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ConcepTest 16.1aConcepTest 16.1a Electric Charge IElectric Charge I
1) one is positive, the other
is negative
2) both are positive
3) both are negative
4) both are positive or both
are negative
Two charged balls are Two charged balls are
repelling each other as repelling each other as
they hang from the ceiling. they hang from the ceiling.
What can you say about What can you say about
their charges?their charges?
ConcepTest 16.1aConcepTest 16.1a Electric Charge IElectric Charge I
The fact that the balls repel each
other only can tell you that they
have the same chargesame charge, but you do
not know the sign. So they can
be either both positive or both
negative.
1) one is positive, the other
is negative
2) both are positive
3) both are negative
4) both are positive or both
are negative
Two charged balls are Two charged balls are
repelling each other as repelling each other as
they hang from the ceiling. they hang from the ceiling.
What can you say about What can you say about
their charges?their charges?
Follow-up:Follow-up: What does the picture look like if the two balls are oppositely What does the picture look like if the two balls are oppositely charged? What about if both balls are neutral?charged? What about if both balls are neutral?
1) have opposite charges
2) have the same charge
3) all have the same charge
4) one ball must be neutral (no charge)
From the picture, From the picture,
what can you what can you
conclude about conclude about
the charges?the charges?
ConcepTest 16.1bConcepTest 16.1b Electric Charge IIElectric Charge II
1) have opposite charges
2) have the same charge
3) all have the same charge
4) one ball must be neutral (no charge)
From the picture, From the picture,
what can you what can you
conclude about conclude about
the charges?the charges?
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 16.1bConcepTest 16.1b Electric Charge IIElectric Charge II
ConcepTest 16.2aConcepTest 16.2a Conductors IConductors I
1) positive
2) negative
3) neutral
4) positive or neutral
5) negative or neutral
A metal ball hangs from the ceiling
by an insulating thread. The ball is
attracted to a positive-charged rod
held near the ball. The charge of
the ball must be:
Clearly, the ball will be attracted if its
charge is negativenegative. However, even if
the ball is neutralneutral, the charges in the
ball can be separated by inductioninduction
(polarization), leading to a net
attraction.
1) positive
2) negative
3) neutral
4) positive or neutral
5) negative or neutral
A metal ball hangs from the ceiling
by an insulating thread. The ball is
attracted to a positive-charged rod
held near the ball. The charge of
the ball must be:
remember the ball is a conductor!
ConcepTest 16.2aConcepTest 16.2a Conductors IConductors I
Follow-up:Follow-up: What happens if the What happens if the metal ballmetal ball is replaced by a is replaced by a plastic ballplastic ball??
QQ QQF1 = 3N F2 = ?
1) 1.0 N
2) 1.5 N
3) 2.0 N
4) 3.0 N
5) 6.0 N
What is the magnitude What is the magnitude
of the force Fof the force F22??
ConcepTest 16.3aConcepTest 16.3a Coulomb’s Law ICoulomb’s Law I
The force F2 must have the same magnitudesame 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 reverseforce of one on the other of a pair is the same as the reverse.
Note that this sounds suspiciously like Newton’s 3rd Law!!Note that this sounds suspiciously like Newton’s 3rd Law!!
QQ QQF1 = 3N F2 = ?
1) 1.0 N
2) 1.5 N
3) 2.0 N
4) 3.0 N
5) 6.0 N
What is the magnitude What is the magnitude
of the force of the force FF22??
ConcepTest 16.3aConcepTest 16.3a Coulomb’s Law ICoulomb’s Law I
ConcepTest 16.3bConcepTest 16.3b Coulomb’s Law IICoulomb’s Law II
1) 3/4 N
2) 3.0 N
3) 12 N
4) 16 N
5) 48 N
If we increase one charge to If we increase one charge to 4Q4Q, ,
what is the magnitude ofwhat is the magnitude of F F11??
4Q4Q QQF1 = ? F2 = ?
QQ QQF1 = 3N F2 = ?
ConcepTest 16.3bConcepTest 16.3b Coulomb’s Law IICoulomb’s Law II
Originally we had:
F1 = k(Q)(Q)/r2 = 3 N
Now we have:
F1 = k(4Q)(Q)/r2
which is 4 times bigger4 times bigger than before.
1) 3/4 N
2) 3.0 N
3) 12 N
4) 16 N
5) 48 N
If we increase one charge to If we increase one charge to 4Q4Q, ,
what is the magnitude of what is the magnitude of FF11??
44QQ QQF1 = ? F2 = ?
QQ QQF1 = 3N F2 = ?
Follow-up:Follow-up: Now what is the magnitude of Now what is the magnitude of FF22??
1) 9 F
2) 3 F
3) F
4) 1/3 F
5) 1/9 F
The force between two charges
separated by a distance d is F. If
the charges are pulled apart to a
distance 3d, what is the force on
each charge?
QQFF
QFF
dd
QQ??
QQ??
33dd
ConcepTest 16.3cConcepTest 16.3c Coulomb’s Law IIICoulomb’s Law III
Originally we had:
Fbefore = k(Q)(Q)/d2 = F
Now we have:
Fafter = k(Q)(Q)/(3d)2 = 1/9 F
1) 9 F
2) 3 F
3) F
4) 1/3 F
5) 1/9 F
The force between two charges
separated by a distance d is F. If
the charges are pulled apart to a
distance 3d, what is the force on
each charge?
QQFF
QFF
dd
QQ??
QQ??
33dd
ConcepTest 16.3cConcepTest 16.3c Coulomb’s Law IIICoulomb’s Law III
Follow-up:Follow-up: What is the force if the original distance is halved? What is the force if the original distance is halved?
ConcepTest 18.1ConcepTest 18.1 Connect the Battery Connect the Battery
Which is the correct way to Which is the correct way to
light the lightbulb with the light the lightbulb with the
battery?battery?
4) all are correct
5) none are correct
1) 3)2)
Current can only flow if there is a continuous connectioncontinuous connection from the
negative terminal through the bulb to the positive terminal. This is
only the case for Fig. (3).
ConcepTest 18.1ConcepTest 18.1 Connect the Battery Connect the Battery
Which is the correct way to Which is the correct way to
light the lightbulb with the light the lightbulb with the
battery?battery?
4) all are correct
5) none are correct
1) 3)2)
1) Ohm’s law is obeyed since the current Ohm’s law is obeyed since the current still increases when V increasesstill increases when V increases
2) Ohm’s law is not obeyedOhm’s law is not obeyed
3) This has nothing to do with Ohm’s lawThis has nothing to do with Ohm’s law
ConcepTest 18.2ConcepTest 18.2 Ohm’s Law Ohm’s Law
You double the You double the voltagevoltage
across a certain conductor across a certain conductor
and you observe the and you observe the currentcurrent
increases three times. What increases three times. What
can you conclude?can you conclude?
1) Ohm’s law is obeyed since the current Ohm’s law is obeyed since the current still increases when V increasesstill increases when V increases
2) Ohm’s law is not obeyedOhm’s law is not obeyed
3) This has nothing to do with Ohm’s lawThis has nothing to do with Ohm’s law
Ohm’s law, V = I RV = I R, states that the
relationship between voltage and current is
linearlinear. Thus for a conductor that obeys
Ohm’s Law, the current must double when
you double the voltage.
ConcepTest 18.2ConcepTest 18.2 Ohm’s Law Ohm’s Law
You double the You double the voltagevoltage
across a certain conductor across a certain conductor
and you observe the and you observe the currentcurrent
increases three times. What increases three times. What
can you conclude?can you conclude?
Follow-up:Follow-up: Where could this situation occur? Where could this situation occur?
ConcepTest 18.4ConcepTest 18.4 Dimmer Dimmer
When you rotate the knob of a When you rotate the knob of a
light dimmer, what is being light dimmer, what is being
changed in the electric circuit?changed in the electric circuit?
1) the power
2) the current
3) the voltage
4) both (1) and (2)
5) both (2) and (3)
The voltage is provided at 120 V from the outside.
The light dimmer increases the resistanceincreases the resistance and
therefore decreases the currentdecreases the current that flows through
the lightbulb.
ConcepTest 18.4 ConcepTest 18.4 Dimmer Dimmer
When you rotate the knob of a When you rotate the knob of a
light dimmer, what is being light dimmer, what is being
changed in the electric circuit?changed in the electric circuit?
1) the power
2) the current
3) the voltage
4) both (1) and (2)
5) both (2) and (3)
Follow-up:Follow-up: Why does the voltage not change? Why does the voltage not change?
ConcepTest 18.5aConcepTest 18.5a LightbulbsLightbulbs
Two lightbulbs operate at 120 V, but one Two lightbulbs operate at 120 V, but one
has a power rating of has a power rating of 25 W25 W while the while the
other has a power rating of other has a power rating of 100 W100 W. .
Which one has the greater resistance? Which one has the greater resistance?
1) the 25 W bulb
2) the 100 W bulb
3) both have the same
4) this has nothing to do with resistance
Since P = VP = V22 / R / R the bulb with the lower lower
power ratingpower rating has to have the higher higher
resistanceresistance.
ConcepTest 18.5a ConcepTest 18.5a LightbulbsLightbulbs
Two lightbulbs operate at 120 V, but one Two lightbulbs operate at 120 V, but one
has a power rating of has a power rating of 25 W25 W while the while the
other has a power rating of other has a power rating of 100 W100 W. .
Which one has the greater resistance? Which one has the greater resistance?
1) the 25 W bulb
2) the 100 W bulb
3) both have the same
4) this has nothing to do with resistance
Follow-up:Follow-up: Which one carries the greater current? Which one carries the greater current?
ConcepTest 18.5bConcepTest 18.5b Space Heaters ISpace Heaters I
Two space heaters in your living Two space heaters in your living
room are operated at 120 V. room are operated at 120 V.
Heater 1 has Heater 1 has twicetwice the resistance the resistance
of heater 2. Which one will give of heater 2. Which one will give
off more heat?off more heat?
1) heater 1
2) heater 2
3) both equally
Using P = VP = V22 / R, / R, the heater with the smaller resistancesmaller resistance will
have the larger powerlarger power output. Thus, heater 2 will give off
more heat.
ConcepTest 18.5bConcepTest 18.5b Space Heaters ISpace Heaters I
Two space heaters in your living Two space heaters in your living
room are operated at 120 V. room are operated at 120 V.
Heater 1 has Heater 1 has twicetwice the resistance the resistance
of heater 2. Which one will give of heater 2. Which one will give
off more heat?off more heat?
1) heater 1
2) heater 2
3) both equally
Follow-up:Follow-up: Which one carries the greater current? Which one carries the greater current?
ConcepTest 19.1aConcepTest 19.1a Series Resistors ISeries Resistors I
9 V
Assume that the voltage of the battery Assume that the voltage of the battery
is is 9 V9 V and that the three resistors are and that the three resistors are
identicalidentical. What is the potential . What is the potential
difference across each resistor?difference across each resistor?
1) 12 V
2) zero
3) 3 V
4) 4 V
5) you need to know the actual value of R
Since the resistors are all equalequal,
the voltage will drop evenlyevenly
across the 3 resistors, with 1/3 of
9 V across each one. So we get a
3 V drop3 V drop across each.
ConcepTest 19.1aConcepTest 19.1a Series Resistors ISeries Resistors I
9 V
Assume that the voltage of the battery Assume that the voltage of the battery
is is 9 V9 V and that the three resistors are and that the three resistors are
identicalidentical. What is the potential . What is the potential
difference across each resistor?difference across each resistor?
1) 12 V
2) zero
3) 3 V
4) 4 V
5) you need to know the actual value of R
Follow-up:Follow-up: What would be the potential difference if What would be the potential difference if RR= = 1
ConcepTest 19.1bConcepTest 19.1b Series Resistors IISeries Resistors II
12 V
R1= 4 R2= 2
In the circuit below, what is the In the circuit below, what is the
voltage across voltage across R1??
1) 12 V
2) zero
3) 6 V
4) 8 V
5) 4 V
ConcepTest 19.1bConcepTest 19.1b Series Resistors IISeries Resistors II
12 V
R1= 4 R2= 2
In the circuit below, what is the In the circuit below, what is the
voltage across voltage across R1??
1) 12 V
2) zero
3) 6 V
4) 8 V
5) 4 V
The voltage drop across RThe voltage drop across R11 has has
to be twice as big as the drop to be twice as big as the drop
across across RR22.. This means that VV11 = =
8 V8 V and V2 = 4 V. Or else you
could find the current I = V/R =
(12 V)/(6 = 2 A, then use
Ohm’s Law to get voltages. Follow-up:Follow-up: What happens if the voltage is doubled? What happens if the voltage is doubled?
ConcepTest 19.2aConcepTest 19.2a Parallel Resistors IParallel Resistors I
In the circuit below, what is the In the circuit below, what is the
current through current through R1??
10 V
R1= 5
R2= 2
1) 10 A
2) zero
3) 5 A
4) 2 A
5) 7 A
The voltagevoltage is the samesame (10 V) across each
resistor because they are in parallel. Thus,
we can use Ohm’s Law, VV11 = I = I11 R R11 to find the
current II11 = 2 A = 2 A.
ConcepTest 19.2aConcepTest 19.2a Parallel Resistors IParallel Resistors I
In the circuit below, what is the In the circuit below, what is the
current through current through R1??
10 V
R1= 5
R2= 2
1) 10 A
2) zero
3) 5 A
4) 2 A
5) 7 A
Follow-up:Follow-up: What is the total current through the battery? What is the total current through the battery?
ConcepTest 19.2bConcepTest 19.2b Parallel Resistors IIParallel Resistors II
1) increases
2) remains the same
3) decreases
4) drops to zero
Points P and Q are connected to a Points P and Q are connected to a
battery of fixed voltage. As more battery of fixed voltage. As more
resistors resistors RR are added to the parallel are added to the parallel
circuit, what happens to the circuit, what happens to the total total
currentcurrent in the circuit? in the circuit?
ConcepTest 19.2bConcepTest 19.2b Parallel Resistors IIParallel Resistors II
1) increases
2) remains the same
3) decreases
4) drops to zero
As we add parallel resistors, the overall
resistance of the circuit dropsresistance of the circuit drops. Since V =
IR, and V is held constant by the battery,
when resistance decreasesresistance decreases, the current current
must increasemust increase.
Points P and Q are connected to a Points P and Q are connected to a
battery of fixed voltage. As more battery of fixed voltage. As more
resistors resistors RR are added to the parallel are added to the parallel
circuit, what happens to the circuit, what happens to the total total
currentcurrent in the circuit? in the circuit?
Follow-up:Follow-up: What happens to the current through each resistor? What happens to the current through each resistor?
ConcepTest 19.3aConcepTest 19.3a Short CircuitShort Circuit
Current flows through a Current flows through a
lightbulb. If a wire is now lightbulb. If a wire is now
connected across the connected across the
bulb, what happens?bulb, what happens?
1) all the current continues to flow through all the current continues to flow through the bulbthe bulb
2) half the current flows through the wire, half the current flows through the wire, the other half continues through the the other half continues through the bulbbulb
3) all the current flows through the wire all the current flows through the wire
4) none of the abovenone of the above
The current divides based on the
ratio of the resistances. If one of the
resistances is zerozero, then ALLALL of the
current will flow through that path.
ConcepTest 19.3aConcepTest 19.3a Short CircuitShort Circuit
Current flows through a Current flows through a
lightbulb. If a wire is now lightbulb. If a wire is now
connected across the connected across the
bulb, what happens?bulb, what happens?
1) all the current continues to flow through all the current continues to flow through the bulbthe bulb
2) half the current flows through the wire, half the current flows through the wire, the other half continues through the the other half continues through the bulbbulb
3) all the current flows through the wire all the current flows through the wire
4) none of the abovenone of the above
Follow-up:Follow-up: Doesn’t the wire have SOME resistance? Doesn’t the wire have SOME resistance?
ConcepTest 19.3bConcepTest 19.3b Short Circuit IIShort Circuit II
Two lightbulbs A and B are Two lightbulbs A and B are
connected in series to a connected in series to a
constant voltage source. constant voltage source.
When a wire is connected When a wire is connected
across B, bulb A will:across B, bulb A will:
1) glow brighter than beforeglow brighter than before
2) glow just the same as beforeglow just the same as before
3) glow dimmer than beforeglow dimmer than before
4) go out completely4) go out completely
5) explode5) explode
Since bulb B is bypassed by the wire,
the total resistance of the circuit total resistance of the circuit
decreasesdecreases. This means that the current current
through bulb A increasesthrough bulb A increases.
ConcepTest 19.3bConcepTest 19.3b Short Circuit IIShort Circuit II
Two lightbulbs A and B are Two lightbulbs A and B are
connected in series to a connected in series to a
constant voltage source. constant voltage source.
When a wire is connected When a wire is connected
across B, bulb A will:across B, bulb A will:
1) glow brighter than beforeglow brighter than before
2) glow just the same as beforeglow just the same as before
3) glow dimmer than beforeglow dimmer than before
4) go out completely4) go out completely
5) explode5) explode
Follow-up:Follow-up: What happens to bulb B? What happens to bulb B?
ConcepTest 19.4aConcepTest 19.4a Circuits ICircuits I
1) circuit 1circuit 1
2) circuit 2circuit 2
3) both the sameboth the same
4) it depends on it depends on RR
The lightbulbs in the circuit below The lightbulbs in the circuit below
are identical with the same are identical with the same
resistance resistance RR. Which circuit . Which circuit
produces more light? (brightness produces more light? (brightness
power) power)
ConcepTest 19.4aConcepTest 19.4a Circuits ICircuits I
1) circuit 1circuit 1
2) circuit 2circuit 2
3) both the sameboth the same
4) it depends on it depends on RR
The lightbulbs in the circuit below The lightbulbs in the circuit below
are identical with the same are identical with the same
resistance resistance RR. Which circuit . Which circuit
produces more light? (brightness produces more light? (brightness
power) power)
In #1, the bulbs are in parallelparallel,
lowering the total resistancelowering the total resistance of the
circuit. Thus, circuit #1 will#1 will draw draw
a higher currenta higher current, which leads to
more light, because P = I VP = I V.
