Physics 1112
Spring 2014
University of Georgia
Instructor: HBSchüttler
PHYS 1112 In-Class Exam #2, Version A
Tue. March 18, 2014, 11:00am-12:15am
This is a closed-book, closed-notes exam, but you are permitted to bring and use a clean copy of the official
Formula Sheet for this exam, which you should have printed out from the PHYS1112 web page.
The exam consists of 15 multiple-choice questions. Each question is worth one raw score point, without
penalty for wrong answers. No partial credit will be given. Read all the questions at the start so that you
can allocate your time wisely. (Do easy questions first!)
You may use a scientific calculator for arithmetic only; your calculator must be non-graphing, nonprogrammable, and non-algebraic. You are not allowed to share your calculator. The use of cell phones,
pagers, PDAs, or any other electronic devices (besides calculators) is forbidden.
• Do not open the exam until you are told to begin. When told to open the exam, you may tear
off the last few sheets, marked WORKSPACE, as scratch paper.
• Make sure the scantron sheet has your name and your UGA Card ID (810-...) number filled
in. Make sure you also have entered your name, UGA Card ID number and signature on the exam
cover page (this page!) below.
• Make sure to enter your exam Version ID, A, B, C, D, or E, on your scantron as your answer
to Problem 16. Without version ID, your exam can not be graded and you will get a score of zero.
• At the end of the exam period you must hand in both your scantron sheet and this entire
exam paper, except for last few sheets of scratch paper, with cover page signed and with your name
and UGA Card ID (810-...) number filled-in on both scantron and cover page.
• Your exam will not be graded, and you will receive a score of zero, if you do not hand in all the
foregoing required materials with name and ID information filled in. You will also receive a score
of zero if you talk to or otherwise communicate with anyone during the exam, except for attending
instructor and teaching assistants. You will also receive a score of zero if you use any unauthorized
materials or devices during the exam.
• You have until the end of the class period (i.e., until 10:45am for Period 2 Class, until 12:15pm for
Period 3 Class) to finish the exam and hand in the required exam materials described above.
By signing below, you indicate that you understand the instructions for this exam and agree to abide by
them. You also certify that you will uphold the university standards of academic honesty for this exam,
and will not tolerate any violations of these standards by others.
Name (please print):
UGACard ID (810-...) #:
Signature:
1
Physics 1112
Spring 2014
University of Georgia
Instructor: HBSchüttler
Conceptual Problems
Problem 1: In Fig. 2.30, Q1 is a positive and Q2 is a negative point charge, both of comparable magni~ generated by Q1
tude. Which arrow drawn at P could correctly represent the total electric field vector E
and Q2 at P ?
(C)
P
(D)
Q2
(E)
(A)
Fig. 2.30
(B)
Q1
(A)
(B)
(C)
(D)
(E)
Problem 2: If two point charges q1 and q2 at some distance r repel each other with a force of 9N, what
force would they exert on each other if r is tripled (×3), q1 is halved (×1/2); and q2 is quadrupled (×4)
and its sign is reversed. The two charges will
(A)
(B)
(C)
(D)
(E)
attract each other with a force of 6N.
repel each other with a force of 2N.
attract each other with a force of 12N.
repel each other with a force of 6N.
attract each other with a force of 2N.
Problem 3: Two positive point charges, each of charge Q > 0, are located at the two upper corners of
a square; and two negative point charges, each of charge −Q < 0, are located at the two lower corners of
that square, as shown in Fig. 2.31. The electric potential, V , generated by these four charges, is zero at
infinite distance from the square. The points A, B, C, and D are the mid-points of the edges and M is
the center of the square; and the potential values at these five points are denoted by VA , VB , VC , VD and
VM , respectively. Which of the following is true?
Q
+
A
−Q
(A)
(B)
(C)
(D)
(E)
Q
D
+
C
M
−
−
B
VA
VA
VC
VB
VC
Fig. 2.31
−Q
= VB = VC = VD = 0 and VM < 0.
= VC = VM = 0 and VB = −VD < 0.
= VD > 0 and VA = VB < 0 and VM = 0.
= VD = VM = 0 and VA = −VC > 0.
= VA = VM = 0 and VD = −VB < 0.
2
Physics 1112
Spring 2014
University of Georgia
Instructor: HBSchüttler
Problem 4: A laser beam of wavelength λ is incident upon a narrow single slit of width W , with λ < W , as
shown in Fig. 2.28. Assume |∆y| is the distance (in cm) between the two dark fringes (intensity minima),
observed closest to the central maximum on a screen at a distance L on the other side of the slit. This
distance |∆y| will
1st dark fringes
Δy
Screen
L
Fig. 2.28
Single Slit
Laser Beam
(A)
(B)
(C)
(D)
(E)
decrease if we increase W (keeping λ and L fixed).
decrease if we increase λ (keeping L and W fixed).
increase if we increase W (keeping L and λ fixed).
increase if we decrease λ (keeping L and W fixed).
decrease if we increase L (keeping λ and W fixed).
Problem 5: A diffaction grating is illuminated with coherent (laser) light with wavelength λ < d and
oscillation period τ = λ/c where d is the spacing between adjacent slits in the grating. The second order
intensity maximum, to the right of the central intensity maximum M , is located at point Q, as shown in
Fig. 2.29. A wave crest A from slit R and a wave crest B from the neighboring slit S, to the right of R,
have both departed at the same time from their respective slits of origin. Therefore, at Q,
M
Q
Fig. 2.29
R S
Diffr. Grating
Laser Beam
(A)
(B)
(C)
(D)
(E)
A
B
B
B
A
will arrive 2 periods before B.
will arrive 2 periods before A.
will arrive 3/2 periods before A.
will arrive 4 periods before A.
and B will arrive at the same time.
3
Physics 1112
Spring 2014
University of Georgia
Instructor: HBSchüttler
Numerical Problems
Problem 6: If a double-slit is illuminated at normal incidence by coherent (laser) light it produces, on a
screen parallel to and far away from the slits, its 4th dark fringe (≡intensity minimum), at an angle of
±76.0o , measured from the central bright fringe (≡intensity maximum). At what angle, θ, measured from
the central bright fringe, will the 3rd dark fringe appear?
(All dark fringes are counted here from the central bright one, going outward. So, the 1st dark fringe is
the one closest to the central bright one.)
(A)
(B)
(C)
(D)
(E)
±46.7o
±54.3o
±43.9o
±57.0o
±70.8o
Problem 7: For the double-slit experiment described in Problem 6, how many bright finges, in total, will
be obervable on the screen? (Include in your count the central bright fringe, as well as every bright fringe
to either side of the central one.)
(A)
(B)
(C)
(D)
(E)
8
9
5
6
7
Problem 8: Assume the double-slit experiment described in Problem 6 (with the 4th dark fringe observed
at ±76.0o from the central bright one) was actually perfomed with the entire apparatus (laser, double-slit,
screen) immersed in air. At what angle, θ, measured from the central bright fringe, will the 4th dark fringe
be observed if that entire apparatus is now submerged in water, with an index of refraction nWater = 1.33?
Hints: For air, nAir = 1.00. The vacuum wavelength of the laser and the distance between the two slits
is the same in both experiments.
(A)
(B)
(C)
(D)
(E)
±46.8o
A 4th dark fringe cannot be observed under water.
±71.7o
±57.14o
±101.1o
4
Physics 1112
Spring 2014
University of Georgia
Instructor: HBSchüttler
Problem 9: Two point charges lie on the x-axis: Q1 = +80C is at x = 0, and Q2 = −80C is at x = 2.0m.
Find the x-component, Ex , of the electric field vector, produced by both charges, Q1 and Q2 , at x = 4.0m.
Also find the x-component, Fx , of the electric force exerted on a test charge, q = −2C, placed on the x-axis
at x = 4.0m.
Hint: Use k ∼
= 9 × 109 Nm2 /C2 and 109 N ≡ 1GN.
(A)
(B)
(C)
(D)
(E)
Ex
Ex
Ex
Ex
Ex
= −135 GN/C
= −225 GN/C
= +135 GN/C
= +225 GN/C
= −135 GN/C
and
and
and
and
and
Fx
Fx
Fx
Fx
Fx
= +270 GN
= +450 GN
= −270 GN
= −450 GN
= −270 GN
Problem 10: Two point charges lie on the x-axis: Q1 = +90µC is at x = 0, and Q2 is at x = 20.0cm. The
electric potential V produced by both charges, is found to be V = 0 both at x = 24.0cm and at very large
distance, x = ∞. Therefore, Q2 is
(A)
(B)
(C)
(D)
(E)
+2.5
−15.
−2.5
+15.
−10.
µC.
µC.
µC.
µC.
µC.
Problem 11: Two point charges, Q and q, spaced 50m apart, attract each other with a force of 100N.
What is the amount of the larger point charge, |Q|, if the larger charge is 2 times the amount of the smaller
charge, |q|?
(A)
(B)
(C)
(D)
(E)
22.2 mC
5.5 mC
8.9 mC
7.5 mC
9.5 mC
Problem 12: A point charge Q, placed at the center of a cube, produces an electric field strength of
E = 158V/m, somewhere on the cube’s surface, at a distance of 11.2cm from the cube’s center. What is
the magnitude of the total electric flux, |Φ|, passing through the cube’s surface?
(A)
(B)
(C)
(D)
(E)
12 Vm
100 Vm
6.2 Vm
48 Vm
25 Vm
5
Physics 1112
Spring 2014
University of Georgia
Instructor: HBSchüttler
Problem 13: A charge of −35.4nC is uniformly distributed over a single, very thin square sheet of
~ and
aluminum foil, of 2.0m sidelength, spread out flat in the x−y-plane. Find the magnitude, E ≡ |E|,
the direction of the electric field vector generated by this charge at some point P ≡ (x, y, z), with z > 0,
close to the center of the foil.
(A)
(B)
(C)
(D)
(E)
E
E
E
E
E
= 500N/C,
= 1000N/C,
= 500N/C,
= 500N/C,
= 1000N/C,
~ pointing in the +z-direction
E
~ pointing in the −y-direction
E
~ pointing in the −y-direction
E
~ pointing in the −z-direction
E
~ pointing in the −z-direction
E
Problem 14: Two large, metallic, planar, parallel, charged capacitor plates have an electric potential
difference of ∆V = V2 − V1 = +2500V, where V1 and V2 are the electric potentials on the left and right
plate, respectively. as shown here:
Plate 1:
Plate 2:
Fig. 2.32
An electron is shot through a small hole in the left plate, into the space between the two plates. The
electron, while traveling from the left to the right plate, ...
(A)
(B)
(C)
(D)
(E)
will lose 4.0 × 10−16 J in kinetic energy between left and right plate.
will gain 2.0 × 10−16 J in kinetic energy between left and right plate.
will gain 8.0 × 10−16 J in kinetic energy between left and right plate.
will gain 4.0 × 10−16 J in kinetic energy between left and right plate.
must have a minimum kinetic energy of 8.0 × 10−16 J, as it passes through the left plate, in order to
reach the right plate.
Problem 15: Answer the same question as in Problem 14, but assuming that the potential difference
between the capacitor plates has been changed to ∆V = +5000V and the particle shot through the hole
of the left plate is a proton, not an electron. The possible answers ...
(A) ...
(B) ...
(C) ...
(D) ...
(E) ...
are exactly the same as for Problem 14.
Problem 16:
Enter your exam Version, A, B, C, D or E, on your scantron as the answer to Problem 16!!
Your exam version is printed in bold face on top of the cover page of this exam paper which you signed.
6
Physics 1112
Spring 2014
University of Georgia
Instructor: HBSchüttler
WORKSPACE
7
Physics 1112
Spring 2014
University of Georgia
Instructor: HBSchüttler
WORKSPACE
8
Physics 1112
Spring 2014
University of Georgia
Instructor: HBSchüttler
WORKSPACE
9
Physics 1112
Spring 2014
University of Georgia
Instructor: HBSchüttler
WORKSPACE
10
Physics 1112
Spring 2014
University of Georgia
Instructor: HBSchüttler
WORKSPACE
11
Physics 1112
Spring 2014
University of Georgia
Instructor: HBSchüttler
WORKSPACE
12