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

Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Solution to PHYS 1112 In-Class Exam #2B
Tue. March 22, 2011, 11:00am-12:15pm
Conceptual Problems
Problem 1: A beam of coherent (laser) light of wavelength λ is incident upon a diffraction
grating with line 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 order maxima
Fig. 2.03
Δy
Screen
θ
L
Diff. Grating
Laser Beam
(A)
(B)
(C)
(D)
(E)
decrease if we increase L (keeping λ and d fixed);
increase if we decrease λ (keeping L and d fixed);
increase if we increase d (keeping λ and L fixed).
decrease if we increase λ (keeping L and d fixed);
decrease if we increase d (keeping λ and L fixed);
Answer: (E)
Solution: At 1st maximum (closest to central maximum): sin(θ) = λ/d. Also, ∆y =
2L tan(θ), by trigonometry (see Fig. 2.03). 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. But sin(θ), hence θ and tan(θ), decrease
if d increases and λ is fixed. Hence |∆y| decreases if d increases, with L and λ being fixed.
So, (E) is correct and (C) 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 attract each other with a
force of 3µN, what force would they exert on each other if Q1 is halved (×(1/2)) in absolute
amount, r is halved (×(1/2)), and the signs of both Q1 and Q2 are reversed ? The two
charges will
(A)
(B)
(C)
(D)
(E)
repel each other with a force of 6µN
repel each other with a force of 3µN
attract each other with a force of 83 µN
attract each other with a force of 3µN
attract each other with a force of 6µN
Answer: (E)
Solution: By Coulomb’s law, the force
F =k
|Q1 |
|Q1 ||Q2 |
∝
.
r2
r2
Hence, changing |Q1 | → |Q01 | = (1/2)|Q1 | and r → r0 = (1/2)r will change
F → F0 =
(1/2)
× F = 2 × F = 2 × 3µN = 6µN .
(1/2)2
Since Q1 and Q2 initially attract each other, they initially have opposite signs. Since the sign
of Q1 is reversed and the sign of Q2 is reversed, the two charges will laos have the opposite
signs, after Q1 and Q2 are changed, and therefore will again 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 positive and Q2 is a negative 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. 2.16
P
(E)
(D)
(B)
(C)
(A)
Q2
Q1
(A)
(B)
(C)
P
(D)
(E)
E1
Fig. 2.16
E
(B)
E2
Q1
Q2
Answer: (B)
Solution: At observation point P , the positive point charge Q1 produces an electric field
~ 1 pointing away from Q1 , i.e., along the diagonal upward and rightward. The
vector E
~ 2 pointing towards Q2 , i.e.,
negative point charge Q2 produces an electric field vector E
downward, as shown in Fig.
~ is the vector sum of E
~ 1 and E
~ 2 , i.e., E
~ =E
~1 + E
~ 2 . The horizontal
The total electric field E
~ must therefore be the same as the horizontal component of E
~ 1 , i.e., rightcomponent of E
~
~
ward, since E2 is vertical, i.e., the the horizontal E2 -component is zero. Only answers (B)
and (E) satisfy this condition.
~
~ must always point into the
As also seen in the Fig., the vertical E-component
is such that E
half-plane to the right of (and below) the diagonal. Only answer (B), but not answer (E) is
consistent with this condition.
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Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Problem 4: In Experiment 1 a proton travels from point A to point B, in the electric field
~ generated by some other charged objects, as shown in the figure below. In Experiment 2
E
an electron 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. 2.18
E
A
E
E
B
E
Assume the proton experiences a rise 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 electron 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: (C)
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 a proton, of charge qproton = +e, traveling A → B,
∆Uproton (A → B) = (+e)∆V (A → B) = +160aJ .
Then, for an electron, of charge qelectron = −e, traveling B → A,
∆Uelectron (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 double-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: (A)
Solution: For a double-slit of spacing d, illuminated by wavelength λ, the mth intensity
minimum is observed at angle θ|m|−1/2 given by:
d sin(θ|m|−1/2 ) = |m| −
For m = 4 and m = 3:
1
λ
2
7λ
5λ
,
sin(θ5/2 ) =
2d
2d
is asked for. Hence:
sin(θ7/2 ) =
where θ7/2 = 76.0o is given and θ5/2
sin(θ5/2 )
=
sin(θ7/2 )
5λ
2d
7λ
2d
=
5
,
7
then :
sin(θ5/2 ) =
"
θ5/2
5
5
sin(θ5/2 ) = sin(76.0o )
7
7
#
5
= arcsin
sin(76.0o ) = 43.9o .
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Problem 6: Two point charges lie on the x-axis: Q1 = +360µC is at x1 = −15.0cm, and
Q2 = −120µC is at some unknown location x2 > 0. The electric potential V produced by
both charges, is found to be V = 0 both at x = 0cm and at very large distance, x = ∞.
Therefore, Q2 is located at:
(A)
(B)
(C)
(D)
(E)
x2
x2
x2
x2
x2
= +105cm
= +1.67cm
= +45cm
= +5.0cm
= +8.66cm
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Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Answer: (D)
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 x = 0 and use x2 > 0:
r1 = |x1 | = 15cm ,
r2 = |x2 | = x2 .
Set V = 0 and solve for x2 :
0=k
Q1
Q2
+k
,
|x1 |
x2
hence
x2 = −
−120 Q2
|x1 | = −
(15cm) = 5cm .
Q1
360
Problem 7: Two point charges, Q and q, spaced 112.4m apart, attract each other with a
force of 140.0N. What is the absolute amount of the smaller point charge q if the absolute
amount of the larger charge Q is 2.5 times the smaller charge q ?
(A)
(B)
(C)
(D)
(E)
8.9mC
5.5mC
7.5mC
9.5mC
22.2mC
Answer: (A)
Solution:
|q|2
|Q||q|
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| = 1/2.5, so:
|Q| = (112.4m)
v
u
u
t
!
1
140.0N
×
2.5
8.99 × 109 Nm2 /C2
6
!
= 8.87 × 10−3 C = 8.87mC ∼
= 8.9mC .
Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Problem 8: If a charge is uniformly spread out over a single, very thin square-shaped
piece of sheet metal, of 20m sidelength, it produces an electric field strength of 10N/C close
to the surface of the metal sheet, with the E-field vector pointing away from the metal sheet.
What is the charge stored on the metal sheet? (Hint: Think of the sheet metal as a single,
uniformly charged planar surface.)
(A)
(B)
(C)
(D)
(E)
−70.8nC
+70.8nC
+35.4nC
−35.4nC
−141.6nC
Answer: (B)
Solution:
E=
|Q|
2o A
and
A = a2 ;
so
|Q| = 2o a2 E .
~ points away from the charged foil, hence
Also, Q > 0, since E
Q = +2o a2 E = +2 (8.85 × 10−12 C2 /Nm2 ) (20.0m)2 (10N/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 an electric field strength at the
surface of the ball of E = 316V/m. What is the total electric flux Φ through the ball’s
surface?
(A)
(B)
(C)
(D)
(E)
20.2Nm2 /C
99.3Vm
56.5Nm2 /C
1767.Vm
27.1MVm
Answer: (C)
Solution:
Q = o Φ
and
E=k
|Q|
,
r2
hence
E=k
o |Φ|
|Φ|
=
r2
4πr2
or
using ko = 1/(4π) in the last equality. So
Φ = 4π (0.1193m)2 (316V/m) = 56.5V/m = 56.5N/C .
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|Φ| = 4πr2 E
Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Problem 10: If an electron is placed between the oppositely charged parallel plates of a
planar capacitor shown below, and the electric force F~ acting on the electron has a magntiude of 32aN and is pointing in downward direction, what is the voltage difference VB − VA
between the upper and the lower capacitor plate, assuming the plates are spaced 1.3cm
apart? (1aN = 10−18 N)
Plate B:
Fig. 2.20
Plate A:
(A)
(B)
(C)
(D)
(E)
+260V
−260V
−2.6V
+2.6V
−1.3V
Answer: (C)
Solution:
E = F/|q| = (32 × 10−18 N)/(1.60 × 10−19 C = 200N/C
hence
|VB − VA | = E/d = (200N/C)/(0.013m) = 2.60V .
~ = F~ /q, must
Since F~ is pointing downward and q = −e < 0, the electric field vector, E
~ always points from high to low potential, the upper plate, B,
be pointing upward. Since E
must have a lower electric potential than the lower plate. That is: VB < VA .
Thus VB − VA < 0 and VB − VA = −2.60V.
Problem 11: A 500nF capacitor in an amplifier is connected to a 12V battery. What is
the charge stored on the positive plate of this capacitor ?
(A)
(B)
(C)
(D)
(E)
41.7nC
6.0µC
18.0mC
6.0C
24.0MC
Answer: (B)
Solution:
Q = CV = (500 × 10−9 F) × (12V) = 6.0 × 10−6 C = 6.0µC
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Physics 1112
Spring 2011
University of Georgia
Instructor: HBSchüttler
Problem 12: What charge would have to be stored in the positive plate of the capacitor,
described in the previous problem in order to store 324µJ of electric field energy in it?
(A)
(B)
(C)
(D)
(E)
18µC
36mC
12C
24C
24kC
Answer: (A)
Solution:
UC =
1 2
Q
2C
Thus:
Q = (2CUC )1/2 = [(2 × (324 × 10−6 J) × (0.5 × 10−6 F)]1/2 = 18 × 10−6 C = 18µC
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