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Winter 2017
Section X
PHYSICS 115 MIDTERM EXAM 2
PRACTICE EXAM SOLUTION
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Seat No_________
First Name (Print): _____________________
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Signature: ______________________________ Student ID: _________________________
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During the exam
When the exam begins, print your name and student ID number on the top of each page. Do this first
when you are told to open your exam.
If you are confused about a question, raise your hand and ask for an explanation.
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Name ____________________________________
Last
First
Student ID: _______________
Score: _______
1. [5 pts] A heat engine operating between energy reservoirs at 20°C and 600°C has 30% of the
maximum possible efficiency. How much energy does this engine extract from the hot reservoir to do
1000 J of work?
A) 1500 J
B) 2500 J
𝑒"#$ = 1 −
𝑄* =
D) 4000 J
E) 5000 J
𝑇)
𝑇*
𝑒+,#- = 0.3𝑒+,#- = 0.3 1 −
𝑒+,#- =
C) 3000 J
𝑇)
𝑇*
𝑊
𝑄*
𝑊
𝑒+,#-
=
𝑊
𝑇
0.3 1 − )
𝑇*
=
1000J
= 5000J
293K
0.3 1 −
873K
2. [5 pts] A heat engine operates between a high-temperature reservoir at 610 K and a low temperature
reservoir at 320 K. In one cycle, the engine absorbs 6400 J of heat from the high-temperature
reservoir and does 2200 J of work. What is the net change in entropy as a result of this cycle?
A) 1.8 J/K
Δ𝑆 =
B) 2.6 J/K
C) 5.9 J/K
D) 10 J/K
E) 13 J/K
𝑄
𝑇
𝑄* = W − 𝑄)
Δ𝑆* =
−6400J
= −10.5J/K
610K
Δ𝑆) =
6400J − 2200J
= 13.1J/K
320K
Δ𝑆A,B = Δ𝑆) + Δ𝑆* = 2.6J/K
Physics 115 – Winter 2017
Practice Exam 2
Name ____________________________________
Last
First
Student ID: _______________
Score: _______
3. [5 pts] Three charges are placed at the vertices of an equilateral
triangle of side a = 0.93 m, as shown. Charges 1 and 3 are
+7.3 µC; charge 2 is -7.3 µC. Find the magnitude of the net force
acting on charge 3.
A) 0.52 N
B) 0.72 N
C) 0.55 N
D) 0.28N
E) 0.36 N
4. [5 pts] Consider a system of three point charges on the x-axis. Charge 1 is at x = 0, charge 2 is at
x = 0.20 m, and charge 3 is at x = 0.40 m. In addition, the charges have the following values
q1 = -19 µC, q2 = q3 = +19 µC. The electric field is zero at some point along the x-axis between
x = 0.20 m and x = 0.40 m. Is the point of zero field:
A) At x = 0.30 m
B) To the left of x = 0.30 m
C) To the right of x = 0.30 m.
D) Not possible to tell.
0
0.20 m
0.40 m
For the electric field to be zero, the electric field of charge 2 must be equal to the sum of the fields
due to charge 1 and 3. At x = 0.30 m, the electric field due to charge 2 and 3 are equal in magnitude
and opposite direction. Thus at x = 0.30 there is a net field to the left due to the field of charge 1. To
the right of x = 0.30 m, the field of charge 3 is greater than that of charge 2, also resulting in a net
field. Therefore, the point must be located to the left of x = 0.30, where the electric field due to
charge 2 is greater than that of charge 3.
Physics 115 – Winter 2017
Practice Exam 2
Name ____________________________________
Last
First
Student ID: _______________
Score: _______
5. [5 pts] A particle with a mass of 3.8 g and a charge of +0.045 µC
is released from rest at point A in the figure at right. The electric
field has a magnitude of 1200 N/C. What speed will the particle
have after moving through a distance of 5 cm? (There are no
gravitational forces exerted on the particle.)
A. 1.4 x10-3 m/s
B. 3.8 x10-2 m/s
C. 12 m/s
D. 2.6 x10-2 m/s
E. 3.4 m/s
6. [5 pts] The closed Gaussian surface shown at right consists of a hemispherical
surface and a flat plane. A point charge +q is outside the surface, and no charge is
enclosed by the surface. Is the magnitude of the flux through the flat portion
ΦE greater than, less than or equal to the magnitude of the flux through the
curved portion ΦF ?
A) Greater than
B) Less than
C) Equal to
D) Not enough information given to answer.
The flux through the flat surface is positive, while the flux through the curved portion is negative.
However, since there is no charge enclosed by the surface, the net flux through the surface must be
zero. It can thus be concluded that the magnitude of the flux through the flat portion is equal in
magnitude and opposite in sign to the flux through the curved portion.
7. [5 pts] You make a capacitor from 15.0-cm diameter aluminum circular plates, separating them by
3.50 mm, and connecting them across a 6.00-V battery. There is air between the two plates
(κair = 1). Determine the maximum amount of charge that could be placed on the plates.
A) 0.27 nC
B) 1.1 nC
C) 0.25 pC
D) 11 nC
E) 0.58 nC
ε0 A (8.85×10−12 C2 /N ⋅ m 2 )( π )(0.075 m)2
=
d
3.5×10−3 m
C = 4.47 ×10−11 F
C=
(
)
Q = CΔVC = 4.47 ×10−11 F (6.00 V)
Q = 0.27 nC
Physics 115 – Winter 2017
Practice Exam 2
Name ____________________________________
Last
First
Student ID: _______________
Score: _______
8. [5 pts] The plates of a parallel-plate capacitor are connected to a battery. If the distance between the
plates is halved, the energy of the capacitor:
A) Increases by a factor of 4.
B) Doubles.
C) Remains the same.
D) Is halved.
E) Decreases by a factor of 4.
Since the capacitor is connected to a battery, the voltage across the capacitor remains constant. A
decrease of the distance between the plates increases the capacitance. The change in energy can be
determined as follows:
I
𝑈H = 𝐶𝑉 J
I
I
J
J
𝑈M = 2𝐶𝑉 J = 2
J
𝐶𝑉 J = 2𝑈H
9. [5 pts] 2.0 x 1013 electrons flow through a transistor in 1.0 ms. What is the current through the
transistor?
A) 1.2 mA
𝐼=
B) 2.7 mA
C) 3.2 mA
D) 4.1 mA
E) 4.8 mA
Δ𝑄 2.0×10IR 1.6×10SIT C
=
= 3.2mA
∆𝑡
1×10SR s
10. [5 pts] The wires in the figure below are all made of the same material; the length and radius of each
wire are noted. Rank in order, from largest to smallest, the resistances R1 to R5 of these wires.
A) R5 > R4 = R3 > R2 = R1
B) R4 > R1 = R5 > R3 > R2
C) R4 > R3 = R5 > R1 > R2
D) R2 > R3 > R1 = R5 > R4
E) R5 = R4 > R1 > R3 = R2
𝑅I = 𝜌
[
\+ ]
𝑅J = 𝜌
[
\^+ ]
I
= 𝑅I
Physics 115 – Winter 2017
^
𝑅R = 𝜌
J[
\^+ ]
I
= 𝑅I
J
𝑅^ = 𝜌
J[
\+ ]
= 2𝑅I
𝑅_ = 𝜌
^[
\^+ ]
Practice Exam 2
= 𝑅I
Name ____________________________________
Last
First
Student ID: _______________
Score: _______
A point charge Q = +87.1 µC is held fixed at the origin. A second point charge, with mass m = 0.0576 kg
and charge q = -2.87 µC, is placed at the location (0 m, 0.320 m). (Ignore any gravitational forces.)
11. [5 pts] What is the magnitude of the force exerted on the charge q by the charge Q when the charges
are located at the values above? Show your work.
𝐹ab
𝑞𝑄
= 𝑘 J = 8.99×10T N. mJ /C J
𝑟
2.87×10Sg C 87.1×10Sg C
= 21.9N
0.32m J
12. [5 pts] If the second charge is released from rest, through what potential difference does it move
through when it reaches the point (0 m, 0.160 m)? Show your work.
𝑉h.Igi − 𝑉h.RJi =
𝑘𝑄
𝑟h.Igi
−
𝑘𝑄
𝑟h.RJi
𝑉h.Igi − 𝑉h.RJi = 8.99×10T N.
= 𝑘𝑄
mJ
CJ
1
𝑟h.Igi
−
1
𝑟h.RJi
87.1×10Sg C
1
1
−
= 2.45×10g V
0.16m 0.32m
13. [5 pts] What is the speed if the charge q when it reaches the point (0 m, 0.160 m)? Show your work.
∆𝑈 = 𝑞∆𝑉 = −2.87×10Sg C 2.45×10g V = 7.02J
∆𝑈 = −∆𝐾
I
∆𝐾 = 7.02J = 𝑚 𝑣MJ − 𝑣HJ
J
𝑣M =
2∆𝐾
=
𝑚
𝑣H = 0m/s
2 7.02J
= 15.6m/s
0.0576kg
14. [5 pts] The same experiment is repeated with a charge that has the same mass as the charge q but has
twice the magnitude of charge (-5.74 µC). Suppose that the speed from question 13 is labeled as v.
Will the speed of the new charge at the point (0 m, 0.160 m) be greater than, less than, or equal to
2v? Show your work.
The work done on the -5.74 µC will be twice as large as that done on the -2.87 µC since the force on
the charge is proportional to its magnitude and the distance traveled for both charges is the same.
Thus the kinetic energy of the -5.74 µC will be twice as large as the -2.87 µC from the work-kinetic
energy theorem. However, to have double the kinetic energy, the speed of the -5.74 µC only needs to
be 1.414 times as large as that of the -2.87 µC charge since both charges have the same mass.
Physics 115 – Winter 2017
Practice Exam 2
Name ____________________________________
Last
First
Student ID: _______________
Score: _______
In case 1, shown at right, a positive point charge, +q is held in place
between two fixed positive point charges +3Q and +6Q.
15. [5 pts] In terms of s, determine the value of x1 that results in a zero
net electric force on the positive point charge +q. Explain
The charge +q is repelled from each of the charges +6Q and +3Q since they have the same sign. In
this case, the force exerted by the +6Q charge is in the positive x direction while the force exerted by
the +3Q is in the negative x direction. The electrostatic force is proportional to the magnitude of the
charges and inversely proportional to the square of the distance between them. The product of the
6Q and +q charge is twice as large as that of the 3Q and the +q, so the distance x1 needs to the Ö2 of
s for the net electric force on the +q charge to be zero.
16. [5 pts] In case 2, shown at right, the +6Q positive point charge has
been replaced by two fixed positive point charges +3Q. In order for
net electric force on the positive point charge +q to be zero, is the
value of x2 greater than, less than, or equal to the value of x1?
Explain.
In case 1, the +6Q charge can be considered as being two +3Q charges at the same point. In case 2,
if x2 were equal to x1, the distance from each of the +3Q charges on the left to the charge +q would
be greater than x1. Hence, the sum of the x-components of the forces exerted by the +3Q charges on
the left would be less than the force exerted by the +6Q in case 1. (The y-components cancel.) As a
result there would be a net force on the +q charge to the left. Therefore, x2 must be less than x1 in
order for the net electric force on the charge +q to be zero.
17. [5 pts] The diagram at right shows a fixed positive point charge +Q,
separated by a distance 2s from a fixed negative point charge –2Q.
Which of the arrows 1-7 best represents the direction of the electric
field at point A? Explain.
The net electric field at point A is given by the sum of the individual
fields as shown at right. This illustrates that arrow 2 best represents the direction of the electric field
at point A.
Physics 115 – Winter 2017
Practice Exam 2
Name ____________________________________
Last
First
Student ID: _______________
Score: _______
A +Q point charge and a -2Q point charge are fixed
to a grid as shown. Points B and C are both a
distance r from the -2Q charge.
18. [5 pts] Is the work done by the electric field on a
positive test charge q0 that travels from point B to
point C greater than, less than, or equal to zero?
Explain.
No work is done on the charge q0 by the -2Q
charge, since the force by the -2Q charge is
always perpendicular to the point charges’ path.
However, the charge +Q does positive work on the charge, since a component of the force by the +Q
charge on the point charge points in the same direction as the displacement of the point charge. The
work is therefore greater than zero.
19. [5 pts] Is the electric potential at point B (VB) greater than, less than, or equal to that at point C (VC)?
Explain.
The electric potential due to the -2Q point charge is the same at points B and C since the points are
the same distance from the -2Q charge. However, the potential at point C due to the +Q charge is
less than that at point B. The potential at point C is thus less than the potential at point B.
20. [5 pts] Which of the following changes alone will result in a greater absolute value of the potential
difference between points B and C.
Circle all that applies and explain your reasoning.
A.
The value of +Q charge is increased
B.
The values of -2Q charge is increased
C.
The value of +q0 charge is decreased
Since the point charge moves in a radial path around the -2Q charge, the work done by that charge
will always be zero (force is always perpendicular to the path). Also, the potential difference is
completely independent of the point charge, and only dependent on the source charges that set up the
potential. Since the +Q charge is the only charge doing work on the point charge, and work is
proportional to the charge magnitude, increasing the value of +Q will increase the potential
difference between points B and C.
Physics 115 – Winter 2017
Practice Exam 2