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Physics 2020
Spring 2008
Stephan LeBohec
COVER SHEET - EXAM 2
SEAT #
Constants:
e=1.602×10−19 C , Q Proton =e & Q Electron=−e
1
9
2 −2
k=
=8.99×10 N m C
& 0=8.85×10−12 C 2 N −1 m −2
4  0
Waves:
Speed of sound: c Sound =

RT
M

F

Wave-length, frequency period relation: =c⋅T =c / f
With A and B wave sources in phase, interferences are
constructive when ∣d A−d B∣=n  , n=0,1,2,3,⋯
destructive when ∣d A−d B∣=n1/ 2 , n=0,1,2,3,⋯
c
Standing waves on a string or in an open pipe f n=n
, n=0,1,2,3,⋯
2L

Diffraction through a slit – first minimum sin =
D

Diffraction through a circular opening – first minimum sin =1.22
D
Speed of a wave on a string c String =
Electric forces and potentials:
One point charge in an electric field
 =q E
 & U =q V
F
Two point charges:
q q
 ∣=k 1 2 & U =k q1 q2
∣F
2
r
r
q
q
 ∣=k
∣E
& V =k
r
r2
Triangle geometry:
(ABD) and (ECD), similar triangles
AB DB AD
=
=
EC DC ED
c
a
µ
E
b
c =a 2b2
cos =b/ c
sin =a /c
tan =a/b
D
C
2
A
B
Physics 2020
Spring 2008
Stephan LeBohec
EXAM 2
Name:_____________________________________
TA (circle one): Aaron
Akiko
Jacque
1
Student ID #:___________________________
Matt
A.
[20 points] Two speakers separated by d=1.2m are operating in phase, producing two f=1432Hz sound waves.
The speed of sound is c=343m/s.
1.
answer)?
At mid distance between the two speakers, are the interferences constructive or destructive (explain your
At mid distance, the difference between the distances the speakers is zero, an integer multiple of the half wavelength so
the interference is constructive.
2.
Leaving the mid distance point and moving along a straight line toward one of the speakers, how many
destructive interference points will you find before reaching the speaker?
The wave length is =c / f =343m⋅s −1 / 1432 Hz=0.2395m . Leaving the mid point,. the first destructive
interference point is reached when the difference between the distances to the speakers equals a quarter of the wavelength.
This occurs after moving by an eighth of the wavelength. Starting from that point, destructive interference regions occur
every quarter of a wavelength so m the number of destructive interference spots is given by
d
d 1
1.2m
m= E 1 −/4 // 2=1E  − =1E 
−0.5=1E 4.51=5
2
0.2395m
 2
Physics 2020
Spring 2008
Stephan LeBohec
EXAM 2
Name:_____________________________________
TA (circle one): Aaron
Akiko
Jacque
2
Student ID #:___________________________
Matt
C.
1.
[18 points] What is the potential energy of the collection of three
charges shown on the figure with q=3¹C and d=0.2m
U=
2
8.99×109 × 3×10−6 2
k⋅q
1
2
1
2
−  −2=
−  −2
d
0.2
 2 5
 2 5
U =−0.7333J
2.
V M=
[14 points] What is the electric potential at point M?
k⋅q
1
1 2
8.99×109×3×10−6
−  − =−
=−134,850.0 V
d
0.2
5  5 2
3.
[8 points] How much work is necessary to move charge q'=1.5¹C from point M to point N (Note: you
can get the answer with a detailed calculation or with a simple geometry argument)?
Both points M and N are at the same distance from the charges q and −q and the contributions from these
charges to V M and V N cancel out. Both points M and N are also at the same distance from the charge −2q .
From this it results that V M =V N . The amount of work to be done against the electric force to go from point M to
N is W =q ' V N −q ' V M =q ' V N −V M =0 .
Physics 2020
Spring 2008
Stephan LeBohec
3
EXAM 2
Name:_____________________________________
TA (circle one): Aaron
Akiko
Jacque
Student ID #:___________________________
Matt
D.
This exercise is based on the same setting as in the previous one. The values are the same,
q=3¹C and d=0.2m. Here, however, we concentrate on point P.
1.
[21 points] What is the electric field at point P? Express the electric field as its
 ∣ and direction angle µ with respect to the positive x axis.
magnitude ∣E
The magnitude of the contributions of charges −q , q & −2q
the electric field at point P
kq
kq
q
−2q 2kq
respectively are: ∣E ∣= 2 , ∣E ∣= 2 and ∣E ∣= 2 . The x and y
2d
d
4d
kq
kq
kq
E−q
E−q
E−q
components are E−q
,
y =
x =
y =−
x =0 ,
2 ,
2 ,
4d
22 d
2  2 d2
2kq
E−2q
and E−2q
x and y component from each contribution,
x =−
y =0 . Adding the
d2
kq 1
1
kq 1
6
−1
4
−1
−2=−1.110×10 V⋅m
=−6.982×10 V⋅m .
we get: E x = 2 
and E y = 2  −
4
22
d
d 2 2
E
y
°
°
∣EP∣=  E 2x  E 2y =1.112×106 V⋅m−1 . For the angle, tan = =0.0629 so =3.6 or =183.6 . In
Ex
°
this case, the electric field is pointing a little below the direction −x and the proper solution is =183.6 .
−q
2.
[7 points] What is the magnitude and direction of the electric force exerted by the collection of charges on
a charge q' =-0.15¹C placed at point P?
The charge q ' 0 so the force and the electric field point in opposite directions so the force will point at
above the x direction.
 ∣=∣q '∣∣EP∣=1.5×10−6 C×1.112×106 N⋅C−1=0.1668N
The magnitude of the force is ∣F
3.
[12 points] For each of the curves A, B, C and D, write “Y” if
it could be an electric field line and “N” if it could not. Each electric charge
consist in a conductive sphere.
A: _Y_
B: _N_
C: _N_
D: _N_
°
=3.6