Transcript Conceptual Review Self Quiz

ConcepTest 16.1a Electric Charge I
Two charged balls are
repelling each other as
they hang from the ceiling.
What can you say about
their charges?
a) one is positive, the
other is negative
b) both are positive
c) both are negative
d) both are positive or
both are negative
ConcepTest 16.1a Electric Charge I
Two charged balls are
repelling each other as
they hang from the ceiling.
What can you say about
their charges?
a) one is positive, the
other is negative
b) both are positive
c) both are negative
d) both are positive or
both are negative
The fact that the balls repel each
other only can tell you that they
have the same charge, but you do
not know the sign. So they can be
either both positive or both
negative.
Follow-up: What does the picture look like if the two balls are oppositely
charged? What about if both balls are neutral?
ConcepTest 16.1b Electric Charge II
From the picture,
what can you
conclude about
the charges?
a)
have opposite charges
b)
have the same charge
c)
all have the same charge
d) one ball must be neutral (no charge)
ConcepTest 16.1b Electric Charge II
From the picture,
what can you
conclude about
the charges?
a)
have opposite charges
b)
have the same charge
c)
all have the same charge
d) one ball must be neutral (no charge)
The GREEN and PINK balls must
have the same charge, since they
repel each other. The YELLOW
ball also repels the GREEN, so it
must also have the same charge
as the GREEN (and the PINK).
ConcepTest 16.2a Conductors I
A metal ball hangs from the ceiling
a) positive
by an insulating thread. The ball is
b) negative
attracted to a positive-charged rod
c) neutral
held near the ball. The charge of
d) positive or neutral
the ball must be:
e) negative or neutral
ConcepTest 16.2a Conductors I
A metal ball hangs from the ceiling
a) positive
by an insulating thread. The ball is
b) negative
attracted to a positive-charged rod
c) neutral
held near the ball. The charge of
d) positive or neutral
the ball must be:
e) negative or neutral
Clearly, the ball will be attracted if its
charge is negative. However, even if
the ball is neutral, the charges in the
ball can be separated by induction
(polarization), leading to a net
attraction.
remember
the ball is a
conductor!
Follow-up: What happens if the metal ball is replaced by a plastic ball?
ConcepTest 16.2b Conductors II
Two neutral conductors are connected
a)
0
0
by a wire and a charged rod is brought
b)
+
–
c)
–
+
d)
+
+
e)
–
–
near, but does not touch. The wire is
taken away, and then the charged rod
is removed. What are the charges on
the conductors?
0
0
?
?
ConcepTest 16.2b Conductors II
Two neutral conductors are connected
a)
0
0
by a wire and a charged rod is brought
b)
+
–
c)
–
+
d)
+
+
e)
–
–
near, but does not touch. The wire is
taken away, and then the charged rod
is removed. What are the charges on
the conductors?
While the conductors are connected, positive
0
0
?
?
charge will flow from the blue to the green
ball due to polarization. Once disconnected,
the charges will remain on the separate
conductors even when the rod is removed.
Follow-up: What will happen when the
conductors are reconnected with a wire?
ConcepTest 16.3a Coulomb’s Law I
What is the magnitude
a) 1.0 N
b) 1.5 N
of the force F2?
c) 2.0 N
F1 = 3N
Q
Q
F2 = ?
d) 3.0 N
e) 6.0 N
ConcepTest 16.3a Coulomb’s Law I
What is the magnitude
a) 1.0 N
b) 1.5 N
of the force F2?
c) 2.0 N
F1 = 3N
Q
Q
F2 = ?
d) 3.0 N
e) 6.0 N
The force F2 must have the same magnitude as F1. This is
due to the fact that the form of Coulomb’s Law is totally
symmetric with respect to the two charges involved. The
force of one on the other of a pair is the same as the reverse.
Note that this sounds suspiciously like Newton’s 3rd Law!!
ConcepTest 16.3b Coulomb’s Law II
F1 = 3N
Q
Q
F2 = ?
b) 3.0 N
If we increase one charge to 4Q,
what is the magnitude of F1?
F1 = ?
4Q
Q
a) 3/4 N
F2 = ?
c) 12 N
d) 16 N
e) 48 N
ConcepTest 16.3b Coulomb’s Law II
F1 = 3N
Q
Q
F2 = ?
b) 3.0 N
If we increase one charge to 4Q,
what is the magnitude of F1?
F1 = ?
4Q
Q
a) 3/4 N
F2 = ?
c) 12 N
d) 16 N
e) 48 N
Originally we had:
F1 = k(Q)(Q)/r2 = 3 N
Now we have:
F1 = k(4Q)(Q)/r2
which is 4 times bigger than before.
Follow-up: Now what is the magnitude of F2?
ConcepTest 16.3c Coulomb’s Law III
The force between two charges
a) 9 F
separated by a distance d is F. If
b) 3 F
the charges are pulled apart to a
c) F
distance 3d, what is the force on
d) 1/3 F
each charge?
e) 1/9 F
F
F
Q
Q
d
?
?
Q
Q
3d
ConcepTest 16.3c Coulomb’s Law III
The force between two charges
a) 9 F
separated by a distance d is F. If
b) 3 F
the charges are pulled apart to a
c) F
distance 3d, what is the force on
d) 1/3 F
each charge?
e) 1/9 F
F
Originally we had:
F
Q
Q
Fbefore = k(Q)(Q)/d2 = F
Now we have:
Fafter = k(Q)(Q)/(3d)2 = 1/9 F
d
?
?
Q
Q
3d
Follow-up: What is the force if the original distance is halved?
ConcepTest 16.4b Electric Force II
Two balls with charges +Q and +4Q are separated by 3R. Where
should you place another charged ball Q0 on the line between
the two charges such that the net force on Q0 will be zero?
+4Q
+Q
a)
b)
c)
d)
2R
R
3R
e)
ConcepTest 16.4b Electric Force II
Two balls with charges +Q and +4Q are separated by 3R. Where
should you place another charged ball Q0 on the line between
the two charges such that the net force on Q0 will be zero?
+4Q
+Q
a)
b)
c)
d)
e)
2R
R
3R
The force on Q0 due to +Q is:
F = k(Q0)(Q)/R2
The force on Q0 due to +4Q is:
F = k(Q0)(4Q)/(2R)2
Since +4Q is 4 times bigger than +Q, then Q0 needs to be
farther from +4Q. In fact, Q0 must be twice as far from +4Q,
since the distance is squared in Coulomb’s Law.
ConcepTest 16.5a Proton and Electron I
A proton and an electron are
held apart a distance of 1 m
and then released. As they
approach each other, what
happens to the force between
them?
p
a) it gets bigger
b) it gets smaller
c) it stays the same
e
ConcepTest 16.5a Proton and Electron I
A proton and an electron are
held apart a distance of 1 m
and then released. As they
approach each other, what
happens to the force between
them?
a) it gets bigger
b) it gets smaller
c) it stays the same
By Coulomb’s Law, the force between the
two charges is inversely proportional to
the distance squared. So, the closer they
get to each other, the bigger the electric
force between them gets!
p
e
Follow-up: Which particle feels the larger force at any one moment?
ConcepTest 16.5b Proton and Electron II
A proton and an electron are held
a) proton
apart a distance of 1 m and then
b) electron
released. Which particle has the
c) both the same
larger acceleration at any one
moment?
p
e
ConcepTest 16.5b Proton and Electron II
A proton and an electron are held
a) proton
apart a distance of 1 m and then
b) electron
released. Which particle has the
c) both the same
larger acceleration at any one
moment?
p
The two particles feel the same force.
Since F = ma, the particle with the smaller
mass will have the larger acceleration.
This would be the electron.
e
ConcepTest 16.5c Proton and Electron III
A proton and an electron
are held apart a distance
of 1 m and then let go.
Where would they meet?
a) in the middle
b) closer to the electron’s side
c) closer to the proton’s side
p
e
ConcepTest 16.5c Proton and Electron III
A proton and an electron
are held apart a distance
of 1 m and then let go.
Where would they meet?
a) in the middle
b) closer to the electron’s side
c) closer to the proton’s side
By Newton’s 3rd Law, the electron and proton
feel the same force. But, since F = ma, and
since the proton’s mass is much greater, the
proton’s acceleration will be much smaller!
p
Thus, they will meet closer to the proton’s
original position.
Follow-up: Which particle will be moving faster when they meet?
e
ConcepTest 17.9a Varying Capacitance I
What must be done to
a) increase the area of the plates
a capacitor in order to
b) decrease separation between the plates
increase the amount of
c) decrease the area of the plates
charge it can hold (for
a constant voltage)?
d) either a) or b)
e) either b) or c)
ConcepTest 17.9a Varying Capacitance I
What must be done to
a) increase the area of the plates
a capacitor in order to
b) decrease separation between the plates
increase the amount of
c) decrease the area of the plates
charge it can hold (for
a constant voltage)?
d) either a) or b)
e) either b) or c)
Since Q = C V, in order to increase the charge
that a capacitor can hold at constant voltage,
one has to increase its capacitance. Since the
capacitance is given by C   0 A , that can be
d
done by either increasing A or decreasing d.
ConcepTest 18.1
Which is the correct way to
light the lightbulb with the
battery?
Connect the Battery
d) all are correct
e) none are correct
ConcepTest 18.1
Which is the correct way to
light the lightbulb with the
Connect the Battery
d) all are correct
e) none are correct
battery?
Current can only flow if there is a continuous connection from the
negative terminal through the bulb to the positive terminal. This is
only the case for (c).
ConcepTest 18.2
Ohm’s Law
across a certain conductor
a) Ohm’s law is obeyed since the current
still increases when V increases
and you observe the current
b) Ohm’s law is not obeyed
increases three times. What
c) This has nothing to do with Ohm’s law
You double the voltage
can you conclude?
ConcepTest 18.2
Ohm’s Law
across a certain conductor
a) Ohm’s law is obeyed since the current
still increases when V increases
and you observe the current
b) Ohm’s law is not obeyed
increases three times. What
c) This has nothing to do with Ohm’s law
You double the voltage
can you conclude?
Ohm’s law, V = I R, states that the
relationship between voltage and current is
linear. Thus for a conductor that obeys
Ohm’s Law, the current must double when
you double the voltage.
Follow-up: Where could this situation occur?
ConcepTest 18.4
Dimmer
a) the power
When you rotate the knob of a
b) the current
light dimmer, what is being
c) the voltage
changed in the electric circuit?
d) both a) and b)
e) both b) and c)
ConcepTest 18.4
Dimmer
a) the power
When you rotate the knob of a
b) the current
light dimmer, what is being
c) the voltage
changed in the electric circuit?
d) both a) and b)
e) both b) and c)
The voltage is provided at 120 V from the outside.
The light dimmer increases the resistance and
therefore decreases the current that flows through
the lightbulb.
Follow-up: Why does the voltage not change?
ConcepTest 18.5a
Lightbulbs
Two lightbulbs operate at 120 V, but
a) the 100 W bulb
one has a power rating of 25 W while
b) the 25 W bulb
the other has a power rating of 100 W.
c) both have the same
Which one has the greater
d) this has nothing to do with
resistance
resistance?
ConcepTest 18.5a
Lightbulbs
Two lightbulbs operate at 120 V, but
a) the 100 W bulb
one has a power rating of 25 W while
b) the 25 W bulb
the other has a power rating of 100 W.
c) both have the same
Which one has the greater
d) this has nothing to do with
resistance
resistance?
Since P = V2 / R the bulb with the lower
power rating has to have the higher
resistance.
Follow-up: Which one carries the greater current?
ConcepTest 18.5b
Two space heaters in your living
room are operated at 120 V.
Space Heaters I
a) heater 1
Heater 1 has twice the resistance
b) heater 2
of heater 2. Which one will give
c) both equally
off more heat?
ConcepTest 18.5b
Two space heaters in your living
room are operated at 120 V.
Space Heaters I
a) heater 1
Heater 1 has twice the resistance
b) heater 2
of heater 2. Which one will give
c) both equally
off more heat?
Using P = V2 / R, the heater with the smaller resistance will
have the larger power output. Thus, heater 2 will give off
more heat.
Follow-up: Which one carries the greater current?
ConcepTest 19.1a
Series Resistors I
a) 12 V
Assume that the voltage of the battery
is 9 V and that the three resistors are
identical. What is the potential
difference across each resistor?
b) zero
c) 3 V
d) 4 V
e) you need to know the
actual value of R
9V
ConcepTest 19.1a
Series Resistors I
a) 12 V
Assume that the voltage of the battery
is 9 V and that the three resistors are
identical. What is the potential
difference across each resistor?
b) zero
c) 3 V
d) 4 V
e) you need to know the
actual value of R
Since the resistors are all equal,
the voltage will drop evenly
across the 3 resistors, with 1/3 of
9 V across each one. So we get a
3 V drop across each.
9V
Follow-up: What would be the potential difference if
R= 1 W, 2 W, 3 W
ConcepTest 19.1b
Series Resistors II
a) 12 V
In the circuit below, what is the
b) zero
voltage across R1?
c) 6 V
d) 8 V
e) 4 V
R1= 4 W
R2= 2 W
12 V
ConcepTest 19.1b
Series Resistors II
a) 12 V
In the circuit below, what is the
b) zero
voltage across R1?
c) 6 V
d) 8 V
e) 4 V
The voltage drop across R1 has
to be twice as big as the drop
across R2. This means that V1 =
R1= 4 W
R2= 2 W
8 V and V2 = 4 V. Or else you
could find the current I = V/R =
(12 V)/(6 W) = 2 A, then use
12 V
Ohm’s Law to get voltages.
Follow-up: What happens if the voltage is doubled?
ConcepTest 19.2a
Parallel Resistors I
a) 10 A
In the circuit below, what is the
b) zero
current through R1?
c) 5 A
d) 2 A
e) 7 A
R2= 2 W
R1= 5 W
10 V
ConcepTest 19.2a
Parallel Resistors I
a) 10 A
In the circuit below, what is the
b) zero
current through R1?
c) 5 A
d) 2 A
e) 7 A
The voltage is the same (10 V) across each
R2= 2 W
resistor because they are in parallel. Thus,
we can use Ohm’s Law, V1 = I1 R1 to find the
R1= 5 W
current I1 = 2 A.
10 V
Follow-up: What is the total current through the battery?
ConcepTest 19.2b
Points P and Q are connected to a
Parallel Resistors II
a) increases
battery of fixed voltage. As more
b) remains the same
resistors R are added to the parallel
c) decreases
circuit, what happens to the total
d) drops to zero
current in the circuit?
ConcepTest 19.2b
Parallel Resistors II
Points P and Q are connected to a
a) increases
battery of fixed voltage. As more
b) remains the same
resistors R are added to the parallel
c) decreases
circuit, what happens to the total
d) drops to zero
current in the circuit?
As we add parallel resistors, the overall
resistance of the circuit drops. Since V =
IR, and V is held constant by the battery,
when resistance decreases, the current
must increase.
Follow-up: What happens to the current through each resistor?
ConcepTest 19.3b
Two lightbulbs A and B are
connected in series to a
constant voltage source.
When a wire is connected
across B, bulb A will:
Short Circuit II
a) glow brighter than before
b) glow just the same as before
c) glow dimmer than before
d) go out completely
e) explode
ConcepTest 19.3b
Two lightbulbs A and B are
connected in series to a
constant voltage source.
When a wire is connected
across B, bulb A will:
Short Circuit II
a) glow brighter than before
b) glow just the same as before
c) glow dimmer than before
d) go out completely
e) explode
Since bulb B is bypassed by the wire,
the total resistance of the circuit
decreases. This means that the current
through bulb A increases.
Follow-up: What happens to bulb B?
ConcepTest 19.4a
Circuits I
The lightbulbs in the circuit below
a) circuit 1
are identical with the same
b) circuit 2
resistance R. Which circuit
produces more light? (brightness
 power)
c) both the same
d) it depends on R
circuit 1
circuit 2
ConcepTest 19.4a
Circuits I
The lightbulbs in the circuit below
a) circuit 1
are identical with the same
b) circuit 2
resistance R. Which circuit
produces more light? (brightness
 power)
c) both the same
d) it depends on R
In #2, the bulbs are in parallel,
lowering the total resistance of the
circuit. Thus, circuit #2 will draw
a higher current, which leads to
more light, because P = I V.
circuit 1
circuit 2
ConcepTest 19.4b
The three lightbulbs in the circuit all have
Circuits II
a) twice as much
the same resistance of 1 W . By how
b) the same
much is the brightness of bulb B greater
c) 1/2 as much
or smaller than the brightness of bulb A?
(brightness  power)
d) 1/4 as much
e) 4 times as much
A
C
B
10 V
ConcepTest 19.4b
The three light bulbs in the circuit all have
Circuits II
a) twice as much
the same resistance of 1 W . By how
b) the same
much is the brightness of bulb B greater
c) 1/2 as much
or smaller than the brightness of bulb A?
(brightness  power)
d) 1/4 as much
e) 4 times as much
A
We can use P = V2/R to compare the power:
C
B
PA = (VA)2/RA = (10 V) 2/1 W = 100 W
PB = (VB)2/RB = (5 V) 2/1 W = 25 W
Follow-up: What is the total current in the circuit?
10 V
ConcepTest 19.5a
More Circuits I
What happens to the voltage
a) increase
across the resistor R1 when the
b) decrease
switch is closed? The voltage will:
c) stay the same
R1
S
R3
V
R2
ConcepTest 19.5a
More Circuits I
What happens to the voltage
a) increase
across the resistor R1 when the
b) decrease
switch is closed? The voltage will:
c) stay the same
R1
With the switch closed, the addition of
R2 to R3 decreases the equivalent
S
resistance, so the current from the
battery increases. This will cause an
R3
V
increase in the voltage across R1 .
Follow-up: What happens to the current through R3?
R2
ConcepTest 19.8
Kirchhoff’s Rules
The lightbulbs in the
a) both bulbs go out
circuit are identical. When
b) intensity of both bulbs increases
the switch is closed, what
c) intensity of both bulbs decreases
happens?
d) A gets brighter and B gets dimmer
e) nothing changes
ConcepTest 19.8
Kirchhoff’s Rules
The lightbulbs in the
a) both bulbs go out
circuit are identical. When
b) intensity of both bulbs increases
the switch is closed, what
c) intensity of both bulbs decreases
happens?
d) A gets brighter and B gets dimmer
e) nothing changes
When the switch is open, the point
between the bulbs is at 12 V. But so is
the point between the batteries. If
there is no potential difference, then
no current will flow once the switch is
closed!! Thus, nothing changes.
Follow-up: What happens if the bottom
battery is replaced by a 24 V battery?
24 V