Download PH202-5D Final Comprehensive Exam (August 10, 2007)

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PH202-5D Final Comprehensive Exam (August 10, 2007)
• You may not open the textbook nor notebook.
• A letter size information may be used.
• A calculator may be used. However, mathematics or physics formula programmed in a calculator may
not be used.
• Write down, reasoning, calculation and answer in the blank space after each problem. Use the backside of the sheet if necessary.
• Answers without reasonable explanation nor mathematical derivation will receive no point. Partial
credit may be given to correct reasoning and mathematical procedures even if your final answers are
wrong.
• Don’t forget units in the final answers!
Important Physical Constants:
Electric charge of electron: e = −1.60 × 10−19 C
Mass of electron: me = 9.11 × 10−31 kg
Mass of proton: mp = 1.67 × 10−27 kg
1
= 8.85 × 10−12 C 2 /(N · m2 ) where k = 8.99 × 109 N · m2 /C 2
Permittivity of free space: ǫ0 = 4πk
Permeability of free space: µ0 = 4π × 10−7 T · m/A
Speed of light in vacuum: c=3.00 × 108 m/s
1
1. [15 pts.] Figure shows three point charges fixed in
place. The charge at the coordinate origin has a value
of q1 = +8.00 µC; the other two have identical charge
of q2 = q3 = −4.00 µC.
(a) Draw electric field lines the figure below. You
must draw a sufficient number of lines so that the
electric field inside and outside of the triangle is
clearly displayed. (At least 8 lines.)
(b) Determine the net force (magnitude and direction) exerted on q1 by the other two charges.
(c) What is the electrostatic potential energy of the
whole system assuming that the eletrostatic potential energy is zero when the charges are infinitely far apart?
(a)
(b) Let F2 and F3 be force exerted by q2 and q3 , respectively. Using Coulomb’s law,
kq1 q2 (8.99 × 109 ) × (8.00 × 10−6 ) × (4.00 × 10−6 )
F2 = F3 = 2 =
= 0.17N
r
(1.30)2
There x and y components are
F2x = 0.170 cos(23◦ ) = 0.156 N,
F2y = 0.170 sin(23◦ ) = 0.066 N ;
F3x = 0.170 cos(23◦ ) = 0.156 N,
F3y = −0.170 sin(23◦ ) = −0.066 N ;
Therfore, the components of the net force are given by
Fx = F2x + F 3x = 2 × 0.156 = 0.312N,
Fy = F2y + F3y = 0.066 − 0.066 = 0.00 N
ANS: magnitude=0.312 N ; direction=+x (0◦ )
(c)
U13 = U12 =
U23 =
(8.99 × 109 ) × (8.00 × 10−6 ) × (−4.00 × 10−6 )
kq1 q2
=
= −0.221 J
r12
1.3
(8.99 × 109 ) × (−4.00 × 10−6 ) × (−4.00 × 10−6 )
kq2 q3
= 0.142 J
=
r23
2 × 1.3 × sin(23◦ )
UN ET = U12 + U13 + U31 = −0.221 − 0.221 + 0.142 = −0.30 J
ANS: -0.300 J
2
2. [15 pts.] Consider a parallel plate capacitor with a capacitance of 2.5 µF . Initially, there is no net
electric charge on each plates. When a certain amount of electric charge is transfered from one plate to the
other, the potential difference between two plates becomes 2.0 V .
(a) How many electrons are transfered between the plates?
(b) What is the minimum energy needed to transfer that amount of charge?
(c) A proton is released from the positively charged plate and travels to the negatively charged plate.
What is its kinetic energy when it arrives at the negatively charged plate?
(a)
Q = CV = (2.5 × 10−6 ) × 2.0 = 5.0 × 10−6 C
N=
5.0 × 10−6
Q
=
= 3.1 × 1013
e
1.6 × 10−19
ANS: .1 × 1013
(b)
1
E = CV 2 = 0.5 × (2.5 × 10−6 ) × (2.0)2 = 5.0 × 10−6 J
2
ANS: 5.0 × 10−6 J
(c) Using the law of mechanical energy conservation,
1
mv 2 = eV = (1.6 × 10−19 ) × 2.0 = 3.2 × 10−19 J
2
ANS: 3.2 × 10−19 J
3
3. [14 pts.] A metal rod of length 15 cm is sliding
on two parallel metal tracks to the right with a velocity
of 1.2 m/s. The tracks are connected at one end so
that they and the rod form a closed circuit as shown in
Figure. The rod has a resistance 10 Ω, and the tracks
have negligible resistance. A uniform magnetic field
of 2.0 T perpendicular to the plane of this circuit (into
the page) is applied.
(a) What is the magnitude and direction (up or down) of the induced current through the rod?
(b) What is the magnitude and the direction (left or right) of the force exerted on the rod by the magnetic
field?
(a)
∆x
∆Φ
= BL
= BLv = 2.0 × 0.15 × 1.2 = 0.36 V
∆t
∆t
0.36
|emf|
=
= 0.036 A
I=
R
10.
Since the magnetic flux into the page is increasing, the direction of the indueced magnetic field should out
of page (Lenz’s law). Therfore, the induced current in the rod is upward (RHL).
ANS: 0.036 A, up
|emf| =
(b)
F = ILB = 0.036 × 0.15 × 2.0 = 0.011N
Using RHL, the direction is left.
ANS: 0.011 N , left
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4. [14 pts] An object (the arrow in Figure) is placed 15.0 cm in front of a spherical concave mirror whose
radius is 20.0 cm.
(a) Find the location of the image using a mirror equation.
(b) If the height of the object is 6.00 cm, what is the image height?
(c) Construct the image in Figure below by drawing. Leave rays and other lines used to construct the
image in the figure so that the instructor knows how you find the image. (At least three rays should
be used.) [Make it sure that the drawing is consistent with the calculation.]
(a) f = R/2 = 10.0 cm and do = 15.0 cm.
1
1
1
1
1
1
= −
=
−
=
di
f
do
10.0 15.0
30.0
Hence, di = 30.0 cm.
ANS: 30.0 cm in front of the mirror
(b)
hi = −
di
30.0
× 6.00 = −12.0
ho = −
do
15.0
ANS: -12.0 cm
(c)
5
5. [15 pts.] A 12-V battery and four 16-Ω light bulbs are connected
as shown in Figure.
(a) How much current is drawn from the battery?
(b) How much power is consumed by the whole system?
(c) Determine the power dissipated in the light bulb L2 .
(a)
1
1
1
1
1
1
+
=
=
+
=
R23
R1 R2
16 16
8
R23 = 8.0 Ω
R123 = R1 + R23 = 16 + 8.0 = 24
1
1
1
1
5
1
=
+
=
+
=
Req
R123 R4
24 16
48
Req = 48/5 = 9.6 Ω
I=
V
12
=
= 1.25 A
Req
9.6
ANS: 1.25 A
(b)
P = IV = 1.25 × 12 = 15 W
ANS: 15 W
(c)
I4 =
12
V
=
= 0.75 A
R4
16
I1 = I − I4 = 1.25 − 0.75 = 0.50 A
V1 = I1 R1 = 0.50 × 16 = 8.0V
V2 = V − V1 = 12 − 8.0 = 4.0 V
P2 =
V22
4.02
=
= 1.0 W
R2
16
ANS: 1.0 W
6
6. [14 pts.] Two long, straight wires carry the
identical currents of I1 =I2 =2.0 A as shown in Figure. A 4.0-µC point charge is moving at 30 m/s
along the wires in direction B shown in Figure.
The distance between the wires are 6.0 cm and
the distance between wire 1 and the charge is 4.0
cm. Answers the following questions. The direction should be specified by a letter (U , D, F , B,
L, R) shown in the figure.
(a) Find the magnitude and direction of the
magnetic field at the position of the charge.
(b) Determine the magnitude and direction of
the force exerted on the charge by the magnetic field
(a)
B1 =
µ0 I1
4π × 10−7 × 2.0
=
= 1.0 × 10−5 T
2πr1
2π0.04
B1 is upward at the position of the charge
B2 =
4π × 10−7 × 2.0
µ0 I2
=
= 2.0 × 10−5 T
2πr2
2π0.02
B2 is downward at the position of the charge. Hence the net magnetic field is B1 − B2 = −1.0 × 10−5 T .
ANS: 1.0 × 10−5 T , direction = D
(b)
F = qvB = 4.0 × 10−6 × 30 × 1.0 × 10−5 = 1.2 × 10−9 N
ANS: 1.2 × 10−9 N , direction=L
7
7. [12 pts.] A 375-nm thick soap film is surrounded on
both sides by air as shown in Figure. The soap film has
an index of reflecetion n = 1.33. When sunlight strikes
the film nearly perpendicularly, two colors look bright in
the reflected light due to constructive interference. Noting that sunlight consists of visible light whose wavelengths extend from 380 nm to 750 nm. Which wavelengths in this range become bright?
2t =
λ=
N+
1
2
λ
n
2 × 375 × 1.33
998
2tn
=
=
N + 21
N + 12
N + 12
where N is non-negative integer. For N = 0, 1, 2, 3, ..., λ = 1996, 665, 399, 285, .... Hence N = 1 and
N = 2 correspond to the visible light.
ANS: 399 nm and 665 nm
8