Transcript Solutions - UCSB CLAS

Physics 6B
Practice Final Solutions
1. Two speakers placed 4m apart produce sound waves with frequency 425Hz. A listener is standing 3m in front
of the left speaker. Describe the sound that he hears. Assume the speed of sound is 340 m/s.
4m
First find the distance from the right speaker (use the
pythagorean theorem or recognize the 3-4-5 right triangle).
Now find the difference in the distances to each speaker:
(5m – 3m) = 2m
We need to compare this to the wavelength of the sound.
v 340 ms
 
 0.8m
f 425 Hz
2m
 2. 5
0. 8m
This means the difference is 2.5 wavelengths, yielding
destructive interference (a quieter sound)
3m
5m
2. Charge q1 = -5.4µC is placed at the origin, and charge q2 = -2.2 µC is on the x-axis at x=1m. Where should a
charge q3 be placed between q1 and q2 so that the net force acting on it is zero?
Here is a diagram, with q3 in between. We just
need to find the forces on q3 due to the other
charges, and make them equal.
q1
F(q1 on q3)
x
q3 F(q2 on q3) q2
1-x
We are not given the sign of charge q3, but it
shouldn’t matter. The forces due to q1 and q2 will
be in opposite directions in either case.
kqq'
r2
kq q
Fq1  q3  12 3
x
kq2 q3
Fq2  q3 
1  x 2
Felec 
Setting these equal, and plugging in the given values:
kq1q3 kq2q3
q1
q2
5.4
2.2





x2
1  x 2 x 2 1  x 2 x 2 1  x 2
2.2  x 2  5.4  1  x 2  1.48  x  2.32  1  x   3.8 x  2.32  x  0.61m
3. An object with a charge of – 3.6µC and a mass of 12g experiences an upward electric force, due to a uniform
electric field, equal in magnitude to its weight. Find the direction and magnitude of the electric field.
Electric force is simply charge times electric field.
mg
Felec  qE  mg  E 
q
E
0.012 kg 9.8 sm2 
6
3.6  10 C
Felec
 33,000 CN
This is the magnitude of the field. The direction must be downward to give an upward
force on a negative charge. One way to remember this is that the direction of the field is
always the direction of the force on a positive charge. Since our charge is negative, it
must be the opposite direction.
mg
4. Two conducting spheres have net charges q1=+8 µC and q2=-2 µC. The spheres touch and some charge is
transferred. How many electrons are transferred, and to which sphere?
BEFORE
AFTER
q1
q2
q1
q2
+8 µC
-2 µC
+3 µC
+3 µC
When the spheres touch, charges are transferred until both spheres have equal charge.
The total charge before they touch is +6µC, so the charge on each sphere after they touch must be +3µC.
So there must be a total of 5µC of charge transferred.
We need to divide by the charge on the electron to find the # of electrons transferred:
5  10 6 C
 3.1  1013 electrons
19
C
1.6  10 electron
Since sphere 1 gets less positive, it must be gaining the negative electrons. Conversely, sphere 2 is getting more
positive, so it is losing electrons. Thus the electrons must be transferred to sphere 1.
5. During a lightning strike, electrons are transferred from the bottom of a thundercloud to the ground. During this
process, the electrons:
a) gain potential energy as they move toward a higher potential
b) lose potential energy as they move toward a lower potential
c) gain potential energy as they move toward a lower potential
d) lose potential energy as they move toward a higher potential
Since the electron is moving downward, the ground must be positively
charged, and the bottom of the cloud must be negative.
- - - - e-
So the electron is moving toward a higher (positive) potential.
Because its own charge is negative, however, the potential energy of the
electron is getting more negative as it moves toward the ground.
So it is losing potential energy.
+ + + + + + + + +
6. During a lightning strike, electrons are transferred from the bottom of a thundercloud to the ground. This
occurs due to dielectric breakdown of the air, when the electric field is greater than 3x106 V/m. The distance from
the ground to the cloud is 1000m. Find the magnitude of the potential difference between the cloud and the
ground.
Assuming a constant electric field, we can use the formula V=Ed


V  3  106 Vm 1000 m  3 109Volts
- - - - 1000m
+ + + + + + + + +
7. An electric dipole consists of two equal charges, +q and –q, a distance d apart. Find the total electric potential
at a point that is a distance of d/2 to the right of the positive charge, as shown.
kq
r
In this formula we include the sign of the charge to tell us the
sign of the potential. All we need to do is find the potential
due to each charge, and then add them together.
The electric potential due to a point charge is given by:  
left 
k  q 
3d
2
right 
total 
k q 
1d
2
k  q  k q   2 kq
kq 4 kq



2

3d
1d
3
d
d
3 d
2
2
-q
d
+q
d/2
8. How much charge is on each plate of the capacitors in the circuit shown? The battery has voltage 12V, and the
capacitances are: C1=3µF, C2=2µF, C3=4µF.
First we need to find the equivalent capacitance for the circuit.
C2 and C3 are in parallel, so we just add them together.
12V
C2
Next we see that C1 is in series with the (C2+C3) combination, so we
need to use the reciprocal formula.
1
1
1
1
1
 
 
 Ceq  2F
Ceq C1 C2  C3 3 2  4
C1
C3
The total charge for the system is given by Q=CV. Qtotal=24µC
Now we need to work backward to find the charge on each capacitor.
First look at capacitor C1. There are no parallel paths connected to it,
so all the charge must land on its plates. Q1=24µC
12V
Next we notice that since the other 2 capacitors are in parallel, the
charge must split up so that the total adds up to 24µC.
Option 1
C2+C3
Option 2
12V
The voltage across C2 and C3
must be equal because they are in
parallel. Using V=Q/C we get
V1=8volts. This leaves 4volts for
the others. Now using Q=CV
This adds up to 24, and Q2 is half
again we get Q2=8 and Q3=16
as much as Q3.
Since C2 has half the
capacitance of C3, it will only get
half the charge. So we need
Q2=8µF and Q3=16µF.
Ceq=2µF
C1
9.
A parallel-plate capacitor is initially charged by a battery with voltage V. The battery is disconnected, and a
dielectric with constant k is inserted between the plates. What happens to the energy stored in the capacitor?
When the battery is disconnected, the charges that have built up on the plates have nowhere to go, so the charge
on the capacitor will remain constant when the dielectric material is inserted. To calculate the amount of charge
we could use the formula Q=CV.
Now what happens when the dielectric is inserted?
The capacitance is increased by a factor of k (the dielectric constant).
So what does that do to the energy?
(1)U cap  12 CV 2
We have 3 versions of the potential energy formula for capacitors:
Let’s use version (3). We already know that the charge (Q) is not
going to change.
(2)U cap  12 QV
(3)U cap  12
Q2
C
Since the capacitance is increased by a factor of k, the energy
must decrease by that factor (the C is in the denominator of the
formula).
So the answer is c) The stored energy decreases by a factor of k
Note: if the battery had remained connected, the energy would have increased by a factor of k instead. Think
about why this happens.
10. How much power is dissipated in each of the resistors in the circuit shown? The battery has voltage 12V, and
the resistances are: R1=4Ω, R2=3Ω, R3=6Ω.
First we need to find the equivalent resistance for the circuit.
12V
R2
R2 and R3 are in parallel, so we use the reciprocal trick.
1
1
1 1 1


   Req  2
Req R2 R3 3 6
Next we see that R1 is in series with the (R2 and R3) combination, so
we just add them together, For a total of 6Ω.
V 12V
 2A
Now we can use Ohm’s law to find the current: I  
R 6
All of this current must go through R1, since there is no other path
parallel to it. We can use the formula for power dissipated by a
resistor: P=I2R = IV = V2/R. In this case the first one works fine:
P1  2 A2 4  16W
R1
R3
12V
The current will split when it gets to the R2/R3 parallel combination:
Option 1
Option 2
Since R2 has half the resistance The voltage across R2 and R3
of R3, it will get twice the current. must be equal because they are in 12V
So we get .I 2  43 A and I3  23 A
parallel. Using V=IR we get
V1=8volts. This leaves 4volts for
This adds up to 2, and I2 is half
the others.
as much as I3.
Now using P=V2/R we get:
Now use P=I2R again:
2
4V 2
P2  43 3  163 W  5.3W
P2  3  163 W  5.3W

2
P3  23  6  83 W  2.7W
2Ω
P3 
4V 2
6
 166 W  2.7W
6Ω
R1
11. In the circuit below all the bulbs have the same resistance.
When the switch is closed, what happens to bulb A?
A
First realize that since we assume that the resistance stays constant, we
know that bulbs get brighter when the current increases.
C
B
Next, see from the diagram that all the current must go through bulb A.
So now think about what happens when the switch is closed. The path
through bulb B is closed, creating a parallel path to bulb C. This has the
effect of reducing the equivalent resistance (adding a parallel path will
always reduce the overall resistance).
Since the resistance is reduced, the current must increase. (V=I/R)
So the current through bulb A increases, and bulb A gets brighter.
Bonus question – what happens to bulb C?
12. Resistor 1 is a solid cylinder with length L and diameter D. Resistor 2 is made of the same material, but it has
length 2L and diameter 2D. Compared to the resistance of resistor 1, resistance 2 is:
Use the formula for resistance: R 
L
A
The resistivity is the same for each resistor.
The length is twice as large.
The area is 4 times as big, because area depends on the diameter squared
Now compare the 2 resistances:
R1 
R2 
L
A
 2 L 
4A
 12
L
A
 12 R1
So the answer is b) half as large
13. A coil with 1000 turns of wire and a cross-sectional area of 0.5m2 is rotating at 60rpm in a uniform magnetic
field of magnitude 2 Tesla. Find the voltage induced in the coil as it rotates through one quarter of a revolution?
There will be an induced EMF (voltage) whenever there is a changing magnetic flux through the coil.
This coil is rotating, so the flux is constantly changing. The coil makes 60 revolutions each minute, so that is 1
revolution each second. So a quarter-turn takes a time of 0.25 seconds.
In a quarter of a revolution the coil will go from maximum flux to zero flux, so the change in flux will be equal to
the maximum flux. Now we can use our formula for induced voltage (N=1000 turns of wire):




B A
2T  0.5m2
Vinduced   N
N
 1000 
 4000V
t
t
0.25s
14. Two singly ionized isotopes of uranium (one electron has been removed) are projected at v=1.5x105m/s into a
region with a uniform magnetic field of strength 0.75T directed into the page, as shown. The particles land at
distances d1=68.2cm and d2=69.2cm from the entry point. Find the difference between the masses of the
particles.
B
Each charge feels a magnetic force directed toward the center of
the circular path. Setting centripetal force equal to magnetic force:
mv2
qBr
 qvB  m 
r
v
The charge, magnetic field, and velocity are the same for both
charges. Only the radius is different. Notice that the distance
given are diameters, so we must divide them by 2.


qB
1.6  10 19 C 0.75T   0.692 m 0.682 m 
r2  r1  
m2  m1 



v
2 
1.5  105 ms
 2
m2  m1  4  10  27 kg
2
1
v
15. Consider the three electric charges shown below. Charge B is equidistant from charges A and C. List the
charges in order of the magnitude of the force they experience, from smallest to largest.
Each charge will feel 2 forces from the other charges.
Remember – same sign repels, opposites attract.
FBonA  
kqq
kqq
kqq
; FConB   2  Fnet , B  2 2
2
r
r
r
kqq
kqq
3 kqq

;
F



F


BonC
net
,
C
4 r2
r2
2r 2
FAonB  
FAonC
kqq
kqq
5 kqq
;
F



F


ConA
net
,
A
4 r2
r2
2r 2
-q
A
r
+q
B
r
+q
C
16. A beam of electrons is passing through a region with a uniform magnetic field directed downward. If the
electrons are initially moving East, which direction are they deflected when they enter the field?
Use a right-hand-rule for this. Start with fingers along the velocity,
then bend them into the page (toward the magnetic field). The
thumb gives the direction of the force on a positive charge.
In this case we get the thumb pointing Northward. However, our
charge is an electron, so the force is in the opposite direction. You
can flip your hand over, or use your left hand if you want.
B
e-
v
Fmag
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
17. A circuit is arranged with a sliding wire that can move vertically, as shown. There is a uniform magnetic field
directed out of the page. The sliding wire is released from rest in the position shown. As the wire falls:
a) induced current is clockwise and the bulb gets brighter until it reaches a steady intensity.
b) induced current is clockwise and the bulb glows with a constant brightness.
c) induced current is counter-clockwise and the bulb gets brighter until it reaches a steady intensity.
d) induced current is counter-clockwise and the bulb glows with a constant brightness.
First note that there is magnetic flux through the loop, directed out of the
page. As the wire slides down, the loop gets smaller, so the amount of flux
decreases.
v
The reaction to this decreasing flux is to induce current in the wire in such a
way that magnetic flux is created to oppose the change – out of the page. A
right-hand rule will tell us that the induced current flows counter-clockwise.
Iinduced
B
The wire is accelerating downward, so the flux is decreasing faster and faster
as the wire falls. This means the induced current must get stronger and
stronger, making the bulb glow brighter and brighter.
Eventually the induced current is strong enough to produce enough magnetic
force to balance the weight of the wire, and the wire falls at a steady speed.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
18. A metal ring is falling toward a wire with a steady current flowing to the right. What direction is the induced
current in the ring?
The magnetic field makes concentric circles around the wire, so
there is some magnetic flux through the ring, out of the page.
As the ring falls, the flux increases because the field is stronger
closer to the wire. The induced current will try to oppose this
change.
To oppose the change, the induced current must create magnetic
flux through the ring, into the page. Using a right-hand-rule, this
means the induced current must be clockwise, as shown.
v
Iinduced
I
19. A metal ring is falling along a wire with steady current flowing upward. What direction is the induced current in
the ring, as viewed from above?
The magnetic field makes concentric circles around the wire, so there is no
magnetic flux through the ring (as long as the ring is flat, as shown). With no
flux, there is no change in flux, thus no induced current.
I
20. In Europe, the standard voltage available from a wall socket is 240V rms. What is the maximum voltage in
this case?
V
Vrms  max
The formula for RMS voltage is
2
So the maximum voltage is Vmax  2 240V   340V