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

Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Solution to PHYS 1112 In-Class Exam #2B
Thu. March 19, 2009, 2:00pm-3:15pm
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 quadrupled (×4) without
change of sign; and r and Q2 are unchanged ? The two charges will
(A) repel each other with a force of 40µN
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Physics 1112
Spring 2009
(B)
(C)
(D)
(E)
University of Georgia
Instructor: HBSchüttler
attract each other with a force of 40µN
attract each other with a force of 100N
repel each other with a force of 2560kN
repel each other with a force of 100N
Answer: (A)
By Coulomb’s law, the force F = k|Q1 ||Q2 |/r2 ∝ |Q1 |. Hence, changing |Q1 | → |Q01 | = 4|Q1 |
will change F → F 0 = 4 × F = 4 × 10µN = 40µN.
Since Q1 and Q2 initially repel each other, they initially have the same sign. Since the
signs of Q1 and Q2 are unchanged, the two charges will still have the same sign, after Q1 is
changed, and therefore still repel each other.
Problem 3: Q in the figure below is a positive point charge. Which arrow drawn at P
~ generated by Q at P ?
could correctly represent the electric field vector E
Fig. 2.10
P
(A)
Q
(C)
(B)
(E)
(D)
(A)
(B)
Answer: (B)
(C)
(D)
(E)
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,
(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
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.
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Fig. 2.13
------------------------B
E
A
E
+++++++++++++++++++++++++++
Assume the electron experiences a drop in electric potential, ∆V = −500V, and a gain of
potential energy, ∆U = +80aJ (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
= −500V and ∆U = −80aJ
= +500V and ∆U = +80aJ
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) = −500V.
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 (proton) = −∆U (electron) = −80aJ.
So, the answer is (D).
Numerical Problems
Problem 5: If a double-slit with a line spacing of 1400nm is illuminated at normal incidence
by coherent (laser) light with a wavelength of 600nm, at what angles θ, measured from the
central (0th order) intensity maxium, will the 2nd order maxima be observed ? (See figure
in Problem 1.)
(A)
(B)
(C)
(D)
(E)
±73.06o
±59.00o
±31.37o
±22.51o
±15.62o
Answer: (B)
At 2nd-order maxima: d sin θ = mλ with m = ±2. So, θ = arcsin(±2λ/d) = arcsin(±2 ×
600/1400) = ±59.00o .
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Problem 6: What is the electric force between two point charges of +9.0mC and −8.0mC,
spaced 15m apart ? Is there attraction or repulsion between the two charges ?
(A)
(B)
(C)
(D)
(E)
82.52N,
479.5N,
779.1N,
2877.N,
5678.N,
repulsion
attraction
repulsion
attraction
repulsion
Answer: (D)
By Coulomb’s law, the force F = k|Q1 ||Q2 |/r2 , i.e., F = (8.99 × 109 Nm2 /C2 )(9.0 ×
10−3 C)(8.0 × 10−3 C)/(15m)2 = 2877N. Also, since the two charges Q1 = 9.0mC and
Q2 = −8.0mC have opposite sign, they will attract each other.
Problem 7: Two identical point charges, spaced 112.4m apart, repel each other with a force
of 40.0N. What is the amount of each point charge?
(A)
(B)
(C)
(D)
(E)
3.5mC
5.5mC
7.5mC
9.5mC
11.5mC
Answer: (C)
By Coulomb’s law, the force F = k|Q1 ||Q2 |/r2 with Q1 = Q2 ≡ Q. Solving for Q:
|Q| = [r2 F/k]1/2 , i.e., |Q| = [(112.4m)2 (40N)/(8.99 × 109 Nm2 /C2 )]1/2 = 7.5 × 10−3 C
Problem 8: If a charge of +36µC is uniformly spread out over a single, very thin square of
copper sheet metal, 4.0m to the side, 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?
(Hint: Think of the copper sheet as a single, uniformly charged planar surface.)
~ pointing away from the sheet
(A) 127.1 × 103 N/C, E
~ pointing away from the sheet
(B) 4952. × 103 N/C, E
~ pointing towards the sheet
(C) 4952. × 103 N/C, E
~ pointing away from the sheet
(D) 32.98 × 103 N/C, E
~ pointing towards the sheet
(E) 32.98 × 103 N/C, E
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Answer: (A)
For a single, uniformly charged, planar surface (see Formula Sheet): E = |Q|/(2o A), so
E = (36 × 10−6 C)/(2 × (8.85 × 10−12 C2 /Nm2 ) × (4m)2 ) = 127.1 × 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 = −226Nm2 /C (assuming outward directed
normal to the surface). What is the total electric charge enclosed inside the ball ?
(A)
(B)
(C)
(D)
(E)
+0.80nC
−1.20nC
+1.60nC
−2.00nC
+2.40nC
Answer: (D)
By Gauss’ Law: Q = o Φ, so Q = (8.85 × 10−12 C2 /Nm2 ) × (−226Nm2 /C) = −2.00 × 10−9 C
Problem 10: The 750µF capacitor in a defibrillator is charged with a 200V battery. What
is the charge stored on the positive plate of this capacitor ?
(A)
(B)
(C)
(D)
(E)
50mC
150mC
250mC
350mC
450mC
Answer: (B)
Q = CV = (750 × 10−6 F) × (200V ) = 0.15C
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)
120J
90J
75J
45J
15J
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Answer: (E)
UE = 12 CV 2 = 12 (750 × 10−6 F) × (200V)2 = 15.0J
Problem 12: If the electric field strength between the capacitor plates in Problem 4 is
522.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)
5.00 × 1010 m/s2 , ~a pointing leftward
5.00 × 1010 m/s2 , ~a pointing upward
12.00 × 1010 m/s2 , ~a pointing leftward
12.00 × 1010 m/s2 , ~a pointing upward
12.00 × 1010 m/s2 , ~a pointing rightward
Answer: (B)
~ 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)/(1.67 ×
(q/m)E,
−27
10
2
10 kg] × (552V/m) = 5.00 × 10 m/s .
~ and E
~ points upward and q = +e > 0 for the proton, the force
Also, since ~a = (q/m)E
~
~
~ i.e., ~a points upward, which means
F = q E, and hence ~a, points in the same direction as E,
that the proton accelerates towards minus-plate and away from plus-plate.
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Formula Sheet
Wave Optics, Interference, Diffraction
(1) Periodic Wave Condition:
v = λf =
λ
τ
(2) Index of Refraction for electromagnetic waves, definition:
n=
λvacuum
c
=
v
λ
with λvacuum ≡ c/f = cτ .
(3) Definition of Path Length Difference for two-source, double-slit or adjacent slits in
multi-slit/diffraction grating:
∆` ≡ `2 − `1
(4) Path Length Difference vs. Angle: ∆` is given approximately in terms of observation
angle θ measured from central axis:
∆` ∼
= d sin θ
if
dL
where L =distance from slits or sources to observation screen, d =spacing of adjacent sources,
slits or lines in double-slit, multi-slit or diffraction grating.
(5) Constructive Interference Condition (≡ intensity maxima, principal maxima, bright
fringes) for two-source, double-slit, multi-slit or diffraction grating:
∆` = mλ
or
d L);
d sin θ = mλ (if
with m = 0, ±1, ±2, ...
where m is the ”order” of the (principal) maximum.
(6) Destructive Interference Condition 1 (≡ intensity minima, dark fringes) for twosource or double-slit experiment:
∆` = m +
1
λ
2
or
d sin θ = m +
1
λ (if d L);
2
with
m+
1
1 3
= ± , ± , ...
2
2 2
(7) Destructive Interference Condition 2 (≡ intensity minima, dark fringes) for singleslit diffration with W =slit width and W L:
W sin θ = mλ
with m = ±1, ±2, ... (but m 6= 0).
(8) Destructive Interference Condition 3 for first intensity minimum of circular aperture
diffration with W =aperture diameter and W L:
W sin θ ∼
= 1.22λ (for 1st circular intensity minimum).
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Charge, Electric Force, Electric Field, Flux, Gauss’s Law
(1) Coulomb’s Law for force F ≡ |F~ | on two point charges Q1 and Q2 at distance r:
F =k
|Q1 ||Q2 |
r2
with k = 8.99 × 109 Nm2 /C2 .
~ in terms of electric force F~ exerted on test charge q
(2) Definition of Electric Field E
~ generated by ”other” charges):
(with E
~
~ ≡F ;
E
q
~ .
hence F~ = q E
~
~ depends
Note E-field
is independent of the test charge q used to detect the electric force F~ : E
~
only on the ”other” charges which generate E.
~ by Point Charge Q, at an observation point P with distance
(3) Electric Field E ≡ |E|
r from Q:
|Q|
E=k 2
r
~ pointing radially away from positive charge, Q > 0; or radially towards negative
with E
charge, Q < 0.
~ by Uniform Surface Charge Density σ ≡ Q/A on a single
(4) Electric Field E ≡ |E|
planar surface of area A with total charge Q, in close proximity to the surface:
E=
|σ|
|Q|
=
2o
2o A
~ ∼
~
with E
= const (uniform E−field)
~ normal to the surface, pointing away
where o ≡ 1/(4πk) = 8.85 × 10−12 C2 /Nm2 ; and E
from positively charged (Q > 0) or towards negatively charged (Q < 0) surface.
~ in Planar Capacitor: between closely spaced, parallel, planar
(5) Electric field E ≡ |E|
plates of charges Q and −Q, and opposing surface areas A, without dielectric (κ = 1):
E=
|Q|
o A
~ ∼
~
with E
= const (uniform E−field)
~ normal to the plate surfaces, pointing from positive towards negative plate.
with E
~ is being generated
(6) Superposition Principle of Electric Field: If an electric field E
~ at any observation point P is the vector
by multiple charged objects (Q1 , Q2 , ...), then E
~ 1, E
~ 2 , ... that would be generated
sum (resultant vector) of the electric field contributions E
by each of the charged objects in isolation at that point P :
~ =E
~1 + E
~ 2 + ...
E
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
~
(7) Electric Flux Φ of constant E-field
through planar surface of area A with the surface
o
~
normal at angle θ from E, with 0 ≤ θ ≤ 180o :
Φ = EA cos θ
~
(8) Gauss’s Law, relating total electric flux Φ(S) of the E-field
through a closed surface S
(with outward-directed surface normal) to the total charge Q(S) enclosed inside S:
Φ(S) =
1
Q(S)
o
Electric Potential, Potential Energy, Capacitance, Electric Energy Storage
(1) Definition of Electric Potential V and electric potential difference ∆V (also known
as ”voltage drop”), in terms of potential energy U and potential energy difference ∆U ,
~
respectively, for a test charge q moving or being moved through E-field:
V = U/q ,
∆V = ∆U/q ;
hence U = qV ,
∆U = q∆V .
~
~
Note that V or ∆V is a property of the E-field,
depends only on the E-field,
and is
therefore independent of the test charge q.
~ = const):
(2) Electric Potential Difference in a Uniform Electric Field (E
∆V = −E ∆s cos θ
where ∆V ≡ VB − VA is the electric potential difference between points B and A; the vector
~ with
∆~s points from A to B with length ∆s ≡ |∆~s|; and θ is the angle between ∆~s and E
~ (θ < 90o ); and
0o ≤ θ ≤ 180o . Hence, ∆V < 0 when moving from A to B in direction of E
o
~ (θ > 90 ).
∆V > 0 when moving from A to B against direction of E
(3) Electric Potential for Point Charge Electric Field (with E = k|Q|/r2 ):
V =k
Q
r
where r is the distance from point charge Q to observation point.
~ is being gener(4) Superposition Principle of Electric Potential: If an electric field E
ated by multiple charged objects (Q1 , Q2 , ...), then its electric potential V at any observation
point P is the scalar sum (sum of numbers) of the electric potential contributions V1 , V2 ,
... that would be generated by each of the charged objects in isolation at that point P ; and
likewise for the electric potential difference ∆V ≡ VB − VA between any points A and B:
V = V1 + V2 + ...
or
∆V = ∆V1 + ∆V2 + ...
(5) Definition of Capacitance: For two oppositely charged metallic objects a and b,
with −Q stored on a and Q stored on b, their electric potential difference V ≡ Vb − Va is
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
proportional to the charge Q. The capacitance of the two metallic objects is then defined
as:
Q
Q
C≡
,
hence
Q = CV
or
V =
V
C
(6) Voltage and Capacitance of a Planar Capacitor: For two oppositely charged,
parallel planar metallic plates, each of opposing surface area A, closely spaced with distance
d, the voltage V and capacitance C are
|V | = Ed =
|Q|d
;
κo A
C≡
|Q|
A
= κo ≡ κ Co
|V |
d
where κ is the dielectric constant of the dielectric (insulating) material between the plates
and κ = 1 for vacuum or air, and Co ≡ o A/d is the capacitance without dielectric.
(7) Electric Field Energy Storage in a Capacitor: The energy UE required to build up
a charge Q and a voltage V = Q/C in a capacitor is stored as electric field energy between
the capacitor plates and it is given by
UE =
1 2 1
Q = CV 2
2C
2
Mechanics Memories: Velocity, Acceleration, Force, Energy, Power
(1) Velocity
~v =
∆~r
∆t
if constant; else ~v = lim
∆~r
∆t→0 ∆t
~a =
∆~v
∆t
if constant; else ~a = lim
(2) Acceleration
∆~v
∆t→0 ∆t
(3) Constant-Acceleration Motion: for ∆~r ≡ ~rf − ~ri and ∆~v ≡ ~vf − ~vi
∆~r =
1
(~vi + ~vf ) t ;
2
∆~r = ~vi t +
1
~a t2 ;
2
∆~v = ~a t .
(4) Newton’s 2nd Law:
m~a = F~
(5) Kinetic Knergy:
1
K = m v2
2
(6) Energy Conservation Law for ∆K ≡ Kf − Ki and ∆U ≡ Uf − Ui :
Ki + Ui = Kf + Uf
or
∆K + ∆U = 0
(7) Mechanical Power: P =rate of work done by force F~ on an object moving at speed
~v , with ~v pointing at an angle θ from F~ and 0o ≤ θ ≤ 180o :
P = F v cos θ
10
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Algebra and Trigonometry
2
az + bz + c = 0
sin θ =
opp
,
hyp
⇒
cos θ =
z=
adj
,
hyp
−b ±
√
b2 − 4ac
2a
tan θ =
opp
sin θ
=
adj
cos θ
sin2 θ + cos2 θ = 1
For very small angles θ (with |θ| 90o ):
sin θ ∼
= tan θ ∼
= θ (in radians)
Numerical Data
Acceleration of gravity (on Earth):
Speed of light in vacuum:
Coulomb’s constant:
Electron mass:
Proton mass:
c = 3.00 × 108 m/s
k = 8.99 × 109 Nm2 /C2
Permittivity of vacuum:
Elementary charge:
g = 9.81m/s2
o ≡ 1/(4πk) = 8.85 × 10−12 C2 /Nm2
e = 1.60 × 10−19 C
me = 9.11 × 10−31 kg
mp = 1.67 × 10−27 kg
Other numerical inputs will be provided with each problem statement.
SI numerical prefixes:
y = yocto =10−24 , z = zepto =10−21 , a = atto =10−18 , f = femto =10−15 , p = pico =10−12 ,
n = nano =10−9 , µ= micro =10−6 , m = milli =10−3 , c = centi =10−2 , d = deci =10−1 ,
da = deca =10+1 , h = hecto =10+2 , k = kilo =10+3 , M = Mega =10+6 , G = Giga =10+9 ,
T = Tera =10+12 , P = Peta =10+15 , E = Exa =10+18 , Z = Zetta =10+21 , Y = Yotta =10+24 .
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