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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