A stationary 5.00 W power source emits a 460 Hz signal isotropically.
An observer stands at rest between the source and a vertical wall which is
perpendicular to the line joining all three.
(a) [6 pts] What intensity is heard by the observer who is located
10.0 m from the source?
I
=
P
A
=
P
4pr2
=
4.00 W
4p(10.0 m)2
=
3.98×10-3 W/m2
(b) [6 pts] Calculate the corresponding intensity level, b, for
part (a).
b
=
(10 dB)log10
æ è
I
Io
ö ø
=
(10 dB)log10
æ è
3.98×10-3 W/m2
10-12 W/m2
ö ø
=
96.0 dB
(c) [6 pts] The observer suddenly takes of running away from the source
toward the wall at 10.0 m/s. What frequency does the observer hear?
fo
=
fs
æ è
c-vo
c
ö ø
=
(460 Hz)
æ è
344 m/s-10.0 m/s
344 m/s
ö ø
=
447 Hz
(d) [7 pts] The observer also experiences the signal reflected from the
wall. What beat frequency is detected?
fb
=
Dfo
=
fs
æ è
c+vo
c
ö ø
-fs
æ è
c-vo
c
ö ø
=
2vofs
c
=
2(10.0 m/s)(460 Hz)
344 m/s
=
26.7 Hz
The figure shows three equally separated static point charges.
q=4.80 mC
a=0.120 m
(a) [5 pts] Does the top charge exert a force on the bottom charge?
ANS: Yes. The electric field around a charge permeates all space. Therefore,
the bottom charge is affected. The middle charge just distorts the surrounding
field.
(b) [5 pts] If the middle charge was released, which way would it move?
ANS: Both top and bottom charges exert downward forces on the middle charge.
Therefore, the net force is DOWN.
(c) [10 pts] Calculate the net force on the middle charge.
Fnet
=
Fq,q+F4q,q
=
(Fq,q+F4q,q) (-
^
j
)
The magnitude of the net force is
Fnet
=
kq2
a2
+
4kq2
a2
=
5kq2
a2
=
5(8.99×109 N·m2/C2) (2.40×10-6 C)2
(0.120 m)2
=
71.9 N
(d) [5 pts] If the middle charge resided on a mass of
2.40×10-12 kg and was released, what would be the charge's
initial acceleration?
a
=
Fnet
m
=
-71.9 N
^
j
2.40×10-12 kg
=
-3.00×1013 m/s2
^
j
Two point charges, q1 and q2, lie on the x-axis at rest.
They are separated by a distance d. A field point is located on the x-axis
at x=2.00 m.
q1=+5.00 mC
q2=-2.00 mC
d=0.800 m
(a) [4 pts] At P, draw in the electric field line for each charge.
There is only one line per charge.
ANS: The electric field line for charge q1 points to the right. The
electric line for charge q2 points to the left.
(b) [8 pts] Calculate the net electric field at P at x=2.00 m.
EP
=
kq1
x2
-
kq2
(x-d)2
=
(8.99×109 N·m2/C2)
é ë
5.00×10-6 C
(2.00 m)2
-
2.00×10-6 C
(1.20 m)2
ù û
=
-1.25×103 N/C
(c) [4 pts] If a point charge q3=-4.20 mC is placed at point
P, what force acts on it?
F
=
q3EP
^
i
=
(-4.20×10-6 C)(-1.25×103 N/C)
^
i
=
+5.24×10-3 N
^
i
(d) [9 pts] The figure shows three regions: A, B, and C. There is
one value of x where the net electric field is zero (infinity NOT included).
In which region does it lie? Find the value of x where the field is zero.
ANS: The position lies in region C.
E(x)
=
kq1
x2
-
kq2
(x-d)2
E(xo)
=
0
kq1
x2o
=
kq2
(xo-d)2
Cancel the k's and rearrange the equation. Do NOT expand the square term.
æ è
xo-d
xo
ö ø
2
=
q2
q1
xo-d
xo
=
±
æ Ö
q2
q1
Solve for xo.
xo
=
d
1-±
Ö
q2/q1
Since x must be in region C, then x > d. Choose the minus sign.
xo
=
0.800 m
1-
Ö
2.00/5.00
=
2.18 m
The figure shows a uniform line charge of total charge +Q and length
L positioned on the x-axis. A field point P is chosen on the y-axis a
distance a above the origin. The differential electric field components
at P due to the point charge dq are
dEx=-dEcosq
dEy=+dEsinq
Available symbols are
k
Q
dq
r
x
a
l
q
L
dx
For parts (b) through (e) below, write the answers in terms of the above
symbols.
(a) [4 pts] At the field point P draw in the direction of dE
due to dq.
ANS: dE points away from dq along r.
(b) [5 pts] Write an expression for the line charge density, l.
l = Q/L
(c) [5 pts] Write an expression for dq.
dq = l dx
(d) [5 pts] Write and expression for dE.
dE
=
kdq
r2
=
kQdx
L(x2+a2)
(e) [6 pts] Write the final expressions for Ex and Ey in
integral form but do NOT integrate them.
cosq = x/r
sinq = a/r
Ex
=
-
kQ
L
ó õ
L
0
xdx
(x2+a2)3/2
Ey
=
kQa
L
ó õ
L
0
dx
(x2+a2)3/2
CONSTANTS
m = 10-6
k=8.99×109 N·m2/C2
FORMULAS
F=kq1q2/r2
E=kq/r2
F=qE
å
F=ma
I=P/A
b = (10 dB) log (I/Io)
fo=fs[(c+vo)/(c-vs)]
fb=Df
dE=kdq/r2
A=4pr2
c=344 m/s
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version 3.85. On 03 Feb 2016, 11:32.