
R. D. Field
Final Exam Solutions
Page 1 of
9
December 9, 2003
PHY 2061 Final Exam Solutions
Problem 1 (10 points):
Circle true or false for following (2 points each).
(a) (True or False)
The vector function
z
x
y
y
x
z
z
y
x
E
ˆ
ˆ
ˆ
)
,
,
(
−
+
=
r
represents a possible
electrostatic field with magnitude given by
2
2
2
)
,
,
(
z
y
x
z
y
x
E
+
+
=
.
Note that
0
ˆ
2
≠
=
×
∇
y
E
r
r
and hence this vector function cannot represent an electrostatic field since
all electrostatic fields have
0
=
×
∇
E
r
r
.
(b) (True or False)
The vector function
z
x
y
y
x
z
z
y
x
B
ˆ
ˆ
ˆ
)
,
,
(
−
+
=
r
represents a possible
magnetic field with magnitude given by
2
2
2
)
,
,
(
z
y
x
z
y
x
B
+
+
=
. Note that
0
1
≠
=
⋅
∇
B
r
r
and hence this vector function cannot represent a magnetic field since all
magnetic fields have
0
=
⋅
∇
B
r
r
.
(c) (True or False)
The electric field is zero on the surface of all conductors in static
equilibrium.
(d) (True or False)
It is possible to construct both electric and magnetic dipole field
configurations, but it is not possible to construct a magnetic monopole field.
(e) (True or False)
Lines of constant electric potential are always perpendicular to
electric field lines.
(f) (True or False)
In relativity, energy and momentum are conserved, but mass is not
conserved.
(g) (True or False)
When light moves from one medium to another both its speed and
wavelength change, but its frequency remains the same.
(h) (True or False)
Very little diffraction occurs when light of wavelength
λ
passes
through a single slit of width W provided
λ
<< W.
(i) (True or False)
The real image created by a concave mirror is always inverted.
(j) (True or False)
Convex mirrors can produce both real and virtual images.
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View Full DocumentPHY2061
R. D. Field
Final Exam Solutions
Page 2 of
9
December 9, 2003
Problem 2 (18 points):
Suppose that a
nonuniform
electric field is present and is given by
r
ar
r
E
ˆ
)
(
=
r
,
where
a = 50 V/m
2
and
r
is the radial distance from the origin,
2
2
2
z
y
x
r
+
+
=
.
The electric field points radially away from the
origin and has a magnitude given by
E(r) = ar
.
(Note that K = 8.99x10
9
Nm
2
/C
2
,
ε
0
= 8.85x10
12
C
2
/Nm
2
)
Part A (6 points):
How much electric charge (
in nanoC
) is contained with a spherical
shell centered at the origin with inner radius
R
1
= 1 m
and outer radius
R
2
= 2 m
?
Answer:
38.9
Solution:
There are two ways to do this problem.
One way is to calculate the net flux
emanating from the spherical shell as follows:
)
(
4
4
)
(
4
)
(
3
1
3
2
2
1
1
2
2
2
1
2
R
R
a
R
R
E
R
R
E
A
d
E
A
d
E
A
d
E
R
R
shell
−
=
−
=
⋅
+
⋅
=
⋅
=
Φ
∫
∫
∫
π
r
v
r
v
r
v
and we know from Gauss’s Law that
nC
C
Nm
m
m
m
V
K
R
R
a
R
R
a
Q
enclosed
9
.
38
/
10
99
.
8
)
)
1
(
)
2
)((
/
50
(
)
(
)
(
4
2
2
9
3
3
2
3
1
3
2
3
1
3
2
0
0
=
×
−
=
−
=
−
=
Φ
=
πε
ε
.
A second method is to calculate the charge density according to
a
z
E
y
E
x
E
E
z
y
x
0
0
0
3
ρ
=
∂
∂
+
∂
∂
+
∂
∂
=
⋅
∇
=
v
r
where I used E
x
= x, E
y
= y, E
z
= z.
Now we calculate the charge in the shell using
)
(
4
3
3
1
3
2
0
0
2
1
R
R
a
dV
a
dV
Q
R
R
−
=
=
=
∫
∫
,
which of course gives the same answer (note that the charge density is constant).
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 Spring '08
 FRY
 Physics

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