Download Solutions to Conceptual Practice Problems PHYS 1112 In-Class Exam #2A+2B

Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Solutions to Conceptual Practice Problems
PHYS 1112 In-Class Exam #2A+2B
Thu. Mar. 19, 2009, 11:00am-12:15pm and 2:00pm-3:15pm
CP 2.01: In a two-source interference experiment two sources are oscillating in phase with
the same period. The first intensity minimum, closest to the central intensity maximum M
and to the left of M, is located at point P, as shown in Fig. 2.01. A wave crest A from source
1 and a wave trough B from source 2 arrive simultaneously at P. Therefore, A and B must
have departed from their respective sources as follows:
Fig. 2.01
P
l1
M
l2
Source 1
(A)
(B)
(C)
(D)
(E)
Source 2
A departed 1/2 period before B.
B departed 1/2 period before A.
A departed 1 period before B.
B departed 1 period before A.
A departed 3/2 period before B.
CP 2.01 Answer: (B)
At P , the pathlength difference ∆l ≡ l2 − l1 (see Fig. 2.01) must be half a wavelength:
∆l = 12 λ, since it’s the first interference minimum to the left of center M (with ∆l = 0 at
M ). Hence, B has 12 λ further to travel than A, which takes an extra travel time of 12 period.
So B must leave 21 period earlier than A to arrive simultaneously at P . [Recall: v = λ/τ ; so
it takes 1 period τ for a crest or trough to travel 1 wavelength λ.]
CP 2.02: A diffaction grating is illuminated with coherent (laser) light with a wavelength
λ < d where d is the spacing between adjacent slits in the grating. The first intensity
maximum, to the right of and closest to the central intensity maximum M, is located at
point Q, as shown in Fig. 2.02. A wave crest A from slit R and a wave crest B from the
neighboring slit S to the right of R, arrive simultaneously at Q. Therefore, A and B must
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
have departed from their respective slits as follows:
Fig. 2.02
M
lR
Q
lS
R S
Diff. Grating
Laser Beam
(A)
(B)
(C)
(D)
(E)
A departed 1/2 period before B.
B departed 1/2 period before A.
A departed 1 period before B.
B departed 1 period before A.
A departed 3/2 period before B.
CP 2.02 Answer: (C)
At Q, the pathlength difference ∆l ≡ lR − lS (see Fig. 2.02) must be one full wavelength:
∆l = λ, since it’s the first interference maximum to the right of center M (with ∆l = 0 at
M ). Hence, A has 1 λ further to travel than B, which takes an extra travel time of 1 period.
So A must leave 1 period earlier than B to arrive simultaneously at Q. [Recall: v = λ/τ ; so
it takes 1 period τ for a crest or trough to travel 1 wavelength λ.]
CP 2.03: 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 Fig. 2.03. 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
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
1st order maxima
Fig. 2.03
Δy
Screen
θ
L
Diff. Grating
Laser Beam
(A)
(B)
(C)
(D)
(E)
decrease if we increase L (at fixed λ and d)
decrease if we increase λ (at fixed L and d)
increase if we decrease λ (at fixed L and d)
increase if we increase d (at fixed λ and L)
decrease if we increase d (at fixed λ and L)
CP 2.03 Answer: (E)
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, (E) is correct; and (B), (C) and (D) are wrong. Because of ∆y = 2L tan θ, |∆y|
increases if L increases: hence (A) is wrong.
CP 2.04: A beam of coherent (laser) light of wavelength λ is incident upon two parallel
slits with a spacing d and λ < d, as shown in Fig. 2.04. Assume |∆y| is the distance (in cm)
between the two adjacent dark fringes (interference intensity minima), observed closest to
the central maximum on a screen at a distance L on the other side of the slits. This distance
|∆y| will
1st dark fringes
Fig. 2.04
Δy
Screen
θ
L
Double Slit
Laser Beam
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Physics 1112
Spring 2009
(A)
(B)
(C)
(D)
(E)
University of Georgia
Instructor: HBSchüttler
increase if we increase L (at fixed λ and d)
increase if we decrease λ (at fixed L and d)
decrease if we increase λ (at fixed L and d)
increase if we increase d (at fixed λ and L)
decrease if we decrease d (at fixed λ and L)
CP 2.04 Answer: (A)
At the dark fringe (interference minimum) nearest to center: sin θ = 21 λ/d. Also, ∆y =
2L tan θ, by trigonometry (see Fig. 2.04). So sin θ, hence θ, hence tan θ, hence |∆y|, will
increase if d decreases or if λ increases. Therefore, (B), (C), (D) and (E) are wrong. Because
of ∆y = 2L tan θ, |∆y| increases if L increases: hence (A) is correct.
CP 2.05: 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); and q1 and q2 are
unchanged ? The two charges will
(A)
(B)
(C)
(D)
(E)
repel each other with a force of 22µN
attract each other with a force of 10N
repel each other with a force of 10N
attract each other with a force of 65000MN
repel each other with a force of 65000MN
CP 2.05 Answer: (C)
By Coulomb’s law, the force F = k|q1 ||q2 |/r2 ∝ 1/r2 . Hence, changing r → r0 = 4r will
change F → F 0 = F/42 = 160N/16 = 10N.
Since q1 and q2 initially repel each other, they both have the same sign. Since the signs of
q1 and q2 are unchanged, the two charges will continue to repel each other after r is increased.
CP 2.06: If two point charges q1 and q2 at some distance r attract each other with a force
of 10N, what force would they exert on each other if q1 is quadrupled (×4) and its sign is
reversed; and q2 and r are unchanged ?
(A)
(B)
(C)
(D)
(E)
repel each other with a force of 40N
attract each other with a force of 2.1mN
repel each other with a force of 2.1mN
attract each other with a force of 56000MN
repel each other with a force of 56000MN
CP 2.06 Answer: (A)
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
By Coulomb’s law, the force F = k|q1 ||q2 |/r2 ∝ |q1 |. Hence, changing |q1 | → |q10 | = 4|q1 | will
change F → F 0 = 4 × F = 4 × 10N = 40N.
Since q1 and q2 initially attract each other, they initially have opposite sign. Since the sign
of q1 is reversed and q2 is unchanged, the two charges will both have the same sign, after q1
is changed, and therefore repel each other.
CP 2.07: If two point charges q1 and q2 at some distance r attract each other with a force
of 6N, what force would they exert on each other if q1 is doubled (×2) without changing its
sign; q2 is tripled (×3) and its sign is reversed; and r is unchanged ? The two charges will
(A)
(B)
(C)
(D)
(E)
repel each other with a force of 1N
attract each other with a force of 1N
repel each other with a force of (1/6)N
attract each other with a force of (1/6)N
repel each other with a force of 36N
CP 2.07 Answer: (E)
By Coulomb’s law, the force F = k|q1 ||q2 |/r2 ∝ |q1 |×|q2 |. Hence, changing |q1 | → |q10 | = 2|q1 |
and |q2 | → |q20 | = 3|q2 | will change F → F 0 = 2 × 3 × F = 6 × 6N = 36N.
Since q1 and q2 initially attract each other, they initially have opposite sign. Since the sign
of q2 is reversed and the sign of q1 is unchanged, the two charges will both have the same
sign, after q1 is changed, and therefore repel each other.
CP 2.08: If two point charges q1 and q2 at some distance r repel each other with a force of
4.5N, what force would they exert on each other if q1 is unchanged; q2 is doubled (×2) and
its sign is reversed; and r is tripled (×3) ? The two charges will
(A)
(B)
(C)
(D)
(E)
repel each other with a force of 1N
attract each other with a force of 1N
repel each other with a force of (1/6)N
attract each other with a force of (1/6)N
repel each other with a force of 36N
CP 2.08 Answer: (B)
By Coulomb’s law, the force F = k|q1 ||q2 |/r2 ∝ |q2 |/r2 . Hence, changing |q2 | → |q20 | = 2|q2 |
and r → r0 = 3r will change F → F 0 = 2 × F/32 = 2 × 4.5N/9 = 1N.
Since q1 and q2 initially repel each other, they initially have the same sign. Since the sign
of q2 is reversed and q1 is unchanged, the two charges will have opposite sign, after q2 is
changed, and therefore attract each other.
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
CP 2.09: Q in Fig. 2.09 is a positive point charge. Which arrow drawn at P could correctly
~ generated by Q at P ?
represent the electric field vector E
Fig. 2.09
P
(A)
E
Q
(A)
(B)
(C)
(D)
(E)
CP 2.09 Answer: (A)
For a single positive point charge Q, at any observation point P , the electric field vector
~ must (a) point along the straight line connecting Q and P and (b) away from Q. Hence
E,
(A) is the correct answer.
CP 2.10: Q in Fig. 2.10 is a negative point charge. Which arrow drawn at P could correctly
~ generated by Q at P ?
represent the electric field vector E
Fig. 2.10
P
E
Q
(A)
(C)
(B)
(C)
(D)
(E)
CP 2.10 Answer: (C)
For a single negative point charge Q, at any observation point P , the electric field vector
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
~ must (a) point along the straight line connecting Q and P and (b) towards Q. Hence
E,
(C) is the correct answer.
CP 2.11: Q1 and Q2 in Fig. 2.11 are positive point charges. Which arrow drawn at P could
~ generated by Q1 and Q2 at P ?
correctly represent the total electric field vector E
P
Q2
E1
Fig. 2.11
(E)
E
E2
Q1
(A)
(B)
(C)
(D)
(E)
CP 2.11 Answer: (E)
~1
At observation point P , the positive point charge Q1 produces an electric field vector E
pointing away from Q1 , i.e., upward; the positive point charge Q2 produces an electric
~ 2 pointing away from Q2 , i.e., rightward; as shown in Fig. 2.11
field vector E
~ is the vector sum of E
~ 1 and E
~ 2 , i.e., E
~ = E
~1 + E
~ 2 . Therefore,
The total electric field E
~
E must have a vertical component pointing upward and a horizontal component pointing
rightward. Hence, (E) is the only possible correct answer.
CP 2.12: Q1 in Fig. 2.12 is a positive point charge, and Q2 is a negative point charge.
~ generated
Which arrow drawn at P could correctly represent the total electric field vector E
by Q1 and Q2 at P ?
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
P
(D)
Q2
Fig. 2.12
E
E1
E2
Q1
(A)
(B)
(C)
(D)
(E)
CP 2.12 Answer: (D)
~1
At observation point P , the positive point charge Q1 produces an electric field vector E
pointing away from Q1 , i.e., upward; 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,
~ is the vector sum of E
~ 1 and E
The total electric field E
~ must have a vertical component pointing upward and a horizontal component pointing
E
leftward. Hence, (D) is the only possible correct answer.
CP 2.13: In Experiment 1, a proton travels from point A to point B, in the electric field
~ generated by/between two oppositely charged capacitor plates, as shown in Fig. 2.13. In
E
Experiment 2, an electron 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. 2.13
------------------------E
B
A
E
+++++++++++++++++++++++++++
Assume the proton experiences a drop 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 electron experience ?
(A)
(B)
(C)
(D)
∆V
∆V
∆V
∆V
= −500V and ∆U = −80aJ
= −1000V and ∆U = −160aJ
= +1000V and ∆U = +160aJ
= −1000V and ∆U = +160aJ
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
(E) ∆V = −250V and ∆U = +40aJ
CP 2.13 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 (electron) = ∆V (proton) = −250V.
However, ∆U = q∆V ; and q = +e for the proton, but q = −e for the electron. Hence, ∆U for
the electron has the opposite sign as for the proton: ∆U (electron) = −∆U (proton) = +40aJ.
So, the answer is (E).
CP 2.14: In Experiment 1, an electron travels from point A to point B, in the electric field
~ generated by/between two oppositely charged capacitor plates, as shown in Fig. 2.14. In
E
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
------------------------E
A
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
= +250V and ∆U = +40aJ
= −100V and ∆U = −16aJ
= +100V and ∆U = +16aJ
= −100V and ∆U = +16aJ
= −50V and ∆U = −8aJ
CP 2.14 Answer: (A)
~
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 (electron) = ∆V (proton) = +250V.
However, ∆U = q∆V ; and q = +e for the proton, but q = −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 (A).
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