Download Solution to PHYS 1112 In-Class Exam #2A

Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Solution to PHYS 1112 In-Class Exam #2A
Tue. March 22, 2011, 9:30am-10:45am
Conceptual Problems
Problem 1: A beam of coherent (laser) light of wavelength λ is incident upon two parallel
slits with spacing d, and with λ < d, as shown in the figure below. Assume |∆y| is the
distance (in cm) between the two bright fringes (interference intensity maxima), observed
closest to the central maximum on a screen at a distance L on the other side of the doubleslits. This distance |∆y| will
1st bright fringes
Fig. 2.21
Δy
Screen
L
Double Slit
Laser Beam
(A)
(B)
(C)
(D)
(E)
increase if we decrease L (keeping λ and d fixed);
decrease if we increase λ (keeping L and d fixed);
increase if we decrease d (keeping λ and L fixed);
increase if we decrease λ (keeping L and d fixed);
decrease if we decrease d (keeping λ and L fixed).
Answer: (C)
Solution: At 1st maximum (closest to central maximum): sin(θ) = λ/d. Also, ∆y =
2L tan θ, by trigonometry (see Fig. 2.21). So sin θ, hence θ, hence tan θ and hence |∆y|,
will all decrease if λ decreases. Also, at fixed λ and d, θ is fixed, thus |∆y| decreases if L
decreases. Therefore, (A), (B) and (D) are wrong. d decreases and L and λ are fixed. But
sin(θ), hence θ and tan(θ), increase if d decreases and λ is fixed. Hence |∆y| increases if d
decreases, with L and λ being fixed. So, (C) is correct and (E) is wrong.
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Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Problem 2: If two point charges Q1 and Q2 at some distance r repel each other with a
force of 3µN, what force would they exert on each other if Q2 is quadrupled (×4) without
change of sign; r is halved (×(1/2)), and the sign of Q1 is reversed ? The two charges will
(A)
(B)
(C)
(D)
(E)
repel each other with a force of 24µN
attract each other with a force of 3µN
attract each other with a force of 48µN
attract each other with a force of 24µN
attract each other with a force of 6µN
Answer: (C)
Solution: By Coulomb’s law, the force
F =k
|Q1 ||Q2 |
|Q2 |
∝ 2 .
2
r
r
Hence, changing |Q2 | → |Q02 | = 4|Q2 | and r → r0 = (1/2)r will change
F → F0 =
4
× F = 16 × F = 16 × 3µN = 48µN .
(1/2)2
Since Q1 and Q2 initially repel each other, they initially have the same sign. Since the sign
of Q1 is unchanged and the sign of Q2 is reversed, the two charges will have the opposite
sign, after Q1 and Q2 are changed, and therefore will attract each other.
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Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Problem 3: In the figure below Q1 is a negative and Q2 is a positive point charge, |Q1 |
and |Q2 | being of comparable magnitude. Which arrow drawn at P could correctly represent
~ generated by Q1 and Q2 at P ?
the electric field vector E
Fig. 2.15
P
(D)
Q2
(E)
(B)
(C)
(A)
Q1
(A)
(B)
(C)
(D)
(E)
Fig. 2.15
P
Q2
E2
E1
E
(A)
Q1
Answer: (A)
Solution: At observation point P , the negative point charge Q1 produces an electric field
~ 1 pointing towards Q1 , i.e., along the diagonal downward and leftward. The
vector E
~ 2 pointing away from Q2 , i.e.,
positive point charge Q2 produces an electric field vector E
rightward, as shown in Fig.
~ is the vector sum of E
~ 1 and E
~ 2 , i.e., E
~ =E
~1 + E
~ 2 . The vertical
The total electric field E
~ must therefore be the same as the vertical component of E
~ 1 , i.e., downward,
component of E
~ 2 -component is zero. Only answers (A) and (C)
since E~2 is horizontal, i.e., the the vertical E
satisfy this condition.
~
~ must always point into
As also seen in the Fig., the horizontal E-component
is such that E
the half-plane to the right of (and below) the diagonal. Only answer (A), but not answer
(C) is consistent with this condition.
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Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Problem 4: In Experiment 1 an electron travels from point A to point B, in the electric
~ generated by some other charged objects, as shown in the figure below. In Experiment
field E
2 a proton travels from point B to point A, in the same electric field. A proton has a charge
+e and an electron has a charge −e.
Fig. 2.17
E
A
E
E
B
E
Assume the electron experiences a drop in electric potential, ∆V = −1000V, and a gain of
potential energy, ∆U = +160aJ (where 1aJ ≡ 10−18 J), in traveling from A to B. What will
the proton experience in traveling from B to A?
(A)
(B)
(C)
(D)
(E)
∆V
∆V
∆V
∆V
∆V
= −1000V and ∆U = −160aJ
= −500V and ∆U = +80aJ
= −1000V and ∆U = +160aJ
= +1000V and ∆U = −160aJ
= +1000V and ∆U = +160aJ
Answer: (E)
Solution: For travel A → B, independent of test particle (electron or proton):
∆V (A → B) = VB − VA = −1000V
Thus, for travel B → A, independent of test particle (electron or proton):
∆V (B → A) = VA − VB = −∆V (A → B) = +1000V .
Then, for an electron, of charge qelectron = −e, traveling A → B,
∆Uelectron (A → B) = (−e)∆V (A → B) = +160aJ .
Then, for a proton, of charge qproton = +e, traveling B → A,
∆Uproton (B → A) = (+e)∆V (B → A) = (−e)∆V (A → B) = +160aJ .
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Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Numerical Problems
Problem 5: If a single, narrow slit is illuminated at normal incidence by coherent (laser)
light it produces 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)
±43.9o
±71.6o
±54.3o
±46.7o
±57.0o
Answer: (D)
Solution: For single slit of width W , illuminated by wavelength λ, the mth intensity minimum is observed at angle θm given by:
W sin(θm ) = m λ
For m = 4 and m = 3:
3λ
4λ
,
sin(θ3 ) =
W
W
o
where θ4 = 76.0 is given and θ3 is asked for. Hence:
sin(θ4 ) =
sin(θ3 ) =
3λ
3 4λ 3
3
= ×
= sin(θ4 ) = sin(76.0o )
W
4
W
4
4
"
#
3
sin(76.0o ) = 46.7o .
θ3 = arcsin
4
Problem 6: Two point charges lie on the x-axis: Q1 = +360µC is at x = 0, and Q2 is at
x = 10.0cm. The electric potential V produced by both charges, is found to be V = 0 both
at x = 12.0cm and at very large distance, x = ∞. Therefore, Q2 is
(A)
(B)
(C)
(D)
(E)
+60µC.
+10µC.
−60µC.
−10µC.
−30µC.
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Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Answer: (C)
Solution: For charge Q1 located at x1 and Q2 located at x2 the electric potential V at
location x is:
V = V1 + V2 = k
Q1
Q2
+k
,
r1
r2
r1 = |x − x1 | ,
r2 = |x − x2 | .
Set V = 0 and solve for Q2 :
0=k
Q2
Q1
+k
,
r1
r2
hence
Q2 = −
r2
Q1 .
r1
Here: x1 = 0cm, x2 = 10cm and x = 12cm, hence r1 = |12 − 0|cm = 12cm and r2 = |12 − 10|cm =
2cm. So
2
Q2 = − × 360µC = −60µC .
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Problem 7: Two point charges, Q and q, spaced 112.4m apart, repel each other with a
force of 140.0N. What is the amount of the larger point charge Q if the larger charge is 2.5
times the smaller charge q?
(A)
(B)
(C)
(D)
(E)
8.9mC
5.5mC
7.5mC
9.5mC
22.2mC
Answer: (E)
Solution:
|Q||q|
|Q|2
F = k 2 = k 2
r
r
|q|
|Q|
!
,
hence
|Q| = r
v
u
u
t
!
|Q| F
.
|q| k
Here, F = 140.0N, r = 112.4m, k = 8.99 × 109 Nm2 /C2 and |Q|/|q| = 2.5, so:
|Q| = (112.4m)
v
u
u
t2.5 ×
140.0N
8.99 × 109 Nm2 /C2
6
!
= 22.2 × 10−3 C = 22.2mC .
Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Problem 8: If a charge is uniformly spread out over a single, very thin square sheet of
aluminum foil, of 4.0m sidelength, it produces an electric field strength of 250N/C close to
the surface of the foil, with the E-field vector pointing towards the foil. What is the charge
stored on the foil? (Hint: Think of the aluminum foil as a single, uniformly charged planar
surface.)
(A)
(B)
(C)
(D)
(E)
−70.8nC
+70.8nC
+35.4nC
−35.4nC
−141.6nC
Answer: (A)
Solution:
E=
|Q|
2o A
and
A = a2 ;
so
|Q| = 2o a2 E .
~ points towards the charged foil, hence
Also, Q < 0, since E
Q = −2o a2 E = −2 (8.85 × 10−12 C2 /Nm2 ) (4.0m)2 (250N/C) = −70.8 × 10−9 C .
Problem 9: A point charge Q, placed the center of an approximately spherical real,
NBA-approved basket ball of 11.93cm radius, produces a total electric flux through the
ball’s surface of Φ = 113Vm. What is the electric field strength at the surface of the ball?
(A)
(B)
(C)
(D)
(E)
632N/C
9.03µV/m
80.8kN/C
80.8MN/C
20.2V/m
Answer: (A)
Solution:
Q = o Φ
and
E=k
|Q|
,
r2
hence
E=k
o |Φ|
|Φ|
=
2
r
4πr2
using ko = 1/(4π) in the last equality. So
E=
113Vm
= 632V/m = 632N/C .
4π (0.1193m)2
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Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Problem 10: If the voltage difference VA − VB between the two oppositely charged parallel
capacitor plates shown below is −2600V, and a proton is placed between the plates, what is
the magnitude and the direction of the electric force F~ on the proton, assuming the plates
are spaced 1.3cm apart? (1aN = 10−18 N, 1fN = 10−15 N.)
Plate A:
Fig. 2.19
Plate B:
(A)
(B)
(C)
(D)
(E)
320aN, F~ pointing upward
32fN, F~ pointing upward
64fN, F~ pointing upward
640aN, F~ pointing downward
32fN, F~ pointing downward
Answer: (B)
Solution:
E=
|V |
|VA − VB |
2600V
=
=
= 200, 000V/m = 2.0 × 105 V/m ,
d
d
0.013m
hence
F = |q|E = (1.6 × 10−19 C) (2.0 × 105 V/m) = 32 × 10−15 N = 32fN .
Since VA − VB < 0, the potential VA on the upper plate is lower than VB on the lower plate.
~ points from high to low potential, E
~ points upwards, towards Plate A. Since q > 0,
Since E
~ points in the same direction as E,
~ i.e., upward as well.
F~ = q E
Problem 11: A 1500µF capacitor in a starter motor is connected to a 12V battery. What
is the charge stored on the positive plate of this capacitor ?
(A)
(B)
(C)
(D)
(E)
0.125mC
18mC
18C
72C
8.0kC
Answer: (B)
Solution:
Q = CV = (1500 × 10−6 F) (12V) = 18 × 10−3 C = 18mC .
8
Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Problem 12: What battery voltage would have to be applied to the starter motor capacitor,
described in the previous problem in order to store 432mJ of electric field energy in it?
(A)
(B)
(C)
(D)
(E)
18µV
36mV
12V
24V
24kV
Answer: (D)
Solution:
1
U = CV 2 ,
2
hence
V =
2U
C
!1/2
=
9
2 × 432 × 10−3 J
1.5 × 10−3 F
!1/2
= 24V .