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

Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
Solution to PHYS 1112 In-Class Exam #2B
Thu. March 18, 2010, 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, with λ < d, as shown in the figure below. Assume |∆y| is the
distance (in cm) between the two 1st-order intensity maxima, observed on a screen at a
distance L on the other side of the grating. 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);
decrease if we increase λ (keeping L and d fixed);
decrease if we increase d (keeping λ and L fixed);
increase if we decrease λ (keeping L and d fixed);
increase if we increase d (keeping λ and L fixed).
Answer: (C)
At 1st-order maximum: sin θ = λ/d. Also, ∆y = 2L tan θ, by trigonometry (see Fig. 2.03).
So sin θ, hence θ, hence tan θ, hence |∆y|, will decrease if d increases or if λ decreases.
Therefore, (C) is correct; and (B), (D) and (E) are wrong. Because of ∆y = 2L tan θ, |∆y|
increases if L increases: hence (A) is wrong.
Problem 2: If two point charges Q1 and Q2 at some distance r repel each other with
a force of 160N, what force would they exert on each other if r is quadrupled (×4); Q1 is
tripled (×3) with unchanged sign; and the sign of Q2 is reversed ? The two charges will
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Physics 1112
Spring 2010
(A)
(B)
(C)
(D)
(E)
University of Georgia
Instructor: HBSchüttler
attract each other with a force of 120N
attract each other with a force of 2560N
attract each other with a force of 30N
repel each other with a force of 2560N
repel each other with a force of 30N
Answer: (C)
By Coulomb’s law, the force F = k|Q1 ||Q2 |/r2 ∝ |Q1 |/r2 . Hence, changing |Q1 | → |Q01 | =
3|Q1 | and r → r0 = 4r will change F → F 0 = (3/42 ) × F = (3/16) × 160N = 30N.
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.
Problem 3: In the figure below, Q1 and Q2 are both negative point charges with |Q1 | and
|Q2 | being of comparable magnitude. Which arrow drawn at P could correctly represent the
~ generated by Q at P ?
electric field vector E
Fig. 2.12
P
Q2
(D)
(E)
(C)
(A)
(B)
Q1
(A)
(B)
(C)
(D)
(E)
Answer: (B)
~1
At observation point P , the negative point charge Q1 produces an electric field vector E
pointing towards Q1 , i.e., downward; the negative point charge Q2 produces an electric
~ 2 pointing towards Q2 , i.e., leftward; as shown in Fig. 2.12.
field vector E
~ 2 , i.e., E
~ =E
~1 + E
~ 2 . Therefore, E
~
~ is the vector sum of E
~ 1 and E
The total electric field E
must have a vertical component pointing downward and a horizontal component pointing
leftward. Hence, (B) is the correct answer.
Problem 4: In Experiment 1 an electron travels from point A to point B, in the electric
~ generated by/between two oppositely charged capacitor plates, as shown in the figure
field E
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Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
below. In Experiment 2 a proton travels from point A to point B, in the same electric field
between the same two charged capacitor plates. A proton has a charge +e and an electron
has a charge −e.
Fig. 2.14
------------------------A
E
B
E
+++++++++++++++++++++++++++
Assume the electron experiences a rise in electric potential, ∆V = +250V, and a loss of
potential energy, ∆U = −40aJ (where 1aJ ≡ 10−18 J), in traveling from A to B. What will
the proton experience ?
(A)
(B)
(C)
(D)
(E)
∆V
∆V
∆V
∆V
∆V
= −25V and ∆U = −4aJ
= −50V and ∆U = −8aJ
= +50V and ∆U = +8aJ
= −250V and ∆U = −40aJ
= +250V and ∆U = +40aJ
Answer: (E)
~
Both proton and electron are traveling in the same E-field;
∆V is completely determined by
~
the E-field;
and it is independent of the test charge traveling through this field. So, ∆V is
the same for both: ∆V (proton) = ∆V (electron) = +250V.
However, ∆U = q∆V ; and q(proton) = +e for the proton, but q(electron) = −e for the
electron. Hence, ∆U for the proton has the opposite sign as for the electron: ∆U (proton) =
−∆U (electron) = +40aJ. So, the answer is (E).
Numerical Problems
Problem 5: If a double slit is illuminated at normal incidence by coherent (laser) light it
produces 5th order maxima at angles ±75o measured from the central (0th order) intensity
maxium. At what angles θ, measured from the central (0th order) intensity maxium, will
the 3rd order maxima be observed ?
(A)
(B)
(C)
(D)
(E)
±45.0o
±22.5o
±18.8o
±70.8o
±35.4o
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Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
Answer: (E)
Since d sin(θm ) = mλ for the m-th order maximum, (λ/d) = sin(θm )/m = sin(θ5 )/5 can be
gotten from the 5th order maximum given, with θ5 = 75.0o . Thus, for the 3rd order maxima,
m = 3, we get sin(θ3 ) = 3(λ/d) = (3/5) sin(θ5 ) = (3/5) sin(75o ) = 0.57956. So θ3 = 35.4o
and θ−3 = −35.4o . Note that only the ratio λ/d is required here to solve this problem; not
actual separate values of both λ and d.
Problem 6: Two point charges lie on the x-axis: Q1 = −60µC is at x = 0, and Q2 is at
x = 15.0cm. A third charge q will be in equilibrium when placed at x = 10.0cm. Therefore,
Q2 is
(A)
(B)
(C)
(D)
(E)
−90µC
−15µC
+30µC
+15µC
−30µC
Answer: (B)
Important: Make a drawing of the locations of all 3 charges on the x-axis! Draw the x-axis
with the +x-direction pointing rightward. In the drawing also indicate the directions of the
~ 1 and E
~ 2 generated by Q1 and Q2 , respectively, at x(q), the location
E-field contributions E
~ 1 must point towards Q1 , i.e., in −x-direction (leftward) at x(q), since
of q. Note that E
Q1 < 0.
~
To have q in equilibrium, the net force on q, exerted by Q1 and Q2 combined, F~ = eE
must be zero. Hence the total E-field generated by Q1 and Q2 at x(q) must also be zero:
~ = E
~1 + E
~ 2 . But then, E
~ 2 = −E
~ 1 which opposes E
~ 1 , i.e., points in +x-direction
0 = E
~ 2 at x(q) points towards Q2 , Q2 being located at x(Q2 ) > x(q),
(rightward) at x(q). So E
i.e., to the right of q. Therefore Q2 must be a negative charge: Q2 < 0.
The distances from Q1 to q and from Q2 to q are, respectively, r1 = |x(q) − x(Q1 )| =
(10.0 − 0.0)cm = 10.0cm and r2 = |x(q) − x(Q2 )| = (10.0 − 15.0)cm = 5.0cm (see your
~ 2 = −E
~ 1 must cancel E
~ 1, E
~ 2 and E
~ 1 must be of equal magnitude.
drawing!). Also, since E
So, by Coulomb’s law:
|Q1 |
~ 1 | = |E
~ 2 | = k |Q2 | .
k 2 = |E
r1
r22
Hence, |Q2 | = (r2 /r1 )2 × |Q1 | = (5.0/10.0)2 × (60µC) = 15µC. So, Q2 = −15µC.
Problem 7: Two point charges, spaced 112.4m apart, repel each other with a force of
96.0N. What is the amount of the larger point charge if the larger charge is 2.4 times the
smaller charge?
(A) 7.5mC
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Physics 1112
Spring 2010
(B)
(C)
(D)
(E)
University of Georgia
Instructor: HBSchüttler
13.2mC
18.0mC
22.8mC
27.6mC
Answer: (C)
By Coulomb’s law, the force is F = k|Q||q|/r2 with Q = 2.4q or q = Q/2.4. So F =
(k/2.4)|Q|2 /r2 . Solving for Q: |Q| = [2.4r2 F/k]1/2 , i.e., |Q| = [2.4×(112.4m)2 ×(96.0N)/(8.99×
109 Nm2 /C2 )]1/2 = 18.0 × 10−3 C.
Problem 8: If a charge of −360nC is uniformly spread out over a single thin, squareshaped sheet of gold foil, of 3.0m sidelength, what is the strength and the direction of the
electric field generated by that charge, very close to the surface and far from the edges of
that foil?
~ pointing towards the foil
(A) 0.1395 × 103 N/C, E
~ pointing towards the foil
(B) 2.260 × 103 N/C, E
~ pointing towards the foil
(C) 28.76 × 103 N/C, E
~ pointing away from the foil
(D) 0.1395 × 103 N/C, E
~ pointing away from the foil
(E) 28.76 × 103 N/C, E
Answer: (B)
For a single, uniformly charged, planar surface (see Formula Sheet): E = |Q|/(2o A), so
E = (360 × 10−9 C)/(2 × (8.85 × 10−12 C2 /Nm2 ) × (3m)2 ) = 2260.N/C.
~ must point towards it, i.e., towards the
Also, since the charge Q = −360nC is negative, E
charged surface, on either side of that surface.
Problem 9: The total electric flux passing through the approximately spherical surface
of a real football (i.e., a soccer ball !!) is Φ = +452Nm2 /C (assuming outward directed
normal to the surface). What is the total electric charge enclosed inside the ball ?
(A)
(B)
(C)
(D)
(E)
−3.20nC
+2.40nC
−1.60nC
−4.80nC
+4.00nC
Answer: (E)
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Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
By Gauss’ Law: Q = o Φ, so Q = (8.85 × 10−12 C2 /Nm2 ) × (+452Nm2 /C) = +4.00 × 10−9 C.
Problem 10: The 900µF capacitor in a defibrillator must be charged to store ±270mC in
its two plates. What charging voltage is required for this ?
(A)
(B)
(C)
(D)
(E)
300V
400V
500V
600V
700V
Answer: (A)
V = Q/C = (270 × 10−3 C)/(900 × 10−6 F) = 300V.
Problem 11: How much electric field energy is stored in the defibrillator capacitor in
Problem 10 when it is fully charged ?
(A)
(B)
(C)
(D)
(E)
22.5J
81.0J
40.5J
62.5J
34.5J
Answer: (C)
UE = 12 Q2 /C = 21 (270 × 10−3 C)2 /(900−6 F) = 40.5J.
Problem 12: If the electric field strength between the capacitor plates in Problem 4 is
57.0V/m, what is the magnitude and the direction of the electron’s acceleration ~a, assuming
the electron is subject only to the electric force between the plates ?
(A)
(B)
(C)
(D)
(E)
1.00 × 1013 m/s2 , ~a
1.00 × 1013 m/s2 , ~a
2.25 × 1013 m/s2 , ~a
2.25 × 1013 m/s2 , ~a
2.25 × 1013 m/s2 , ~a
pointing
pointing
pointing
pointing
pointing
leftward
downward
rightward
leftward
downward
Answer: (B)
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Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
~ So: ~a =
By Newton’s 2nd Law: ~a = F~ /m; and by definition of electric field: F~ = q E.
−19
~ and a = |q/m|E where a ≡ |~a| and E ≡ |E|.
~ So: a = [(1.60 × 10 C)/(9.11 ×
(q/m)E,
−31
13
2
10 kg] × (57V/m) = 1.00 × 10 m/s .
~ and E
~ points upward and q = −e < 0 for the electron, the force
Also, since ~a = (q/m)E
~ and hence ~a, points in the opposite direction as E,
~ i.e., ~a points downward. This
F~ = q E,
means that the electron accelerates towards the plus-plate and away from the minus-plate.
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