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

Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
Solution to PHYS 1112 In-Class Exam #2A
Thu. March 18, 2010, 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 dark fringes (interference intensity minima), 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 dark fringes
Fig. 2.04
Δy
Screen
θ
L
Double Slit
Laser Beam
(A)
(B)
(C)
(D)
(E)
decrease if we decrease L (keeping λ and d fixed);
decrease if we increase λ (keeping L and d fixed);
decrease if we decrease 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: (A)
At 1st minimum (closest to central maximum): sin θ = (1/2)λ/d. Also, ∆y = 2L tan θ,
by trigonometry (see Fig. 2.04). So sin θ, hence θ, hence tan θ, hence |∆y|, will decrease
if d increases or if λ decreases. Therefore, (B), (C), (D) and (E) are wrong. Because of
∆y = 2L tan θ (and θ is fixed if λ and d are fixed), |∆y| decreases if L decreases: hence (A)
is correct.
Problem 2: If two point charges Q1 and Q2 at some distance r repel each other with
a force of 10µN, what force would they exert on each other if Q1 is tripled (×3) without
change of sign; r is doubled (×2), and the sign of Q2 is reversed ? The two charges will
(A) attract each other with a force of 13.33µN
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Physics 1112
Spring 2010
(B)
(C)
(D)
(E)
University of Georgia
Instructor: HBSchüttler
attract each other with a force of 15µN
repel each other with a force of 15µN
repel each other with a force of 7.5µN
attract each other with a force of 7.5µN
Answer: (E)
By Coulomb’s law, the force F = k|Q1 ||Q2 |/r2 ∝ |Q1 |/r2 . Hence, changing |Q1 | → |Q01 | =
3|Q1 | and r → r0 = 2r will change F → F 0 = (3/22 ) × F = (3/4) × 10µN = 7.5µN.
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 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. 2.11
P
Q2
(D)
(E)
(C)
(A)
(B)
Q1
(A)
(B)
Answer: (D)
(C)
(D)
(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 negative point charge Q2 produces an electric
~ 2 pointing towards Q2 , i.e., leftward; 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
~ must have a vertical component pointing upward and a horizontal component pointing
E
leftward. Hence, (D) 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
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
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Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
has a charge −e.
Fig. 2.13
------------------------B
E
A
E
+++++++++++++++++++++++++++
Assume the electron experiences a drop in electric potential, ∆V = −1500V, and a gain of
potential energy, ∆U = +240aJ (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
= −50V and ∆U = −8aJ
= −100V and ∆U = −16aJ
= +100V and ∆U = +16aJ
= −1500V and ∆U = −240aJ
= +1500V and ∆U = +240aJ
Answer: (D)
~
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) = −1500V.
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) = −240aJ. So, the answer is (D).
Numerical Problems
Problem 5: If a diffraction grating is illuminated at normal incidence by coherent (laser)
light it produces 3rd order maxima at angles of ±72.0o measured from the central (0th
order) intensity maxium. At what angles θ, measured from the central (0th order) intensity
maxium, will the 2nd order maxima be observed ?
(A)
(B)
(C)
(D)
(E)
±24.0o
±15.2o
±48.0o
±18.5o
±39.3o
Answer: (E)
Since d sin(θm ) = mλ for the m-th order maximum, (λ/d) = sin(θm )/m = sin(θ3 )/3 can be
gotten from the 3rd order maximum given, with θ3 = 72.0o . Thus, for the 2nd order maxima,
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Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
m = 2, we get sin(θ2 ) = 2(λ/d) = (2/3) sin(θ3 ) = (2/3) sin(72o ) = 0.63404. So θ2 = 39.3o
and θ−2 = −39.3o . 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 = +360µC is at x = 0, and Q2 is at
x = 10.0cm. A third charge q will be in equilibrium when placed at x = 12.0cm. Therefore,
Q2 is
(A)
(B)
(C)
(D)
(E)
+60µC
+10µC
−60µC
−10µC
−30µC
Answer: (D)
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 away from Q1 , i.e., in +x-direction (rightward) 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
~
(leftward) at x(q). So E2 at x(q) points towards Q2 , Q2 being located at x(Q2 ) < x(q), i.e.,
to the left 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 )| =
(12.0 − 0.0)cm = 12.0cm and r2 = |x(q) − x(Q2 )| = (12.0 − 10.0)cm = 2.0cm (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 | = (2.0/12.0)2 × (360µC) = 10µC. So, Q2 = −10µC.
Problem 7: Two point charges, spaced 112.4m apart, repel each other with a force of
140.0N. What is the amount of the smaller point charge if the larger charge is 3.5 times the
smaller charge?
(A) 3.5mC
(B) 5.5mC
(C) 7.5mC
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Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
(D) 9.5mC
(E) 26.3mC
Answer: (C)
By Coulomb’s law, the force is F = k|Q||q|/r2 with Q = 3.5q. So F = 3.5k|q|2 /r2 . Solving
for q: |q| = [r2 F/(3.5k)]1/2 , i.e., |q| = [(112.4m)2 (140N)/(3.5 × 8.99 × 109 Nm2 /C2 )]1/2 =
7.5 × 10−3 C
Problem 8: If a charge of +72µC is uniformly spread out over a single, very thin square
of titanium sheet metal, of 4.0m 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 sheet?
(A)
(B)
(C)
(D)
(E)
9904. × 103 N/C,
254.2 × 103 N/C,
40.46 × 103 N/C,
9904. × 103 N/C,
40.46 × 103 N/C,
~
E
~
E
~
E
~
E
~
E
pointing
pointing
pointing
pointing
pointing
away from the sheet
away from the sheet
away from the sheet
towards the sheet
towards the sheet
Answer: (B)
For a single, uniformly charged, planar surface (see Formula Sheet): E = |Q|/(2o A), so
E = (72 × 10−6 C)/(2 × (8.85 × 10−12 C2 /Nm2 ) × (4m)2 ) = 254.2 × 103 N/C.
~ must point away from it, i.e., away from
Also, since the charge Q = +36µC is positive, E
the charged surface, on either side of that surface.
Problem 9: The total electric flux passing through the approximately spherical surface
of a real, NBA-approved basket ball is Φ = −113Nm2 /C (assuming outward directed
normal to the surface). What is the total electric charge enclosed inside the ball ?
(A)
(B)
(C)
(D)
(E)
−1.00nC
−0.60nC
+0.80nC
+0.40nC
+2.20nC
Answer: (A)
By Gauss’ Law: Q = o Φ, so Q = (8.85 × 10−12 C2 /Nm2 ) × (−113Nm2 /C) = −1.00 × 10−9 C
Problem 10: The 1500µF capacitor in a defibrillator is charged with a 200V battery.
What is the charge stored on the positive plate of this capacitor ?
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Physics 1112
Spring 2010
(A)
(B)
(C)
(D)
(E)
University of Georgia
Instructor: HBSchüttler
100mC
300mC
500mC
700mC
900mC
Answer: (B)
Q = CV = (1500 × 10−6 F) × (200V ) = 0.30C
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)
20J
90J
60J
80J
30J
Answer: (E)
UE = 21 CV 2 = 12 (1500 × 10−6 F) × (200V)2 = 30.0J
Problem 12: If the electric field strength between the capacitor plates in Problem 4 is
2610.V/m, what is the magnitude and the direction of the proton’s acceleration ~a, assuming
the proton is subject only to the electric force between the plates ?
(A)
(B)
(C)
(D)
(E)
25.0 × 1010 m/s2 , ~a
25.0 × 1010 m/s2 , ~a
60.0 × 1010 m/s2 , ~a
60.0 × 1010 m/s2 , ~a
60.0 × 1010 m/s2 , ~a
pointing
pointing
pointing
pointing
pointing
rightward
upward
leftward
upward
rightward
Answer: (B)
~ So: ~a =
By Newton’s 2nd Law: ~a = F~ /m; and by definition of electric field: F~ = q E.
~ and a = |q/m|E where a ≡ |~a| and E ≡ |E|.
~ So: a = [(1.60 × 10−19 C)/(1.67 ×
(q/m)E,
−27
10
2
10 kg] × (2610V/m) = 25.00 × 10 m/s .
~ and E
~ points upward and q = +e > 0 for the proton, the force
Also, since ~a = (q/m)E
~ and hence ~a, points in the same direction as E,
~ i.e., ~a points upward. This means
F~ = q E,
that the proton accelerates towards the minus-plate and away from the plus-plate.
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