Physics II - Final Exam

A system of charges is labeled below. The values are given in coulombs. Which of the following arrows most closely shows the net force on a?

Vector a

In the picture, what is the ratio of the magnitude of the charges, absolute value of QA over absolute value of QB?

2

1. A student connects a battery to the ends of a tiny light bulb and measures the current flowing through the bulb to be . If the student had instead used a battery with TWICE as much voltage, the current in the bulb would have been:

2I_0 This is as simple as Ohm's Law. If the circuit consists of a single battery and a single light bulb (a resistor), then the voltage across the battery will have to be the same difference in potential across the bulb, but the voltage across the bulb is known to be due to Ohm's Law. Since , then if we had used , we would've had . Since doesn't change, it must mean that , or in words, the current doubled.

Three circuit segments are shown in the figure at the right, labeled one, two, three. All capacitors are shown with equal capacity and see. Each circuit segment is connected to an identical battery as each of the other circuit segments. Rank the circuit segments according to their total charge. Note the total charge stored within a circuit segment is the sum of all the charges stored by all capacitors within a single circuit segment.

3=1<2 Number two stores without having to borrow from its neighbors.

Two electric charges (𝑞 and 3𝑞) are separated by a distance 𝑑 and the force of repulsion between them is 𝐹. If the charges are LATER moved to a distance 0.5𝑑, how strong is the NEW force of repulsion between them?

4𝐹 Because the strength of the Coulomb force (the electric force that charged particles exert on each other) depends on 1/𝑑2, the inverse square of the distance between the charged particles, moving the charges twice as close causes the force they feel to increase by a factor of four: 𝐹𝑛𝑒𝑤=𝑘𝐸|𝑞||3𝑞|𝑑𝑛𝑒𝑤2 =𝑘𝐸|𝑞||3𝑞|(0.5𝑑)2 =𝑘𝐸|𝑞||3𝑞|0.25⋅𝑑2 =4∙𝑘𝐸|𝑞||3𝑞|𝑑2 =4∙𝐹

Two electric charges (𝑞 and 3𝑞) are separated by a distance 𝑑 and the force of repulsion between them is 𝐹. Now suppose the charges are again separated by the original distance 𝑑, but the 𝑞 charge is REPLACED by a 4𝑞 charge. What is the force of repulsion on the NEW 4𝑞 charge?

4𝐹 The original force of repulsion by these charges was 𝐹=𝑘𝐸|𝑞||3𝑞|𝑑2. But then, we replaced the 𝑞 with a 4𝑞 charge, so the new force is 𝐹𝑛𝑒𝑤=𝑘𝐸|4𝑞||3𝑞|𝑑2 =4∙𝑘𝐸|𝑞||3𝑞|𝑑2=4∙𝐹. Also, remember that this is the force of repulsion that BOTH charges feel. They can't push on each other with different forces - they will both feel this force. So, even though we replaced only one of the charges, both charges now feel a stronger repulsion to each other.

Excess electric charge is deposited at a location called 𝑃 on a rubber ball. If we check on the excess charge a few seconds later, which will be true? a. All of the excess charge will remain at 𝑃 b. The excess charge will evenly distribute itself over the outside surface of the ball c. The excess charge will evenly distribute itself over the inside AND outside surfaces of the ball d. Most of the charge will remain at 𝑃, but some of it will have spread out over the surface of the ball e. There will no longer be any excess charge remaining on the ball

All of the excess charge will remain at 𝑃 Rubber is not a conductive material. It's highly resistive, and a great insulator. Any charge placed at a certain location on the rubber basically won't move at all, because it can't. Insulators don't really allow charge to freely flow. You could (maybe) argue in favor of choice D in realistic terms (a little bit), but I think it's going to be minimal if not non-existent given how good of an insulator rubber is. Where you put the charge is pretty much the only place it's going to be. Note that this discussion is totally untrue for conductors like copper, steel, iron, etc. Those materials allow the charge to completely (almost instantaneously) spread out evenly over the outside surface of the object.

Two circuits are shown below. Each circuit contains three identical light bulbs, and an identical battery. (pic) Which light bulb is brightest?

Bulbs D, E, and F are brightest (equally) The equivalent resistance of the circuit on the left is . The equivalent resistance of the circuit on the right is . If both equivalent resistors (in the left circuit and right circuit) are connected to the same battery potential , then the current running through each battery will be and . The current on the right has to split (equally, because all three resistors are identical), but it's still 9x as large as the current on the left, so even though each bulb only takes one third of the available current in the right circuit, the bulbs will be plenty brighter than the bulbs on the left, which all receive the same weak trickle of a current.

What are the following best illustrate The electric field vectors at point a and B?

C are correct

Which of the following is false regarding the hills/valleys analogy for electric potential: A. The steeper the electric potential hill in a region, the stronger the electric field B. Positive charges create electric potential hills and negative charges create electric potential valleys C. Higher electric potential can be thought of as hills from which both positive and negative charges fall down to lower electric potential D. Electric field of actors always point toward the fastest descent downhill E. Equipotential curves represent lines of constant electric potential, around which the potential never changes, just like contour lines on topographical maps or curves with constant height on a hill

C is FALSE

1. Voltmeters have extremely high internal resistance. Ammeters have extremely low internal resistance. Which of the following is true? a) Most of the current should flow through a voltmeter when measuring b) Very little of the current should flow through an ammeter when measuring . c) Voltmeters should be placed in series with circuit elements when measuring . d) Ammeters should be placed in parallel with circuit elements when measuring e) None of these answers is correct.

Everything about this is wrong. Most of the current should NOT flow through a voltmeter when measuring a voltage. First of all, how could it? The resistance of the voltmeter is in the millions of Ohms - almost no current will go through it. Second of all, why would you want that anyway? Imagine trying to measure the voltage across, say, a resistor, where is the thing you want to measure. If most of the current goes through the meter and not the resistor, then through the resistor is way too small, and is way too small, and your measurement is ruined. Also, never place voltmeters in series with circuit elements, because they have mega-Ohm resistance. That will wreck your circuit and kill all the current down the line. Conversely, ammeters have to be placed in series so that the current is forced to go through the circuit element (like a resistor) and ammeter (both), because if you put the ammeter in parallel with a resistor, it has such a small resistance that almost none of the current will go through the resistor at all, and once again: circuit = ruined.

A circuit, labeled "Circuit 2" is shown with a 10 V battery and three resistors connected in series, each having an identical resistance . If the two leads of a voltmeter are clipped onto the points "C" and "D" shown on the circuit, what will the voltmeter read? (pic)

Exactly 0 V Connecting the two leads of the voltmeter between points C and D will result in us comparing the potential at point C to the potential at point D. If the voltmeter is looking for a difference in potential between those two points, it's not going to find one: they're the same wire with nothing in between. The voltage will be shown as 0 V.

A circuit is shown with multiple batteries and multiple resistors and a single capacitor on the far left side between points k and h. The potential differences of the three batteries are V1=6V, V2=9.3V, and V3=3V. Assume that the circuit has been connected and running for a VERY LONG TIME already. How much current flows through the 3 V battery? Explain how you know.

Fun fact: it actually already says "I=0" A right on the diagram between points and . Did you notice? Go back and look - I promise it was always labeled that way. And it's true: there will be no current between points and as long as we wait long enough for the capacitor to totally fill up with charge. That effectively eliminates that entire branch from the circuit, because no current will flow there anyway (at least not after enough time passes, which the question says it has). Capacitors start out (empty) like a closed switch/normal wire - they have almost no resistance initially and they don't impede current at all. But they end their life (full of charge) like an open switch/broken wire - they have impenetrable impedance and don't permit any current to flow at all.

Two identical capacitors in PARALLEL have an equivalent capacitance _______ a single (identical) capacitor

Greater than The capacitance of the circuit actually goes up when you add more identical capacitors in parallel: 𝐶𝑒𝑞𝑢𝑖𝑣=𝐶1+𝐶2=𝐶+𝐶=2𝐶. This is more capacitance than a single capacitor with capacitance 𝐶. Remember that the reason this occurs is because it effectively makes one really wide capacitor with a lot of surface area; this increases the efficiency of the whole set.

Two circuits are shown below. Each circuit contains three identical light bulbs, and an identical battery. 1. Now suppose that a NEW identical light bulb is added to the left circuit AFTER Bulb C. Additionally, a new identical light bulb is added in PARALLEL with bulbs D, E, and F. Which of the following is true? I. Bulbs A, B, and C grow dimmer II. Bulbs D, E, and F grow dimmer III. The current flowing through the battery in the RIGHT circuit is NOT changed

I only Adding a new bulb in series on the left will increase the equivalent resistance, so the current in the entire circuit slows down a bit, and the bulbs will not be as bright as before. So (I) is correct. But, on the right, adding a new bulb in parallel actually reduces the equivalent resistance, so the current running through the battery can actually speed up (because it has more room to run spread out now). So (III) cannot be correct. But (II) is also incorrect, because although there's another bulb with whom D, E, F have to share current, there's also just more current overall to share now, because of the previous sentence in this paragraph. So the two effects (more bulbs that have to share current between them vs. more current available to share due to lower equivalent resistance) actually balance out and the bulbs stay the same brightness, but now the new fourth bulb also has that brightness too.

Which of the following actions would result in the storage of more charge with a parallel plate capacitor connected to a 3 V battery? I. Separate the metal plates farther apart II. Increase the area of the metal plates III. Replace the 3 V battery with a 9 V battery

II and III only Separating the plates farther apart only increases the size of the potential hill between the plates, but the battery can't support a hill that's bigger than the one it provides, so some of the charge no longer feels compelled to stay on the opposing plates, and will run back through the battery to the other plate. This does the opposite of storing more charge, and actually releases some. It makes the capacitor less efficient. Increasing the area of the metal plates is good because it allows the charge already on the plates to spread out more, which causes less repulsion from same-sign charges on the same plate, and actually allows more charge to join. This stores more charge and makes the capacitor more efficient. Introducing a bigger battery is an easy way to store more charge when the capacitor is the only thing connected to it. The definition of capacitance is 𝐶≡𝑄Δ𝑉𝑐, the charge stored divided by the voltage across the plates. If the capacitor is the only thing connected to the battery, then the voltage across the battery will be equal to the voltage across the plates (if we wait long enough for all the charge to settle down and reach electrostatic equilibrium), so Δ𝑉𝐶→Δ𝑉𝐵𝑎𝑡𝑡𝑒𝑟𝑦. Then, since the stored charge must be 𝑄=𝐶∙Δ𝑉𝐶=𝐶∙Δ𝑉𝐵𝑎𝑡𝑡𝑒𝑟𝑦, it stands to reason that attaching a bigger battery will allow the plates to separate charge to a higher potential difference across the plates, which encourages more charge to separate onto the plates. This is also visible from the expression from 𝑄=𝐶∙Δ𝑉𝐶=𝐶∙Δ𝑉𝐵𝑎𝑡𝑡𝑒𝑟𝑦, in which a bigger value of Δ𝑉𝐵𝑎𝑡𝑡𝑒𝑟𝑦 directly causes a proportional increase in 𝑄.

An RC circuit is shown with a battery attached. Which of the following is true IMMEDIATELY AFTER the switch is closed if the capacitor is initially NOT charged? (pic) I. The current will be zero II. The voltage across the capacitor will be zero III. The voltage across the resistor will be zero

II only When the capacitor is empty, the current in the RC circuit can be maximized, because why not? However, as the capacitor fills up with charge, it starts to impede current flow more and more until it eventually shuts down the current altogether. The voltage across the resistor (according to Georg Ohm) must be , so if A at the beginning, then neither can the voltage across the resistor be zero at the beginning either (of course, V eventually, but not at first as the question asks - not until the current dies because the capacitor fills up with charge).

Two circuits are shown below. Each circuit contains three identical light bulbs, and an identical battery. 1. Suppose the battery has a voltage of 10 V. Which of the following is true? I. The voltage across Bulb B is 10 V II. The voltage across Bulb E is 10 V III. The voltage across Bulb C is more than 0 V, but less than 10 V

II, III only The voltage across any of the bulbs on the left cannot be as much as 10 V, because that's the entire difference in potential across all of them (and it has to be, because the high side of Bulb A touches the high side of the battery, which must be at 10 V, and the low side of Bulb C touches the low side of the battery, which must be at 0 V). The voltage across any one of the bulbs on the left will be some amount along the stairs of resistors that take us down from 10 V to 0 V, but not all in one step. Therefore, (III) is correct and (I) is wrong. By contrast, the high side of the battery is touching the high side of every bulb in the right circuit, and the low side of the battery is touching the low side of every bulb in the right circuit. Therefore, the step down in potential from one side of each bulb to the other is actually the full 10 V drop.

Which of the following actions would EXACTLY DOUBLE the flow of charge per second through a metal cylinder connected to a 3 V battery? I. Replace the metal cylinder with one TWICE as long II. Replace the metal cylinder with one TWICE as wide (in diameter) III. Replace the 3 V battery with a 6 V battery

III only If you want to double the flow of charge per second, then you want to double the current. There are two ways to do that through this metal object attached to a battery: 1.) you can lower its resistance by a factor of two, or 2.) you can increase the voltage across it by a factor of two. If we do (I) from the list above, that will double the resistance, so that won't help. If we do (II), that will increase the diameter of the metal cylinder by a factor of two, but that increases the cross-sectional area by a factor of FOUR, which cuts the resistance down to one-fourth of its original, and that QUADRUPLES the current. That's too much, so that doesn't work either. Finally, if we do (III), that actually will work, because it doubles the voltage across the metal cylinder, which, according to Ohm's Law, should increase the current by a factor of exactly two.

A LITHIUM NUCLEUS travels along the path shown and passes through a region of magnetic field. There are no other charges present, and the magnetic field is ZERO everywhere EXCEPT in the gray region. What is a possible direction for the magnetic (B) field?

Into the page I don't care what this object is - as soon as I hear that it's the nucleus of an atom, I know for a fact that it has a positive charge. That means that if I use the RHR to determine the direction of the force it feels, nothing will need to be reversed at the end, because it's not a negative charge. The object had velocity up-right, so the first Right Hand Rule (RHR1, for finding magnetic forces) dictates that our arm point up right. Our thumb, representing the force that the object feels, should point roughly to the left. This only leaves our fingers to bend towards the screen/page to indicate the magnetic field direction. We can be guaranteed that, among the choices shown, the only one that would actually result in this force is the field that points into the page. In real life, it might not point directly into the page, but it would definitely have to have at least some vector component that points into the page, or this force would never happen as it did.

Two identical capacitors in SERIES have an equivalent capacitance _______ a single (identical) capacitor.

Less than The capacitance of the circuit actually goes down when you add more identical capacitors in series: 𝐶𝑒𝑞𝑢𝑖𝑣=1/1𝐶1+1/𝐶2+⋯=1/(1𝐶+1𝐶)=12𝐶=𝐶2. This is less equivalent capacitance than a single capacitor with capacitance 𝐶. Remember that the reason this occurs is because it effectively keeps the outermost plates (each storing +𝑄 and −𝑄) even farther apart because there are other capacitors in between them; this lowers the efficiency of the whole set.

A free proton is placed in a uniform electric field that points DOWN. Later, the same experiment is repeated with a free electron. Protons have a mass that is about 2000 times heavier than electrons. Which of the following statements is correct? Explain. a. BOTH the proton and electron will feel a force that points DOWN b. The PROTON will a feel a force that is about 2000 times stronger than the electron c. The PROTON will accelerate about 2000 times MORE quickly (and in the opposite direction) compared to the electron d. The ELECTRON will feel a force that points DOWN while accelerating UP e. The ELECTRON will accelerate about 2000 times MORE quickly (and in the opposite direction) compared to the proton

Let's examine each one of these statements and determine why they're all wrong except for E. a. BOTH the proton and electron will feel a force that points DOWN The proton will feel an electric force that points down, but the electron will feel an electric force that points up, in the opposite direction of the electric field lines. b. The PROTON will a feel a force that is about 2000 times stronger than the electron 8 Both objects have the same size charge, |𝑒|=1.609×10−19 C. Since the magnitude of the electric force they feel will be 𝐹𝐸=𝑞∙𝐸, there is no reason to think they will feel forces of different strengths. c. The PROTON will accelerate about 2000 times MORE quickly (and in the opposite direction) compared to the electron This is exactly backwards: the proton is much heavier, so even when they feel the same size force, the proton picks up speed much more slowly. This is predicted by Newton's 2nd Law, which says Σ𝐹 𝑚=𝑎 , the sum of all forces on an object causes that object's acceleration, but this is inversely related to the object's mass. d. The ELECTRON will feel a force that points DOWN while accelerating UP This doesn't make any sense. Since the sum of all forces on an object causes that object's acceleration (Σ𝐹 =𝑚⋅𝑎 ), why would it accelerate in a different direction than it's being pushed? e. The ELECTRON will accelerate about 2000 times MORE quickly (and in the opposite direction) compared to the proton This is the only one that is true. Newton's 2nd Law, which says Σ𝐹 𝑚=𝑎 , the sum of all forces on an object causes that object's acceleration, but this is inversely related to the object's mass. Since the electron has a tiny mass, it will feel a much greater acceleration from the same magnitude of force felt by the proton. Also, the direction of that force will be determined by 𝐹 𝐸=𝑞∙𝐸⃗ , and since 𝑞<0 for the electron, the direction of the force is reversed from that of the proton's.

A proton and an electron are placed at Reston electric field. The proton moves towards higher OR lower electric potential. The electron moves towards higher OR lower electric potential.

Lower Higher

A proton and an electron are placed at rest and an electric field. The proton moves so that it has a higher OR lower electric potential energy. The electron moves so that it has a higher OR lower electric potential energy.

Lower Lower

Two concentric wire loops carry currents I1 and I2 in opposite directions as shown in the diagram. The wire loops have radii r1 and r2, respectively, where r1>r2. What is the direction of the total magnetic field at the center of the two loops?

More info needed It's impossible to know. They both create magnetic fields that point either into the page (as RHR2 predicts for Wire 1) or out of the page (as RHR2 predicts for Wire 2), and the total magnetic field will be one of those directions, but without knowing the magnitude of each field contribution, we can't know which direction will win. Now, you might recognize that the center of the loops is closer to Wire 2, so maybe the field created by Wire 2 is a little bit stronger at the center of the loop. But what if I1>I2 (i.e., what if the current in Wire 1 is just so much stronger than the current in Wire 2 that the field it creates vastly overpowers the field created by Wire 2)? We can't know, because we don't know how the currents compare to each other.

A system of charges is labeled below. The values are given in Coulombs. Which of the following arrows shows the net force on charge 𝐴? (pic)

None of these -Charge A feels a force directly towards the right (towards Charge B), and a second force down and to the right due to Charge C. The superposition/sum of these two forces (as vectors) will point mostly towards the right, and a little bit down. None of the vectors shown accurately shows the net force on Charge A, since none of them points down and to the right, so the answer is E.

An electric potential map is shown (in Volts) with several regions labeled. (pic) A helium nucleus He2+ is released from rest in Region B. If we want to find this object later, we should look for it:

North of Region B A positive charge loses potential energy by falling toward lower electric potential. The lower potential in this picture is the bluer parts, and the electric field in Region B is pointing almost due North (toward lower potential; straight downhill). We should expect the charge to feel a Coulomb force pointing North, and if we want to find it later, we should look in that direction. Of course, there is also low potential at the bottom of the map, but a positive charge would have to first climb uphill toward Region A to make its way back down toward the bottom of the map, and it won't have the energy to do that. When released from rest, it has no kinetic energy to exchange for the necessary potential energy it would gain by moving uphill toward Region A; it lacks the funding necessary to do this, so that's not an option.

A proton is INITIALLY MOVING RIGHT in a uniform magnetic field pointing to the RIGHT as seen in the figure below. There are no other charges present. The proton is now given a DOWNWARD velocity. What is the direction of the magnetic force on the proton now?

Out of Page Point your arm downward to show the velocity of the proton ("Where arm you going?" Get it? It's funny. Laugh. Fine, it wasn't that funny. But if it helps you remember it, then I'm happy.) Now, your fingers should bend to point towards the magnetic field (right). Your thumb can only point out of the screen/page.

Two concentric wire loops carry currents I1 and I2 in opposite directions as shown in the diagram. The wire loops have radii r1 and r2, respectively, where r1>r2. 1. Suppose the currents are adjusted so that I1=I2. What is the direction of the total magnetic field at the center of the two loops now?

Out of the Page If the currents are equal, then just defer to Wire 2, because it's closer to the center than Wire 1, and the field that it creates will overpower the field created by Wire 1 with superior strength due to proximity.

A circuit is shown with multiple batteries and multiple resistors and a single capacitor on the far left side between points k and h. The potential differences of the three batteries are V1=6V, V2=9.3V, and V3=3V. Assume that the circuit has been connected and running for a VERY LONG TIME already. Write the appropriate Kirchhoff's equation for the middle loop, bcfgb. Write the appropriate Kirchhoff's Rule equation for the junction c.

Pic

When connected to a certain battery, a single capacitor with capacitance 𝐶 stores charge 𝑄. When connected in series with an identical capacitor to that same battery, what is the TOTAL amount of charge stored by both capacitors (i.e., the charge of the first capacitor plus the charge of the second capacitor)? --||--||-- C C a. 𝑄 b. 2𝑄 c. 𝑄/2 d. 2/𝑄 e. 1/𝑄 f. This question cannot be answered without the battery voltage

Q If the single capacitor stores charge 𝑄 when the battery is connected, then (assuming nothing else is in the circuit other than a battery), the battery voltage Δ𝑉𝐵 must be the same as the capacitor voltage Δ𝑉𝐶, so Δ𝑉𝐵=Δ𝑉𝐶. When a second capacitor is connected in series, the overall equivalent capacitance is 𝐶𝑒𝑞𝑢𝑖𝑣=11𝐶+1𝐶=12𝐶=𝐶2, and this equivalent capacitor would store an amount of charge equal to 𝑄𝑒𝑞𝑢𝑖𝑣=𝐶𝑒𝑞𝑢𝑖𝑣∙Δ𝑉𝑒𝑞𝑢𝑖𝑣. But since it's an equivalent capacitor representing all of the capacitors in this circuit, the potential difference across it must be the potential difference across the battery too, so Δ𝑉𝑒𝑞𝑢𝑖𝑣=Δ𝑉𝐵. Therefore, the charge stored by the equivalent capacitor is 𝑄𝑒𝑞𝑢𝑖𝑣=𝐶𝑒𝑞𝑢𝑖𝑣∙Δ𝑉𝐵=(𝐶2)∙Δ𝑉𝐵=12𝐶Δ𝑉𝐵=12𝑄. Series equivalent capacitors store the same amount of charge as each of their individual constituent capacitors, so that means each individual capacitor also stores 𝑄2. Since there are two of them, the total amount of charge stored is 𝑄2+𝑄2=𝑄. This is not a surprise: these series capacitors just divide up an amount of charge that only one of them would've stored and then force each other to share. They don't let you store extra charge, because they just borrow from each other.

Two charges are shown below in a field line diagram picture. What are the signs of QA and QB

QA is greater than 0. QB is less than zero

An electric potential map is shown (in Volts) with several regions labeled. (pic) In which of the labeled regions does the electric field appear to be the strongest in magnitude?

Region D The strongest magnitude of electric field would occur in regions where the equipotential lines are closest together (because electric field strength measures the "steepness" of the electric potential's decline; the quicker the drop in electric potential from one meter to the next, the stronger the electric field points in that region and in that direction. The location where the equipotential curves are closest together appears to be Region D. Everywhere else, the equipotential curves are farther apart.

The circuit shown contains a battery and a switch. Suppose the switch is closed. The circuit now creates a magnetic field inside its cardboard tube. In which direction does this magnetic field point? (pic)

Right The second right hand rule is very versatile. You can use your fingers to represent the current around a loop/coil of wire, and your thumb will show you the magnetic field inside it. But you can also use your thumb to represent current through a straight wire, your fingers will curl around to show you the magnetic field surrounding it. You really can't go wrong - curl your fingers towards something that can curl in the context of whatever problem you're doing, and your thumb (by being straight) will point towards the other thing. If the current can curl (because it's moving through a circular wire), then curl your fingers to make the current. If the magnetic field can curl (because it encircles a long straight wire - not a wire loop), then curl your fingers to make the magnetic field. Whichever quantity can curl in a given question (B or I), curl your fingers towards that, and your thumb will automatically show you the other one that doesn't curl. It's really a win-win no matter the situation. In our case, we have to curl our fingers around the coil of wire to represent its current, and point our thumb to see the magnetic field inside it. Conventional current flows from the positive terminal of the battery (on the right side underneath the circuit) to the negative terminal of the battery. That makes it look like our fingers will be following the current in the coil if we curl them up and behind, and out of the screen/over the top of it. The thumb can only point right in this scenario.

Jessica wears rough-textured leggings that are made mostly of polyester and cotton. When she gets off of the sofa and touches the doorknob, she experiences a slightly painful shock. This minor electrocution is most likely explained by what? Explain. She had a net _________ charge before touching the _______ doorknob, and electrons left the metal and were deposited onto her body. Was it protons or electrons?

She had a net POSITIVE charge before touching the NEUTRAL doorknob, and electrons left the metal and were deposited onto her body. -She has a positive charge because her pants have been scraping on the couch while she was seated. The movement strips off electrons from her body and causes her to have a net positive charge when she gets up off of the couch. Because the doorknob is neutral (and conductive), it can freely donate some electrons back to her body, and this causes a shock.

When connected to a battery, capacitors in series maintain equal _____?

Stored energy NOT voltage

A hollow metal sphere is electrically neutral (with no excess charge). A small amount of charge is then placed on the sphere at a point called P. If we check on the excess charge a few minutes later, what will happen?

The excess charge will evenly distribute itself over the outside surface of the sphere

Two long, straight wires cross each other at right angles, and each carries the same current as in the figure below. Which of the following statements is TRUE regarding the TOTAL magnetic field at each labeled point? a) The field is out of the page at point B and into the page at point D. b) The field strength has the same magnitude at all four points. c) The field is strongest at points A and C. d) The field is out of the page at point C. e) None of the above is correct

The field is out of the page at point B and into the page at point D. It's true that the field is out of the page at Point B and into the page at Point D. Those are the two labeled locations in which both wires create magnetic fields that actually help each other, instead of trying to cancel each other out in opposite directions. The RHR2 will show you this if you point your thumb towards the current (for either wire) and watch your fingers curl around it to illustrate the magnetic field

Suppose we can turn on our own electric external field in the laboratory in which this helium atom is present. Which of the following best represents an external electric field that would be used to strip off the right most electron towards the right. The electric field would be pointing in which direction

To the left

An electron (𝑒−) enters a vertically-oriented electric field (𝐸⃗ ). Which of the trajectories shown in the figure most closely matches the electron's motion after encountering the field? (pic)

Trajectory D Remember that electrons feel electric forces in the direction opposite electric field lines - in other words, they fall "uphill" (toward higher potential) in order to lose potential energy. The field lines point upward in this scenario, so the electron must feel a force downward. A downward force would cause its trajectory to bend downward, much like a ball being thrown under the influence of gravity.

When connected to a battery, capacitors in parallel maintain equal _____?

Voltage NOT stored energy

A proton is INITIALLY MOVING RIGHT in a uniform magnetic field pointing to the RIGHT as seen in the figure below. There are no other charges present. What is the direction of the magnetic force on the proton?

Zero The best place to hide a charged particle from a magnetic field is to aim the charge right at the field (or directly against it). In other words, if the angle between and for a charged particle is or , then necessarily.

Two charges ("C" and "D") are shown below in a 𝐸-field line diagram. (pic) What is the ratio of the magnitudes of the charges, |𝑞𝐶|/|𝑞𝐷| ?

|𝑞𝐶|/|𝑞𝐷| =3 There are 18 field lines emanating from Charge C, but only 6 terminating into Charge D. That indicates that Charge C is probably 3x as large in magnitude as Charge D.

Two charges ("C" and "D") are shown below in a 𝐸-field line diagram. (pic) What are the signs of 𝑞𝐶 and 𝑞𝐷?

𝑞𝐶>0, 𝑞𝐷<0 The 𝐸-field lines emanate from Charge C and terminate into Charge D. This indicates that Charge C is positive and Charge D is negative, because electric field lines always point in the direction that a positive charge (not pictured in the diagram) would feel a push/pull.