General Physics – PH 213
Midterm II (Ch 28 - 32)
August 15, 2013
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Name:____________
Exam is closed book, closed notes. Use only your note card with 10 formulae on it.
Write all work and answers in the space provided. Use color papers for additional space or scratch.
Show all your work and explain your reasoning (except for true/false or multiple choices ones)
Partial credit will be given. No credit will be given if no work is shown (except for true/false or
multiple choices ones)
Useful Constants:
N ⋅m2
C2
− 31
m e = 9.11× 10 kg
C2
N ⋅m2
− 27
m p = 1.67 × 10 kg
mili (m) = 10 −3
micro ( µ ) = 10 − 6
k = 9.0 × 10 9
ε 0 = 8.85 × 10 −12
q e = 1.6 × 10 −19 C
µ 0 = 4π × 10 − 7
T ⋅m
A
1eV = 1.6 × 10 −19 J
nano (n) = 10 −9
Part I – (3 pts each)\True or False: For questions 1 – 8, state whether each statement is true or
false. You do not need to show any work for T/F questions
1. __ True __ Volts per meter, V/m, are equivalent to newtons per coulomb, N/C.
2. __True___ To produce higher voltages, the cells of a battery are connected in series.
3. __True___ The current in a circuit flows in the opposite direction to the direction in which the
electrons move.
4. __True___ If R is resistance and C is capacitance, then the units associated with RC are seconds.
5. __False___ The magnetic field near a current carrying wire is directly proportional to the distance
from the wire. (B-field is inversely proportional to distance from the wire)
6. __True___ For any displacement of a charged particle, the work done by the magnetic force on the
charged particle moving in a magnetic field is zero J..
7. __False___ Cutting a bar magnet near its north end results in a smaller mostly north pole magnet and
a larger mostly south pole magnet.
8. __True___ Capacitors in series share the same charge and capacitors in parallel share the same voltage.
Part II – (3 pts each) Multiple choice questions: For questions 9 – 18 select the one alternative that best
completes the statement or answers the question.
9. If a current of 2.0 A is flowing from point a to point b, the potential difference between the points is
A) 6 V
D) 8 V
B) 14 V
E) 20 V
C) 22 V
10. The electric potential inside a capacitor
A) is constant.
B) increases linearly from the negative to the positive plate.
C) decreases linearly from the negative to the positive plate.
D) decreases inversely with distance from the negative plate.
E) decreases inversely with the square of the distance from the negative plate.
11. If the voltage on a capacitor is doubled, then the energy stored by the capacitor
A) stays the same
B) doubles
C) quadruples
D) halves
12. For the capacitor shown, the potential difference V between points a and b is
A) 3 V
B) 3 tan30o V
C) 3 sin30o V
D) 3 cos30o V
E) 3/sin30o V
F) 3/cos30o V
13. The switch in the circuit is initially in the open position (as shown). After the
switch is closed, what happens to the current through the resistor?
A) It starts out at zero and gradually builds up to a maximum.
B) It suddenly rises to a maximum and gradually decreases to zero.
C) It starts out at zero and gradually builds up to a maximum and drops
sharply down to zero.
V
D) It remains at a constant nonzero value.
E) It is zero at all times.
R
C
14. A proton is released from rest at point A where the potential is 0
After release, the proton
A) moves toward C with a constant speed.
B) moves toward C with an increasing speed.
C) moves toward B with a constant speed.
D) moves toward B with an increasing speed.
E) remains at rest.
15. Three charged metal spheres of different radii are connected by a thin copper wire. Which of the
following is true for the electric field and potential at the surface of each sphere
A) V1 < V2 < V3 and E1 = E2 = E3
B) V1 > V2 > V3 and E1 = E2 = E3
1
C) V1 = V2 = V3 and E1 = E2 = E3
V1 , E1
D) V1 = V2 = V3 and E1 > E2 > E3
V2 , E2
E) V1 > V2 > V3 and E1 > E2 > E3
V3 , E3
F) V1 < V2 < V3 and E1 > E2 > E3
G) Both potential and electric field are zero at each surface.
16. A moving electron can be accelerated without changing its speed by
A) an electric field .
D) This situation can not be possible.
B) a magnetic field.
E) none of these.
C) magnetic and electric field.
V.
17. When a bar magnet is broken in two pieces as shown, then the magnetization of each piece
A) is twice as strong as the original piece.
B) is half as strong as the original piece.
N
S
C) is as strong as the original piece.
D) no longer exists.
E) is a random value and have no relation to one another.
18. A positively charged particle is moving northward in a magnetic field. The magnetic force on the
particle is northeast. What is the direction of the magnetic field?
A) Up
D) South
B) Down
E) This situation can not exist.
C) West
19. A rectangular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil
with its south pole pointing down. What is the direction of the induced current in the coil?
A) counterclockwise
C) There is no current in the coil.
B) clockwise
D) None of the other answers is correct.
20. In the plane of the paper, a group of equipotential lines are shown
at right. At which location will an electron experience a force
directed toward the top of the page (along y-axis)?
At B. Electric field is pointing in direction of decreasing
potential and perpendicular to equipotential surfaces (lines).
At location B, electric field is pointing straight down (–y
direction). An electron will accelerate in opposite direction of
the field.
21. In the circuit shown at right, resistors R1 = R2 = R3. How do power
dissipated in each resistor compare?
A) P1 = (½)P2 =(½)P3
D) P1 = P2 + P3
B) P1 = P2 = P3
E) P1 = 2P2 + 2P3
C) 4P1 = P2 = P3
22. In the circuit shown at right, the current in the 6.0Ω resistor is 3A.
The currents in other two resistors are:
A) I1 = 1A and I2 = 1.5A
B) I1 = 6A and I2 = 9A
C) I1 = 9A and I2 = 6A
D) The answer cannot be obtained without knowing
the emf ε of the battery.
23. The phenomenon of magnetism is best understood in terms of
A) the existence of magnetic poles.
B) the magnetic fields associated with the movement of charged particles.
C) gravitational forces between nuclei and orbital electrons.
D) electrical fluids.
E) None of these is correct.
24. What is the direction of the magnetic field around a wire carrying a current directly into this page?
A) The field is parallel to and in the same direction as the current flow.
B) It is parallel to but directed opposite to the current flow.
C) It is counterclockwise around the wire in the plane of the page.
D) It is clockwise around the wire in the plane of the page.
E) None of these is correct.
25. An electron traveling horizontally enters a region where a uniform magnetic field is directed into the
plane of the paper as shown. Which one of the following phrases most accurately describes the motion
of the electron once it has entered the field?
A) upward and parabolic
B) upward and circular
C) downward and circular
D) upward, along a straight line
E) downward and parabolic
Part III - Questions and Problems
Please explain or show your work clearly and completelyfor each of the following questions or
problem.
26. (10 pts) The graph of electric potential versus position
in the x-direction for a region is shown at right. Six
positions in this region are labeled.
A) How do you calculate the electric field if potential
function is given?
�⃗𝑉, in one dimension this equation
Recall 𝐸�⃗ = −∇
dV
will be: 𝐸�⃗ = − dx 𝚤̂. From the graph the slope of
dV
the curve provides the dx and appling the slope
dV
values to 𝐸�⃗ = − dx 𝚤̂ will show E-field.
B) Rank the electric field at the labeled positions from largest (greatest positive) to smallest (greatest
negative), using successive > sign or = sign where applies
Again First calculate slopes at points A – F, and then multiply the values by –1 and the rank the
result. ED > EC = EE > EF > EB = EA
27. (5 pts)Rank in order, from largest to smallest, the energies U1 to U4 stored in each capacitor shown
1
below. Recall: 𝑈𝐶 = 2 𝐶𝑉 2 <
1
𝑈1 = 2 𝐶𝑉 2
Ranking:
1
1
𝑈2 = 2 (4𝐶)(3 𝑉)2
2
𝑈2 = 9 𝐶𝑉 2
U4 > U1 > U3 > U2
1
1
𝑈3 = 2 (2𝐶)(2 𝑉)2
1
𝑈3 = 4 𝐶𝑉 2
1 1
𝑈4 = 2 �2 𝐶� (2𝑉)2
𝑈4 = 𝐶𝑉 2
28. (15 pts)Two bulb and a capacitor are connected to a battery as
shown. Immediately after the switch is closed:
A) Rank the currents IA , IB , IC , and Ibat. Explain your reasoning.
At t = 0 (immediately after the switched is closed) capacitor is
empty and acts like a short (meaning all current goes through
capacitor and almost none in bulb B.
Ibat = IA = IC > IB ≈ 0
B) A long time after the switch has been closed:
Rank the currents IA , IB , IC , and Ibat. Explain your reasoning.
Long time after switch is closed (t = = ∞), capacitor is fully
charged and acts like open switch (no current will pass throw it). All current goes throw IB.
Ibat = IA = IB > IC ≈ 0
C) A long time after the switch has been closed: Rank the potential difference across bulb A (VA),
bulb B (VB), across the capacitor (VC), and the battery (Vbat)? Explain.
Voltage of battery is divided on each balb equally and since capacitor is in parallel to bulb B,
hence will have same voltage as bulb B.
Vbat > VA = VB = VB
29. (10 pts) Consider two bar magnets placed at right angles to each other, as shown.
A) If a small compass is placed at point P, what direction does the painted end of the compass needle
point? Show the vector diagram.
Outside of a magnet, magnetic field lines are leaving north pole and
enter south pole. The compass needle will be diffelected in the
direction of the net B-field. Assuming both Magnets are identical, the
deflection of compass will be 45o clockwise.
B) If the compass needle instead pointed 15 degrees clockwise of where
you predicted in A), what could you qualitatively conclude about the
relative strengths of the two magnets?
From the vector addition of the fields of magnet a and b we can
write
𝑩𝑏
tan 30𝑜 =
= 0.58
𝑩𝑎
Therefore magnet b is weaker than magnet a. Magnet b is about
58% as strong as magnet a.
30. (10 pts) An electron is initially traveling at 3.0 x 106 m/s to the left at point A when it enters a uniform
electric field. The equipotential surfaces associated with the field is shown at right. The equipotential
surfaces are 25.0 cm apart from each other
A) Is the speed of electron increased, decreased or remains constant when enters the electric
field?
Electric field lines are perpendicular to equipotential surfaces
and in direction of decreasing potentials as shown. An electron
entering field is accelerated in opposite of field direction.
Since electron had an initial velocity to the left (opposite of
field direction), it will be accelerated by the field (speed
increases).
B) What is the velocity of the electron after it has traveled 10.0 cm in the field?
∆𝐾𝐸 = ∆𝑈 = 𝑞∆𝑉
Let’s assume the equipotential surfaces are separated by distance of 25 cm. therefore:
0 + 50𝑉
𝑉0 − 𝑉−50
� ∙ 10𝑐𝑚 =
= +20𝑉
∆𝑉 = �
25𝑐𝑚
2.5
1
1
𝑚𝑣22 − 𝑚𝑣12 = 𝑒∆𝑉
2
2
1
𝑚 2
1
(9.11 × 10−31 𝑘𝑔)𝑣22 − (9.11 × 10−31 𝑘𝑔) �3.0 × 106 � = 1.6 × 10−19 𝐶(+20𝑉)
2
𝑠
2
v2 =4.0×106 m/s
31. (10 pts) Use the circuit at right to answer the following:
A) What is the resistance between points a and b for the resistor combination shown? Show
the new equivalent circuit after every resistors reduction.
The three resistors at the end are in series and will add up to 12Ω. This 12Ω and other 12Ω already
in the circuit are in parallel and combine to make and equivalent of 6Ω. The pattern repeats in
similar way. The net resistance is 11Ω. n(see the sequence of circuit reduction)
B) If a 6.0V battery is placed between point a and b, what will be current throw resistor R1 =2 Ω
The current through the 2Ω is the total current that is drawn from the battery which is
determined by the net resistance (11Ω) in circuit. 𝐼 =
𝑉
𝑅𝑒𝑞
=
6𝑉
11 𝑜ℎ𝑚
= 0.55𝐴
32. (10 pts) Consider the circuit shown at right. Let I1 = 2.0A, I2 = 4.0A, and I3 = 5.0A.
A) Calculate the current values of I4 and I5.
Apply the junction rule at point b:
I1 + I2 = I3 + I4  I4 = 2.0A + 4.0A – 5.0A  I4 = 1.0A
Now apply the junction rule at point e:
I1 + I5 = I3  I5 = 5.0A – 2.0A  I3 = 3.0A
B) If ε1 = 20V and Let R1= 5.0Ω, find the value of R3 .
Applying the loop rule for the loop abe while going CW:
ε1 – I1R1 – I3R3 = 0
20 + 20V – 2.0A . 5.0 Ω = 5.0A R3  R3 = 2.0 Ω
33. (20 pts)Suppose in the circuit shown the switch has been closed long enough for the capacitor to be
fully charged.
A) Find the current (steady) in each resistor.
A fully charged capacitor acts like an open switch, hence
no current flows in capacitor branch. All currents goes
through 6Ω and 12Ω resistors, Since they are in series,
the current through each is the same.
9.0𝑉
𝐼6𝐾 = 𝐼12𝐾 =
= 0.50 𝑚𝐴
6kΩ + 12kΩ
B)
C)
D)
Find he charge on the capacitor.
Q = CV where the voltage across capacitor is the same as that of across 12.0Ω resistor.
𝑉𝐶 = 𝑉12𝐾 = 0.50𝑚𝐴 ∙ 12000Ω = 6.0V
𝑄 = 12.0𝜇𝐹 ∙ 6.0𝑉 = 72𝜇𝐶
Now switch is opened at t = 0. What is the time constant?
When the switch is open the battery and the 6.0kΩ resistor are out of the circuit. The capacitor
will discharge into the 12.0kΩ resistor and 4.0kΩ resistor that are in series (i.e. only the right
loop will be energized.)
Time constant τ = ReqC =(4.0kΩ +12.0kΩ)(12.0µF)  τ = 0.192 s = 192 ms
How long will it take for the charge on capacitor to fall to one-fourth of its original charge?
𝑡
𝑄(𝑡) = 𝑄0 𝑒 −𝜏

1
𝑡
𝑄 = 𝑄0 𝑒 −0.192 𝑠
4 0

1
𝑡
ln �4� = − 0.192𝑠
 t = 0.27 sec
34. (15 pts)A long rigid conductor lying along the x-axis carries a current of 5.0A in the –x direction. A
magnetic field B = 2i + 7x2j is present, where x is in meters and B in mT. Calculate the force on the
1.6-m segment of the conductor that lies between x = 1.4 m and x = 3.0 m.
�⃗ where dl is along the negative x-axis (same as direction of current) and can be
𝐹⃗ = ∫ 𝐼𝑑𝑙�⃗ × 𝐵
replaced by dx
𝚤̂ �𝚥
�⃗ = 𝐼 �−𝑑𝑥 0
𝐼𝑑𝑙�⃗ × 𝐵
2 7𝑥 2
𝑘�
2
0 � = −7𝐼𝑥 𝑑𝑥 𝑘�
0
3.0
𝐹⃗ = � � −7𝐼 𝑥 2 𝑑𝑥� 𝑘� = (−7(5.0𝐴)
1.4
3
𝑥3
� )𝑘� = −283 𝑁 𝑖𝑛 𝑘� 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
3 1.4
35. (20 pts)A thin rod of length 2L is centered on the x axis as shown. The rod carries a uniformly
distributed charge Q distributed along its length 2L with linear charge density λ = Q/2L.
A) Determine the electric potential for any point x > L along the x axis.
The electric potential at point P due small
charge dq a distance x away is given by:
𝑑𝑞
𝑑𝑉 = 𝑘 (𝑥−𝑋) where dq = λ dX
Total potential due to entire rod is:
+𝐿
𝑉= � 𝑘
−𝐿
Recall λ = Q/2L. 
𝜆𝑑𝑋
𝑥−𝑋
𝑉 = −𝑘𝜆 ln(𝑥 − 𝑋)|+𝐿
−𝐿
𝑉 (𝑥 ) =
𝑘𝑄
2𝐿
ln
𝑥+𝐿
𝑥−𝐿
�⃗𝑉 to determine the electric field due to the charged rod at a point x > L
B) Use the fact that 𝐸�⃗ = −∇
along x axis.
�⃗ = − 𝑑𝑉 𝚤̂
�⃗𝑉 reduces to 𝐸
In one dimension 𝐸�⃗ = −∇
𝑑𝑥
𝐸�⃗ = −
𝑑 𝑘𝑄 𝑥 + 𝐿
� 𝚤̂
𝐸�⃗ = − � ln
𝑑𝑥 2𝐿 𝑥 − 𝐿
1
1(𝑥 − 𝐿 ) − 1(𝑥 + 𝐿 )
𝑘𝑄
�
��
� 𝚤̂
(𝑥 − 𝐿 ) 2
2𝐿 𝑥 + 𝐿
𝑥−𝐿
𝑘𝑄
𝐸�⃗ = 2
𝚤̂
𝑥 − 𝐿2
C) How much work is done to move a particle of mass m and charge +q from the point x = 3L to the
point x =2L?
The work done to move a particle of charge q from one point to another is just the potential difference
between the points multiplied q by definition of potential. Thus we have
𝑊 = −𝑞Δ𝑉 = −(+𝑞)(𝑉𝑥=3𝐿 − 𝑉𝑥=2𝐿 )
𝑊 = −𝑞(
𝑘𝑞𝑄
𝑘𝑄 3𝐿 + 𝐿 𝑘𝑄 2𝐿 + 𝐿
ln
−
ln
)=
(ln 3 − ln 2)
2𝐿
2𝐿 3𝐿 − 𝐿 2𝐿 2𝐿 − 𝐿
𝑊=
𝑘𝑄𝑞 3
ln
2𝐿
2