Physics II - Test 1

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A student is trying to calculate the quantity 1/((1/C1)+(1/C2)) on their calculator. Which of the following quantities is algebraically equivalent to this expression?

(C1∙C2)/(C1+C2)

How do capacitors store charge?

(pair of metal plates) secret ingredient that you have to have is a potential difference that can compel charges to go to a place they wouldn't otherwise go in order to establish this difference in potential across the plates you have to physically move charge off of one plate and onto another the way we do it is this "battery or some other voltage" but you have to have something something to provide a difference in potential so that electrons on one plate suddenly feel like they have a lot of potential energy to fall to the other side

(vector) E measures the__________________

(vector) E measures the rate at which the potential declines (i.e., minus the slope of V) ±Any location where the slope of the potential is very steep and changing rapidly, there will be a very strong electric field in that region.

Why should reducing the separation distance between the plates of a parallel plate capacitance increase its capacitance?

+Anything that makes electrons less likely to kick each other off the negative plate should increase the capacitance of this device +That includes bringing them closer to the positive charges on the opposite plate, which should increase the force of attraction between the plates.

What is needed in order to cause capacitors to store charge?

+Capacitors need a voltage source, such as a battery +They need something to cause charge to accumulate on opposing plates where it otherwise never would +The battery initially causes the plates to develop a difference in potential

Why should increasing the surface area of the parallel plate capacitor increase its capacitance?

+More surface area allows more room for more charges to join the plates without requiring a larger battery voltage to convince them +With the same density of charges on the surface, more area means more charge +This is effectively a more efficient capacitor

What is the fundamental purpose of capacitors?

+The fundamental purpose of capacitors is to store electric charge oTo "store charge" is a short way of saying to "store charge [of the same sign in a common location where it doesn't want to be]." +In effect, this does more than just "store charge," because you're also storing all the potential energy that those charges currently possess and would love to lose by being crammed in together. oThis makes a capacitor a device that stores charge - but in doing so, also a device that stores electric potential energy oYou can unleash this energy and put it to work by releasing the charges

Why do negative charges move from one plate onto another while a capacitor is in the process of storing charge?

+The positive terminal of the battery represents a high potential hill that electrons climb until it's no longer preferable to leave their plate. +Compared to the negative terminal of the battery, the neutral plate also looks like a potential hill. +They move in each case because they suddenly feel that they have potential energy to lose by falling ("uphill," as electrons do).

Why is there a difference in electric potential between the plates Δ V_C even when the battery is removed?

+There is a difference in potential from one plate to the next because of the stored charged on those plates. oRemember: individual electric charges make their own potentials (i.e., their own hills/valleys) oWhen charged, the top plate begins to look like one giant potential hill, and the bottom plate begins to look like one deep potential valley +The battery causes the charges to separate, but their very presence/absence causes the plates to have a voltage between them.

Why are the charges of each capacitor plate always equal (but opposite)?

+This is just basic charge conservation +Every electron that leaves or joins a capacitor plate either leaves or negates one extra proton that makes up the atoms/ molecules of the metal +It's inevitable that the charges will be the same on each plate (but opposite signs, ±Q)

What would an electric field look like between neutron and proton

-A neutron sits next to a proton, but they don't really know how to talk to each other. L -The proton has an electric field, but because a neutron's charge is zero, it neither creates nor responds to an Electric Field. -No E-field lines enter into or leave from the neutron.

If more capacitors are added into the parellel capacitors, will the equivalent capacitance be greater, the same, or smaller than before?

-Adding more capacitors in parallel causes the total to look like: C_equiv^parallel=C_1+C_2+C_3+... -Mathematically: It just adds more terms; no additional term will ever make C_equiv smaller. -Physically: It effectively increases the surface area of capacitor plates Adding capacitors in parallel only increases C_equiv

If more capacitors are added into the series capacitors, will the equivalent capacitance be greater, the same, or smaller than before?

-Adding more capacitors in series causes the total to look like: C_equiv^series=1/(1/C_1 +1/C_2 +1/C_3 +...) -Mathematically: Adding more terms just makes the denominator bigger -Physically: Only the outermost plates are storing any useable charge (+Q and -Q) in this scenario. -All a series does is effectively keep those outer plates far apart, thereby reducing their equivalent capacitance. -Adding capacitors in series only decreases C_equiv

_________________capacitors have the same drop in _______________ across their plates as each other (e.g., ΔV_(C_1 )=ΔV_(C_5 ), both having the same voltage of red to yellow.) -But additionally, they also have the same voltage as their ____________ equivalent (i.e., ΔV_(C_15 )=ΔV_(C_1 )=ΔV_(C_5 ))

-Parallel capacitors have the same drop in potential across their plates as each other (e.g., ΔV_(C_1 )=ΔV_(C_5 ), both having the same voltage of red to yellow.) -But additionally, they also have the same voltage as their parallel equivalent (i.e., ΔV_(C_15 )=ΔV_(C_1 )=ΔV_(C_5 ))

What is unique about series capacitors

-SAME CHARGE -NOT guaranteed to have the same voltage as each other electrons on the left plate of c1 have to eventually wind up on the right plate of c2 and the same is true for charge on the left plate of c2 it has to end up on the right plate of c1 in other words if you rip charge off of one of these capacitors there's nowhere to put it except on its next door neighbor. --because these two capacitors cannot take charge from anywhere except their next door neighbor they must store the same amount of charge as each other it's a guarantee they will always store equal charge

What is the charge on C_equiv?

-The charge on C_equiv is that of ANY capacitor: Q=C⋅ΔV_C -This is just due to the definition of capacitance, as we made it up, and every capacitor (actual or equivalent) obeys this relationship by definition: C≡Q/(ΔV_C ) (series and parellel have individual)

-The potential decreases toward the center of the circle, so the electric field vectors must point _________, perpendicular to the _________. -The equipotential curves are far apart at the outer edges, but they're much closer toward the center -This means there's a ________ closer toward the center, and a ________ field farther away. -This is exactly the type of equipotential map and electric field vectors you'd see from a _________

-The potential decreases toward the center of the circle, so the electric field vectors must point inward, perpendicular to the equipotential curves -The equipotential curves are far apart at the outer edges, but they're much closer toward the center -This means there's a strong electric field closer toward the center, and a weaker field farther away. -This is exactly the type of equipotential map and electric field vectors you'd see from a single negative charge (e.g., electron)

**Now that there exists a leftward, uniform, external electric field in this region, in which of the following regions could a third charge be placed and remain motionless upon release? Use your answer to the previous question in order to help you decide.

-To the left of the sulfate ion (YES) -Between the sulfate ion and the hydrogen ion (NO) -To the right of the hydrogen ion (YES) It's possible in that left region, but it has to be perfect: you have to put a charge on the horizontal axis exactly where that hump in potential is located, and you can't nudge it at all or else it'll fall left (due to the external field) or right (due to the sulfate ion). It's not a stable equilibrium, but there is a single location on the left side where equilibrium is theoretically possible, although it's basically completely unmanageable in practice. A stable resting point is visible on the right side of the ions too. We saw it before, but it's even more obvious in this new graph. In both cases, the equilibrium point is defined by its flat potential, a location with a completely horizontal slope. Where the total electric potential isn't changing, even if only at one point, the total electric field must be zero.

What is the SI (metric) unit for measuring amounts of electric charge?

1 Coulomb

current in a straight wire, where the current is going to the right. Magnetic field above and below the wire is pointing?

2nd RHR if you are above the wire you should see a magnetic field that points OUT of the screen, if you're below the wire you should see a magnetic field that points INTO the screen

**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|/𝑑𝑛𝑒𝑤^2 =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.

**In which of the following regions could a third charge be placed and remain motionless upon release? (SO4^2-)---(4E-9m)---(H+)---(⋅) where: <---(𝐸⃗SO4^2-)--(⋅)--(𝐸⃗H+)--->

A new charge can remain at rest somewhere in this region: -To the left of the sulfate ion (NO) -Between the sulfate ion and the hydrogen ion (NO) -To the right of the hydrogen ion (YES) This can be deduced in a couple of ways, and I want you to appreciate that both result in the same answer. From the perspective of "Forces and Fields," it's pretty reasonable to conclude that nowhere is safe for a new, third charge to rest except somewhere to the right of both ions. The reason this makes sense is because, in that far right region, the sulfate ion's large charge but relatively far distance will make an electric field vector that points to the left, and at least somewhere, it'll be exactly balanced out an electric field vector that points to the right created by the proton's small charge but relatively close distance. These two competing influences (size of the charge vs. proximity to the field point) mean that a field point exists far off to the right that will cause the exact cancellation of both 𝐸-field contributions. For a diagram, see below. Note that this is impossible at any location between the two charges, because at any field point you choose in that region, both charges' 𝐸-field contributions will be pointing to the left, so cancellation will never happen. And it's also impossible to the left of the two charges, because in that region, the sulfate ion is both close and possesses a big charge relative to the proton, which is both far away and small by contrast. It would never be a fair contest to the left; the sulfate ion has two advantages when creating its electric field in that region. Only to the far right is there a point where charge can balance proximity and produce equally opposing electric field vectors. At that one location in space where the electric field exactly cancels to zero, no Coulomb force would be felt by any new third charge placed there. Alternatively, from the perspective of "Potentials and Energies," we can actually zoom in on that 𝑉 vs. 𝑥 graph from the previous question and visually locate the one stable position that a new charge could rest (marked in red on the graph; see below). You can't really see it clearly without zooming in, but in the figure on the right, you can easily locate the one place on the horizontal axis where the total electric potential has a flat slope; it looks like the bottom of a valley, as if a marble rolled around in that region would naturally come to rest at that location. As we've already discussed, this location is the exact same place mentioned above in the "Forces and Fields" discussion; we're just seeing it from the perspective of electric potential now. Wherever the electric potential is sloped, a non-zero electric field exists, pushing any nearby charges. But in this one place marked in red, the potential is briefly flat, which means the electric field is zero there. By looking at the graph of 𝑉 vs. 𝑥 in the space surrounding these two ions, it's visually apparent that no other location could sustain a resting charge. Anywhere else results in a slope, which means an electric field, which means a Coulomb force on a new charge if placed there. If you think about a marble trying to find rest on that graph, you can see no other safe place to leave it that it wouldn't "roll down" (or "up" if "negative" marbles exist?) or "roll away."

**A helium nucleus He2+ is released from rest in Region B. If we want to find this object later, we should look for it where?

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.

Which of the following electric charges naturally fall to reduce their electric potential energy? Positive Charges Negative Charges

All electric charges (much like objects experiencing gravity), naturally move to regions where their electric potential energy

**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?

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.

The currents in these two wires are the same, but in opposite directions. At point b, the total magnetic field points? a (.02m) ------> (current/wire I1) b (0.04m) <------ (current/wire I2) c (0.02m)

B_total=B_1+B_2 I1: into screen I2: into screen b:into screen

The currents in these two wires are the same, but in opposite directions. At point c, the total magnetic field points? a (.02m) ------> (current/wire I1) b (0.04m) <------ (current/wire I2) c (0.02m)

B_total=B_1+B_2 I1: into screen(XX) I2: out of screen c: out of screen

The currents in these two wires are the same, but in opposite directions. At point a, the total magnetic field points? a (.02m) ------> (current/wire I1) b (0.04m) <------ (current/wire I2) c (0.02m)

B_total=B_1+B_2 I1: out of screen I2: into screen a:out of screen

What is true regarding this circuit? 17μF and 31μF parellel capacitors

Both capacitors have identical voltages

What is true regarding this circuit? 17μF and 31μF series capacitors

Both capacitors store identical charge

What is equation of capacitance and why is it defined like that? Why is this quantity more useful than simply writing the charge Q a device can store?

C ≡ Q /Δ V(c) •Because capacitors don't simply store one amount of charge. •They're not like buckets that have a finite size; they can accommodate more charge if you're willing to work harder to push it onto the plates. •Capacitance is defined as a ratio of charge per volt, because the larger the difference in potential you place across the plates, the more charge you can compel to separate (and therefore store). •Why are wages paid in dollars per hour instead of a fixed amount of dollars? Because if you work longer hours, you earn more dollars.

What is capacitance? Equation?

C ≡ Q /Δ V(c) -define c to be the amount of charge you store divided by the difference in potential needed to keep them apart -Capacitance C measures how much charge Q a device can separate (and therefore store) per volt of potential difference (Δ V_c) from one side of the device to the other (needed to keep the opposite-sign charges separated). How to store electric charge -it measures how much charge a device can separate and when i say separate i mean store (per volt) because remember pulling them apart against their will positives and negatives is effectively storing them

Equation to calculate its capacitance from purely geometric features

C_plates= E0 (A/d) if you know the area (in m^2) of the 2 parellel plates and you know how far apart (in m) they are and you know one other constant called vacuum permittivity a constant that's used to determine how easily electric field lines can be set up in a vacuum then you can already determine what the capacitance of those plates will be by the way the vacuum permittivity (E0=8.85E-12 F/m) sometimes called epsilon zero or epsilon naught (constant)

capacitor holds 7.5 µC (7.5times10^{-6} C) of charge when the potential difference across its plates is 0.9 V. How much charge would it hold with 1.5 V?

Capacitors Don't Just Store One Amount of Charge +This capacitor stores 7.5 µC of charge when there is a voltage of 0.9 V across its plates. +This means its capacitance is C=Q/(ΔV_C )=(7.5 µC)/(0.9 V)=8.3×〖10〗^(-6) F=8.3 µF +If we increased the voltage across the plates to 1.5 V, we'd compel even more charge to join them We would then store Q=C⋅ΔV_C=(8.3 µF)⋅(1.5 V)=1.25×〖10〗^(-6) =▭(12.5 µC)

**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? A(-1) B(+1) C(+1)

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 none of above.

The two capacitor plates differ in their electric potential by ~Ve = 0.8 V (from the positive plate to the negative plate). If the plates are separated by 6 mm (0.006 m), about how strong* is the electric field between the plates (in V/m)?

Electric field measures the steepness/slope of decline in electric potential, Ex = -(ΔV/ΔX).Since the potential between the plates drops by 0.8 V across a distance of 0.006 m, the electric field strength between the plates must be about IEI = I 0.8V / 0.0006m I ~ 133 v/m.

Electric field vectors always point _________________

Electric field vectors always point most efficiently toward lower potential

current in a straight wire, where the current is going down. --A---I---D-- ----B-I------- --C---I------- (I is current) (A-D are points) (true or false):The magnetic field at Point A is the same strength as the magnetic field at Point B.

FALSE

current in a straight wire, where the current is going down. --A---I---D-- ----B-I------- --C---I------- (I is current) (A-D are points) (true or false):The magnetic field vector at Point A points in the same direction as the magnetic field vector at Point D.

FALSE

Unit for capacitance

Farad (Coulombs per volt)

**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.

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

Heads up and then southeast) 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.

A solenoid is a wire that is tightly coiled into loops. In which direction will the magnetic field inside this solenoid point when viewed from the side. loops going slightly to left on top loops.

In front of solenoid: fingers: into screen thumb to left Behind solenoid: fingers out of screen Thumb to left

on two parallel capacitor plates, what would increase the amount of charge stored on the capacitor shown below?

Increase the surface area of the metal plates Increase the battery voltage

**This is kind of a random, throwaway question that isn't really related, but I've always wanted to ask it anyway. Suppose a uniform, external electric field (pointing to the left) is turned on in the presence of these ions. Which of the following graphs roughly illustrates the electric potential in the region along the 𝑥-axis?

It's exactly like the 𝑉 vs. 𝑥 graph from before, but now, there's a constant/steady downward slope leading to the left due to the external electric field. Once again, imagine the marble analogy, representing the behavior of a positive charge. The slope has to lead down to the left everywhere, because that's where a positive charge would feel a push; similarly, that's the direction that the total potential should decrease, because the total electric field always points in that direction. Anywhere far away from those two fixed ions will just look like a gradual/steady, downward slope to the left, dominated by the influence of that external 𝐸⃗ -field. Obviously, when you're really close to those ions, their influence on the slope of 𝑉 is much more apparent, as their field contributions tend to dominate very near themselves. But everywhere else? It's all sloping down and to the left. If the correct answer had been answer choice H, as it originally was before the external field was turned on, placing a marble (i.e., a positive charge) anywhere on the left side of both ions would eventually cause it to roll down into the well created by that sulfate. That's not the case anymore; now, in answer choice E, you have to be very close to the sulfate ion to fall into its potential well so that its influence overpowers the external field pushing positive charges back to the left. You can even see the tipping point in the left region, where, if you're close enough to the sulfate ion, a positive charge would get pulled toward it, but if you're any farther away to the left of that visible hump, the electric field pushes a positive charge away to the left forever. Notice also that the stable resting point we just calculated is now closer to the proton in the right region, and a lot more obvious. The proton would've otherwise pushed a positive charge away more readily, but now that an external electric field is here to help push a hypothetical positive charge back toward the left, it can hang out at rest at a location closer to the proton due to this extra help.

**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+1𝐶2+⋯=1/1𝐶+1𝐶=1/2/𝐶=𝐶/2. 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.

**Which actions would result in the storage of more charge with a parallel plate capacitor connected to a 3 V battery? What would not?

NOT: -Separate the metal plates farther apart YES: -Increase the area of the metal plates -Replace the 3 V battery with a 9 V battery 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.

On the left side, the potential is very shallow, so the field is ________ wherever equipotentials are __________.

On the left side, the potential is very shallow, so the field is weaker wherever equipotentials are farther apart.

A circular wire loop (o) carries current counterclockwise. Several points in space inside and outside the wire loop. What is the direction of the magnetic field vector inside and outside the loop.

Outside loop: Into the screen Inside: Out of the screen

Equivalent capacitance equation

Parallel capacitors Cequiv = C1 + C2 + C3 series capacitors 1/Cequiv = 1/C1 + 1/C2 + 1/C3

**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|---

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=1/1𝐶+1𝐶 =1/2/𝐶=𝐶/2, 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 =(𝐶/2)∙Δ𝑉𝐵 =(1/2)𝐶Δ𝑉𝐵 =(1/2)𝑄. 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.

What is the charge stored on the equivalent series capacitors, equation? Explain

Qequiv=Q1=Q2=... series capacitors store EQUAL amounts of charge as each other because they have no choice and their equivalent capacitor also stores that same amount of charge why would they do this? because adding more capacitors in series doesn't allow you to store extra charge, it just further subdivides the amount that was already being stored all you can do is steal from your neighbor adding more capacitors in series causes them to continue to steal from each other more and more

the current in the straight wire is moving up and to the right and the wire by having current has a magnetic field that encircles it. Where does the magnetic field point?

RHR 2 -If directly behind the wire the magnetic field appears to point up -if my fingers are directly over top of the wire above the wire they seem to be pointing kind of out -if my fingers are in front of the wire they're pointing down -continuous loops

What is unique about parallel capacitors

SAME VOLTAGE any single wire and it doesn't matter how long it is or how many ways or times it branches out it acts like an equipotential what i mean is anything touching the positive terminal of this battery any piece of this wire will be at nine volts, if the battery is a nine volt battery that means all of this is 9 volts let's suppose that the negative terminal of the battery is 0 volts then that means that anything touching the negative terminal of the battery is 0 volts. everything up to the front door of these capacitors is 9 volts and everything up to the back door of these capacitors is 0 volts --a necessary consequence of that is that when you trace your finger across either capacitor you must have encountered the same gap in electric potential from 9 volts to 0 volts from the highest peak to the bottom of the valley just like the battery and it didn't matter which capacitor i went across I encountered the same drop even if the capacitors have different capacitances even if one of them is more efficient than the other it is unavoidable that the drop in potential will be the same across both because of the way they're arranged in parallel

What are capacitor combinations?

Series Parellel

**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 which of the following?

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.)

Explain components of: Equation: B(wire)=(μ_0/2π)⋅(I/d)

Strength of a magnetic field surrounding a wire. I is the current in the wire d is the distance from the wire to where we measure the magnetic field (B). μ_0 is 4πx10^-7 T⋅m/A 'permeability of free space'

current in a straight wire, where the current is going down. --A---I---D-- ----B-I------- --C---I------- (I is current) (A-D are points) (true or false): The magnetic field vector at Point A points in the same direction as the magnetic field vector at Point C.

TRUE

That the equivalent capacitor of a parallel set stores _________ charges of its constituent capacitors

That the equivalent capacitor of a parallel set stores the sum of the charges of its constituent capacitors

That the equivalent capacitor of a series stores _________ charges as its constituent capacitors

That the equivalent capacitor of a series stores equal charge as its constituent capacitors

**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?

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. WRONG: 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.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? CORRECT: 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.

If the difference in potential across the metal plates of a capacitor is doubled, how much has the capacitance of the device changed?

The capacitance remains unchanged Changing the difference in potential across the capacitor plates allows you to store more or less charge with the same device. But it doesn't fundamentally change the efficiency of that device, which is what C ≡ Q /Δ V(c) measures. If you want to change that, you'll have to change the plate separation distance or the plates' surface area.

two capacitor plates, with a difference in potential of d Ve = 0.8 V, have a capacitance of C = 1400 μF (microFarad, 10- 6 F). How much charge is currently stored on the positive plate?

The definition of capacitance is C ≡ Q /Δ V(c) . It's kind of like an "efficiency," asking, "How well can you separate/store charge per volt required to coax them apart?" The charge stored on this capacitor is therefore: _J C · ΔVc = Q = (1400μF) · (0.8V) = 1120μC

**How much charge is stored on the first 8 μF capacitor

The equivalent capacitor for a series set stores the same amount of charge as EACH of its constituent series capacitors. Therefore, the charge stored by the equivalent capacitor is 𝑄𝑒𝑞𝑢𝑖𝑣=𝐶𝑒𝑞𝑢𝑖𝑣∙Δ𝑉𝐶𝑒𝑞𝑢𝑖𝑣. The voltage across the equivalent capacitor is the same as that of the battery, because it would be the only thing connected to the battery's two terminals anyway. So, 𝑄𝑒𝑞𝑢𝑖𝑣=𝐶𝑒𝑞𝑢𝑖𝑣∙Δ𝑉𝑐𝑒𝑞𝑢𝑖𝑣=(2.7 μF)∙(9 V)=(2.7×10−6CV)∙(9 V)=24×10−6 C=24 μC. Therefore, since the equivalent capacitor replacing a series stores the same amount of charge as each of its individual capacitors, the first 8 μF capacitor will store the same amount of charge as the amount we just calculated: 24 microCoulombs. Note that because the other two capacitors (the other 8 μF capacitor and the equivalent capacitor representing the parallel set, 𝐶6,2) are also in series, they'll also store the same amount of charge, because series capacitors always store the same amount of charge as each other. So, in total, this entire circuit stores 24 μC+24 μC+24 μC=72 μC of charge. Another fun fact: because series capacitors are a stupid arrangement for trying to store more charge (although they're great at reducing equivalent capacitance if that's what you want), notice that this is also exactly how much charge you'd store if you had only one single 8 μF capacitor instead of three series 8 μF capacitors: If the entire circuit had just been one single 8 μF capacitor, it would've had a potential difference across it equal to that of the battery, so it would've stored 𝑄8=𝐶8∙Δ𝑉8=(8 μF)∙(9 V)=72 μC. That's exactly how much all three series caps store combined, and that's because, ultimately, all they did was split this amount of charge up among each other by stealing from their neighbor.

**Two ions, H+ and SO4^2−, are shown on the 𝑥-axis with the sulfate ion situated at the origin. The separation distance between the two objects is about 4×10−9 m. (SO4^2-)---(4E-9m)---(H+) If the H+ ion is released from its current position and allowed to move freely, what will be its change in potential energy Δ𝑃𝐸 when it is at a distance of 2×10−9 m from the sulfate ion (in J)? Assume the sulfate ion remains at a fixed position.

The proton (which is all an H+ ion really is anyway) possesses a potential energy 𝑃𝐸=𝑞⋅𝑉 any time it's placed at a location with an electric potential 𝑉. In this case, it's been placed in the vicinity of a negative charge, which creates a potential well. The proton will "fall into this well," in the sense that it will lose potential energy by falling closer to the sulfate ion, gaining kinetic energy as it falls. How much kinetic energy it gains is exactly equal to the amount of potential energy it loses while falling. We can calculate this change because we happen to know what the electric potential created by a single, point-like electric charge looks like anywhere in space. The sulfate ion creates an electric potential that looks like 𝑉𝑐ℎ𝑎𝑟𝑔𝑒=𝑘𝐸⋅𝑞/𝑑, and it's the potential on which the proton sits (and will "roll," like a marble, downhill). Therefore, the proton loses an amount of potential energy equal to the following: Δ𝑃𝐸=𝑃𝐸𝑓−𝑃𝐸𝑖 =(𝑞𝐻+⋅𝑉𝑆𝑂4^2−𝑓)−(𝑞𝐻+⋅𝑉𝑆𝑂4^2−𝑖) =(𝑞𝐻+⋅𝑘𝐸𝑞𝑆𝑂4^2−/𝑑𝑓)−(𝑞𝐻+⋅𝑘𝐸𝑞𝑆𝑂4^2−/𝑑𝑖) =((1.6×10−19 C)⋅(8.99×109 N⋅m2/C2) (−2)⋅(1.6×10^−19 C)/(2×10−9 m)) =((1.6×10−19 C)⋅(8.99×109 N⋅m2C2)(−2)⋅(1.6×10−19 C)/(4×10−9 m)) =−1.151×10−19 N⋅m,or −1.151×10−19 J This is a negative change in potential energy. Apparently, the proton lost potential energy by moving closer toward the sulfate ion, which means, in the absence of any other forces, the proton had to have gained kinetic energy as it sped up and fell closer. This is totally expected because positive charges always lose potential energy by falling toward lower potential whenever possible. The proton had negative potential energy at the beginning, and then it had even more negative potential energy by the end, which means a net loss in potential energy and a net gain in kinetic energy. This is exactly why chemical bonds occur: it is energetically favorable for objects to lose electric potential energy by approaching oppositely charged neighbors.

**Which of the following graphs roughly illustrates the electric potential (on the vertical axis, position on the horizontal axis) in the region along the 𝑥-axis surrounding these two electric charges?

The proton makes an infinitely tall electric potential hill, and the sulfate makes an infinitely deep potential well. The sulfate ion's well should be larger/steeper at the same distance away due to its larger charge.

The strength of the Coulomb force exerted by these two objects on each other looks like: 𝐹𝐸=𝑘𝐸⋅(|𝑞𝐻+||𝑞𝑆𝑂4^2−|) /𝑑2 =(8.99×10^9 N∙m2/C2)⋅((|1.6×10^−19 C|)(|−2⋅(1.6×10−19 C)|))/(4×10−9 m)^2 =2.9×10−11 N ,or 29 pN

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

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 most fundamental purpose of a capacitor is:

To separate and store electric charge to be released when needed

If the difference in potential across the metal plates of a capacitor is doubled, how much charge does it now store?

Twice as much charge If you increase the gap in potential from the positive plate to the negative plate, you will have given some charges potential energy they didn't previously have, and then more of those charges will separate and rush onto the opposing plate. In other words: the reason capacitance is given in a ratio is because the charge stored is proportional to how hard you work to separate it: C · ΔVc = Q. Unlike a bucket of water, which has a finite size and can't expand, capacitance is more like an air tank: if you're willing to work harder to pump air in, you can simply store more. Likewise, if you're willing to use a bigger battery to compel charges to separate, you can compel more of them to do so and store more charge on the plates.

**Calculate the horizontal position along the 𝑥-axis that a single electric charge could be placed and remain at rest. (SO4^2-)---(4E-9m)---(H+)---(⋅) where: <---(𝐸⃗SO4^2-)--(⋅)--(𝐸⃗H+)--->

Without employing techniques from calculus, which PHY 2054 doesn't permit, the Forces and Fields strategy is our only hope of identifying this location. We don't know where this location is on the 𝑥-axis, so I'm going to call its distance (from the origin) simply "𝑥" until we solve for it algebraically. If the total electric field must cancel at this location 𝑥, then we should be able to write the total electric field at this location and manually require that the vector sum equal zero. The signs will be prescribed on the basis of the direction of each vector arrow (leftward pointing vectors will be negative and rightward pointing vectors will be positive). 𝐸𝑡𝑜𝑡𝑎𝑙 = −𝐸𝑆𝑂4 2− + 𝐸𝐻+ ≡ 0 Now we just have to fill in what each term looks like. In general, the electric field created by any single charge 𝑞 (at a field point a distance 𝑑 away from that charge) looks like 𝐸𝑐ℎ𝑎𝑟𝑔𝑒=𝑘𝐸|𝑞|𝑑2. So in our case, that means.... (ALOT)

the electric field between two charged plates in capacitor is ___________

almost perfectly uniform they're almost the same strength everywhere they point in almost the same direction

What is a series capacitor

and series is defined as whenever charge has to flow through both capacitors before making it back to the battery (cannot trace your finger through one capacitor without tracing your finger through the next one.

based on the current going clockwise in this picture it looks like everywhere inside the loop no matter which part of the wire you pick to examine the current you're going to find that your fingers point _____________ inside the loop and outside the loop no matter which part of the wire loop you choose to put your thumb at you're going to find the magnetic field seems to point ___________ outside the loop and it turns out this is true for any wire loop

based on the current going clockwise in this picture it looks like everywhere inside the loop no matter which part of the wire you pick to examine the current you're going to find that your fingers point INTO the screen inside the loop and outside the loop no matter which part of the wire loop you choose to put your thumb at you're going to find the magnetic field seems to point OUT of the screen outside the loop and it turns out this is true for any wire loop

______________ guarantees that it will always be the same amount so if you have capacitors in ____________ you automatically know that the charge stored on each of them is equal

charge conservation guarantees that it will always be the same amount so if you have capacitors in series you automatically know that the charge stored on each of them is equal

an individual electric charge it moves say to the left and you can predict what the magnetic field is going to look like in the neighborhood of this moving charge and the direction that the field is oriented depends on?

depends entirely on the sign of the charge and which way it's moving

current in a straight wire, where the current is going to the right. Magnetic field directly above the wire is pointing?

directly above the wire my fingers are pointing out of screen 2RHR

Why is it fiction when talking about charge making a trip around a circuit and coming back to the battery as if it's a complete loop that all charge marches all the way around going through a wire through a capacitor

electrons may actually accumulate on the positive terminal the battery they may leave the negative terminal of the battery many or all or some may never make it all the way around the circuit if you keep your finger on one individual electron you may never see it go all the way around but i'm going to talk about them all collectively as if they go around the entire circuit like their cars on a race track that come back to the starting line in fact they won't really do that necessarily but it's convenient

current in a straight wire, where the current is going down. --A---I---D-- ----B-I------- --C---I------- (I is current) (A-D are points) note that Point A is twice as far from the wire as Point B. Therefore, the magnetic field at Point A is _______ as strong as the magnetic field at Point B.

half

First right hand rule is for?

helping you discover what a Lorentz force looks like when a charge gets pushed by a magnetic field

A negative charge, which is free to move, is released from rest in an electric field. Neglect non-electrical forces. It will always move to a position with _______

higher potential.

second Right hand rule does NOT show you? Explain

how strong the field is around the wire or what it depends on Equation: B(wire)=(μ_0/2π)⋅(I/d)

What is high and low capacitance

if you are separating a lot of charge and you're doing so with a small difference of potential holding them apart you probably have a high capacitance the numerator is large the denominator is small. If you think in terms of efficiency you're holding a lot of charge apart with a very small voltage needed to separate them. we would call that a high capacitance (large Q, small ΔVc) by contrast if you have a very small amount of charge and you're working really hard to keep them apart, it's not very efficient and we would call that a low capacitance (small Q, Large ΔVc) -SOMETIMES, we want low capacitance

delta V_c means

little subscript c on this quantity is how i indicate to you that this voltage refers to the difference in potential from one side of this device (capacitor) that stores charge aside holds the positive charge compared to the side that holds the negative charge C ≡ Q /Δ V(c)

lorentz forces are magnetic forces where magnetic fields push electric charges at ________ to where the charges are already going well it turns out that moving charges also make their own magnetic fields at __________ to where they're already moving

lorentz forces are magnetic forces where magnetic fields push electric charges at right angles to where the charges are already going well it turns out that moving charges also make their own magnetic fields also at right angles to where they're already moving

magnetic fields seem to be oriented around charges and currents when they move almost like _______ remember that magnetic fields only make _________ loops, there's no such thing as magnetic monopoles

magnetic fields seem to be oriented around charges and currents when they move almost like rings remember that magnetic fields only make continuous loops, there's no such thing as magnetic monopoles

Define conductor

material that allows charge to move around freely you can put excess charge on it you can take excess charge off conductors do that once any excess charge has sort of settled down once it gets where it's going

moving charges make their own ______1_____, we actually didn't know that wires that ______2______ or just moving charges in general made ____1_______

moving charges make their own magnetic fields we actually didn't know that wires that carry current or just moving charges in general made magnetic fields

current in a straight wire, where the current is going to the right. Magnetic field directly in front of the wire is pointing?

my fingers are pointing down 2RHR

What are the two ways that magnetic fields relate to electric charges

one is that moving charges make their own magnetic fields and two external magnetic fields those not created by the charge itself

What is a parallel capacitor

parallel capacitors are those that force charge to choose either go through one of the capacitors or the other and then come back to the battery but not both

What is the charge stored on the equivalent parellel capacitors, equation? Explain

parallel capacitors, their equivalent capacitor stores the same amount of charge as all of them put together (SUM) Qequiv=Q1+Q2+... capacitor of a parallel set is exactly the amount of charge stored by all of the parallel capacitors put together and the reason that happens is because they don't have to borrow charge from each other they can independently fill up so their equivalent capacitor accurately reflects this by showing that it stores more put more capacitors in parallel you can store more charge

current in a straight wire, where the current is going down. Magnetic field right and left of the wire is pointing? --A---I---D-- ----B-I------- --C---I------- (I is current) (A-D are points)

right: out of the screen left: into the screen

Second right hand rule is for? Explain components

second one has nothing to do with forces. All it does is tell you either which way the magnetic field points around moving charges or it could also tell you which way the charges are moving if you know about the magnetic field they create so it only has two components thumb: current (your thumb needs to point in the direction that the charges are moving wherever the charge is moving whatever the current is whichever way the charges are marching) magnetic field which is where your fingers bend (right angle) (L shape)

how we predict the direction of this magnetic field created by moving charges

second right hand rule

take a loop of wire (o), the current on the lower left of the loop wire. current is flowing up and to the left at that location. what does the magnetic field look like

second right hand rule -i'm going to take my thumb and point it in the direction of the current and i'm going to make sure my fingers are bent at a right angle so i can see that inside the loop my fingers are pointing into the screen. -if fingers are behind the wire loop they seem to be pointing down into the left -if i'm outside the wire loop still with my thumb pointing the direction of the current my fingers are pointing out of the screen

take a loop of wire (o), the current on the bottom right of the loop wire. current is flowing sort of down and to the left at that location. what does the magnetic field look like

second right hand rule pointing my thumb in the direction of the current, i can see that if i make a right angle with my fingers, inside the loop they point into the screen, behind the loop they seem to be pointing kind of down to the right and outside the loop they're pointing directly out of the screen

2 Right hand rules shows?

second right hand rule will show you which direction the magnetic field points around any current carrying wire and we know that it depends on which side of the wire you're looking because the magnetic field encircles the wire if you're in front of it or behind it or above it or below it

current in a straight wire, where the current is going to the right. Magnetic field behind the wire is pointing?

the current is going to the right, once again if you point your thumb towards the current wherever your fingers are located will show you the magnetic field in that region so where i am right here is behind the wire. my fingers are behind the wire so they seem to be pointing up that means the magnetic field points up behind the wire 2 RHR

Why is it fiction when talking about charge flowing through a capacitor and when you hear the word through it might lead you to believe the charge flows through the space between capacitor plates through the air or through a vacuum in between the plates

the fact is charge should never actually jump through the air between the plates of a capacitor that's called dielectric breakdown it's how you get a spark when you touch a doorknob it's how lightning causes a huge deposition of electric charge from cloud to ground charge should not be jumping through the air between capacitor plates dielectric breakdown should never be happening if charge jumps through the air between the capacitor plates that's a broken capacitor that's not what you want

the pattern that you find is the magnetic field everywhere inside the loop wire points ____________ and everywhere outside the loop points _____________

the pattern that you find is the magnetic field everywhere inside the loop points in ONE direction and everywhere outside the loop points *in the OPPOSITE direction* IT DOES NOT HAVE TO BE A CIRCLE

Why is it fiction when talking about charge as if it's going from high potential to low potential that would make sense if moving charges in a circuit were electrically positive

they're not by far most or all moving charges in any circuit are electrons although i'm going to talk about them as if they fall from high potential too low potential as if they go from a high electric elevation to a low electric elevation the truth is it's really backwards electrons fall from low potential to high potential and that's how they lose potential energy

current in a straight wire, where the current is going top right. Magnetic field on top/left and bottom/right is pointing?

top/left: out of screen bottom/right: into screen (XXX)

current in a straight wire, where the current is going to into the screen. Magnetic field on top, right, bottom, left is pointing?

top: down right: left bottom: pointing up left: pointing right (turning clockwise arround)

current in a straight wire, where the current is going left. Magnetic field on top and bottom is pointing?

top: into screen (XXXX) bottom: out of screen

current in a straight wire, where the current is going to out of the screen. Magnetic field on top, right, bottom, left is pointing?

turning counter clockwise

How to find out the strength of a magnetic field surrounding a wire? Explain? Equation?

two things: it depends on the current and the distance away from the wire that you measure the field, the closer you are to the wire, the stronger the magnetic field should be, if you go really far away from the wire you should measure very little magnetic field. If the current is really high the field around it is very strong, if the current is really low charges aren't moving very quickly to the wire, you should expect a very weak magnetic field Equation: B(wire)=(μ_0/2π)⋅(I/d)

**How do you find magnitude?

|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.

Now suppose we disconnect the battery from the plates. What value will the voltmeter read for ΔV_C (the potential difference across the capacitor plates)? separation 6mm plate area 200mm^2

ΔV_C at Electrostatic Equilibrium +Apparently, once charge has accumulated on the plates, ΔV_C does not magically become zero when you remove the battery entirely. +Even though the battery is not connected, the same difference in potential exists across the plates.

When we measure the potential difference across this battery, the voltmeter reads "1.5 V." Assuming no more charge will join the plates, what value will the voltmeter read for Δ V_C (the potential difference across the capacitor plates)?

ΔV_C at Electrostatic Equilibrium +With only a single capacitor connected to this battery, charge will flow and accumulate on the metal plates until ΔV_C=ΔV_Battery +At this moment, any further flow of charge appears unappealing because the potential of each plate is identical to the nearest terminal of the battery +With no difference in potential, there will be no potential energy to induce falling/movement. Electrostatic equilibrium happens when ΔV_C=ΔV_Battery

The currents in these two wires are the same, but in opposite directions. Calculate the magnitude of the total magnetic field at point a? a (.02m) ------> (current/wire I1) b (0.04m) <------ (current/wire I2) c (0.02m)

•Calculate the magnitude of the total magnetic field at each of the three points listed. üPoint a: B ⃗_total=B ⃗_1+B ⃗_2=B_1-B_2 =(μ_0 (10 A))/(2π)( 0.02 m) -(μ_0 (10 A))/(2π)( 0.06 m) =67 µT" (out)"

The currents in these two wires are the same, but in opposite directions. Calculate the magnitude of the total magnetic field at point b? a (.02m) ------> (current/wire I1) b (0.04m) <------ (current/wire I2) c (0.02m)

•Calculate the magnitude of the total magnetic field at each of the three points listed. üPoint b: B ⃗_total=B ⃗_1+B ⃗_2=-B_1-B_2 =-(μ_0 (10 A))/(2π)( 0.02 m) -(μ_0 (10 A))/(2π)( 0.02 m) =-200 µT" (in)"

The currents in these two wires are the same, but in opposite directions. Calculate the magnitude of the total magnetic field at point c? a (.02m) ------> (current/wire I1) b (0.04m) <------ (current/wire I2) c (0.02m)

•Calculate the magnitude of the total magnetic field at each of the three points listed. üPoint c: B ⃗_total=B ⃗_1+B ⃗_2=-B_1+B_2 =-(μ_0 (10 A))/(2π)( 0.06 m) +(μ_0 (10 A))/(2π)( 0.02 m) =67 µT" (out)"

Two ways to use 2nd RHR

•During the Pre-Lecture, we learned that we can use the RHR2 to see the magnetic field inside and outside a wire loop that carries current. •But it's kind of tedious. •You point your thumb with the current at various points around the wire, and then bend your fingers to see the magnetic field inside and outside the loop. •There is an easier way! RHR2 is flexible! •If the wire is not straight (i.e., if it bends), you can let your fingers be the current and let your thumb show you the B ⃗ field inside the loop.

Know that electric field measures ________________________

•Know that electric field measures the rate at which electric potential decreases per meter.

Know that electric fields only exist in regions where ____________________

•Know that electric fields only exist in regions where the electric potential changes

**Two charges ("C" and "D") are shown below in a 𝐸-field line diagram. 3. What are the signs of 𝑞𝐶 and 𝑞𝐷? (arrows leaving C and arrows going into D)

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.

electric field vectors symbol and they must?

(vector) E 1. always point in whichever direction a positive charge would feel a force 2. point toward lower electric potential in whichever direction potential decreases wherever it declines electric field vectors point that way 3. it only exists where potential is changing you won't find an electric field where the potential is flat you only find electric field in places where the potential is sloped 4. electric field vectors are always perpendicular to their nearest equipotential curve and that's because they show you the way downhill they show you the direction towards lower potential but they don't just show you that direction they show you the fastest direction downhill the way to get to lower potential in the most efficient route

following is equivalent to the expression, (ΔV/R_1 )⋅(R_2+R_3/2)?

(ΔV/R_1 )⋅(R_2+R_3/2) (ΔV/R_1 )⋅R_2+(ΔV/R_1 )⋅R_3/2 ((ΔV⋅R_2)/R_1 )+((ΔV⋅R_3)/(R_1⋅2))= ((ΔVR_2)/R_1+(ΔVR_3)/(2R_1 ))

how can you look at an electric potential and know where it's changing the fastest or slowest?

- fastest where the equipotential curves are closest together (remember equipotential curves show you the gap between one potential and the next) - - that is where you will always find the strongest electric fields

how do electric charges (influence) attract or repel each other?

- forces - the force that electric charges exert on each other we call it the coulomb force (sometimes called electric force) (f with a subscript e for electric)

What is the correct energy sequence of this ({PO}_4^{3-}) ion when released from rest at Point C (0V)? -100V -50V 0V(C) +50V +100V

- the charge falls to the right, speeding up along the way. -The charge starts with no kinetic energy, and no potential energy (PE=q⋅V) -As it "falls," its potential energy immediately turns negative to make room for additional kinetic energy. -Once again, its total energy (KE+PE) remains constant the entire time

What would an electric field look like between electron and proton with same number of electric field lines

-A proton sits next to an electron, and there's definitely a connection. -Field lines leave from the proton and aim outwards (positive charge) -Field lines aim into the electron (negative charge) -Equal number of lines coming out of proton as going into electron (same magnitude of charge for each) -Field lines are most dense right in between, because a positive charge feels the strongest pull

•What is the electric charge of each ion in Coulombs? H30+ --- 3 picometers ---- OH- (1e=1.602e-19)

-An H_3 O^+ ion lacks one extra electron that it should have, and so it has a +1e charge -The OH^- ion has one extra electron that it shouldn't, and so it has a -1e charge. -In terms of Coulombs, these will be q_1=+1.6×〖10〗^(-19) C and q_2=-1.6×〖10〗^(-19) C

Why must the E-field lines outside a conductor be perpendicular to the surface of the conductor? Why?

-Because if they didn't, they'd cross lines with a charge next door. -Since electric field lines can never cross each other, the only way to guarantee that is to point perpendicularly out from the surface of a conductor. Why? -They can't cross because an electric field line intends to show you the direction that a positive charge would feel a Coulomb Force at some location in space. -A charge can't feel the same electric force in two different directions. -The total electric field can only push it one way - not two different directions simultaneously.

Explain why excess charge always and immediately distributes itself on the outermost surface of a conductor, as evenly/far-apart as possible

-Because they're in a conductor, so they can move freely -And they hate each other because they're same-sign charges -So they Coulomb repel each other to be as far away as possible, which means the outside edge of the conductor

Suppose we place a phosphate ion ({PO}_4^{3-}) at 0V (at rest). Describe the motion of this charge. -100V -50V 0V(C) +50V +100V

-Negative charges lose potential energy by falling toward higher electric potential -The higher electric potential can be found on the right, so the charge falls in that direction, speeding up along the way.

Suppose we place a calcium ion (Ca^{2+}) at Point E (at rest). Describe the motion of this charge. -100V -50V 0V +50V +100V (E)

-Positive charges lose potential energy by falling toward lower electric potential -The lower electric potential can be found on the left, so the charge falls in that direction, speeding up along the way.

Explain how/why Coulomb's Law correctly describes the following prediction:If even just one of the two objects has no electric charge, then there can be no electric force between either of them.

-Suppose the 2nd charge has no charge, so q_2=0 C -Then the force between the two charges is F=k_E |q_1 ||q_2 |/d^2 →k_E |q_1 ||0|/d^2 =0 N -So there is no force on either charge--Both objects must have some electric charge to exert and feel Coulomb Forces due to each other.

How strong is a proton's Electric Field at distance 5x10^-11 (in Newtons/Coulomb)?

-The Electric Field created by a single charge q at a distance d away from the charge looks like: E=k_E |q|/d^2 -At that specific distance away from a proton, we have: E_proton=(8.99×〖10〗^9 N m^2/C^2 ) |〖1.062×10〗^(-19) C|/(5×〖10〗^(-11) m)^2 ≈(4×〖10〗^11 N/C) -That sounds like an enormous electric field.

What is coulomb's law

-The Electric Force between two charged particles -coulomb's law describes how strong electric forces between electric charges are F_(E)=K_(E) (|q_1| |q2|)/d^2 -F(E) is the strength of the electric force between two electric charges the charges we call q1 and q2 they're measured in coulombs each one q1 is measured in coulomb q2 is measured in coulombs the force overall measured in newtons depends on the product of those two charges how hard q1 pushes on q2

If the electric field is strong enough, what will happen to this compound? -------------> (cl-)(Na+) -------------> (electric field) ------------->

-The electric field is pulling these two ions in opposite directions. -If it pulls them harder than they can cling to each other, it will rip them into separate ions and break the ionic bond they share like an intrusive third wheel in a soap opera. -This is exactly what water molecules do to NaCl when dissolved.

A helium nucleus (He^(2+) ) moving to the right enters a uniform electric field pointing down. Where does it go?

-The electric force on this object is F _E =q∙E where q>0, so it feels a force in the direction of E . -It's moving to the right already, and it doesn't lose that motion. -It just gains downward velocity due to being pushed down by the E ⃗-field. (Southeast -mostly south)

A sulfate ion (SO_4^(2-) ) moving to the right enters a uniform electric field pointing left. Where does it go?

-The electric force on this object is F ⃗_E=-q∙E ⃗ where -q<0, so it feels a force opposite to E ⃗. -It's moving to the right and it also feels a force to the right -This just causes it to accelerate to faster and faster speeds while continuing rightward

An electron is placed at rest in a uniform electric field pointing to the right. Where does it go?

-The electric force on this object is F ⃗_E=-q∙E ⃗ where -q<0, so it feels a force opposite to E ⃗. -It's stationary at the beginning, so the only motion it has is what it gains from its interaction with the field. -It feels a force to the right.

A neutron is placed at rest in a uniform electric field pointing up. Where does it go?

-The electric force on this object is F ⃗_E=q∙E ⃗ where q=0, so it feels no force. -It's stationary, and since it can't feel electric forces, it's going to stay that way. -Nothing happens. Ever.

How much potential energy (in Joules) does the calcium ion (Ca^{2+}) have when placed at Point E? -100V -50V 0V +50V +100V (E)

-The electric potential energy possessed by any electric charge looks like PE=q⋅V -This ion has: q=+2⋅(1.602×〖10〗^(-19) C) -When placed at V=100 V: PE= +2⋅(1.602×〖10〗^(-19) C)⋅(100 V) =(3.204×〖10〗^(-17)J)

How much potential energy (in Joules) does the phosphate (PO4^-3) ion have when placed at Point C (0V)? -100V -50V 0V(C) +50V +100V

-The electric potential energy possessed by any electric charge looks like PE=q⋅V -This ion has: q=-3⋅(1.602×〖10〗^(-19) C) -While briefly passing through Point D, where V=+50 V: PE= -3⋅(1.602×〖10〗^(-19) C)⋅(50 V) =(-2.403×〖10〗^(-17) J)

If the potential decreases to the left at steady distance intervals, then?

-The potential decreases to the left at steady distance intervals -This means the potential has a constant slope as it declines -Therefore, the electric field points to the left everywhere and is of constant size

How to know which has more charge based on Electric Field Line Diagrams?

-There are more field lines passing through the same amount of area at the same distance away in the case of the second diagram -This means that it must have more charge than that of the first diagram -There are twice as many field lines passing through the same area at the same distance away from the charge in the second diagram -That doesn't mean that the second charge is "+2e" -It only tells us that the second charge is twice as big as the first - without telling us what those charges are. -Take your hand (or a finger, or anything of constant size): place it at any point on the diagram -Wherever more field lines are running through your hand (densely packed), you have stronger E-field there.

Identify the directions of all forces acting on the Cl- ion -------------> cl- F_Na+ -------------> (electric field) ------------->

-There are two forces acting on this ion. -It feels a Coulomb attraction to the Sodium ion to its right, but it also feels a Coulomb force pointing to the left due to the electric field. -The electric field provides a Coulomb force in the opposite direction as the field vectors because the charge is negative.

Calculate the strength of the Coulomb Force (in N) between these two ions H30+ --- 3 picometers ---- OH- (1e=1.602e-19)

-These two ions exert a Coulomb Force on each other. -According to Coulomb's Law, it looks like: F_E=k_E⋅(|q_1 |⋅|q_2 |)/d^2 =(8.99×〖10〗^9 N m^2/C^2 )⋅(|+1.6×〖10〗^(-19) C|⋅|-1.6×〖10〗^(-19) C|)/(3×〖10〗^(-12) m)^2 ≈(2.6×〖10〗^(-5) N) -This is the force of attraction they'll use to pull each other closer.

Explain how/why Coulomb's Law correctly describes the following prediction:The force that charge A applies to charge B should be the same strength as the force that charge B applies back on to charge A.

-This equation is the force BETWEEN the two charges; by its definition, it's the same force that they both exert on each other -The force that charge A exerts on charge B looks like F_AB=k_E |q_A ||q_B |/d^2 -But then the force that charge B exerts back on charge A is F_BA=k_E |q_B ||q_A |/d^2 -Notice that it's the exact same magnitude! -All we did was swap the labels "A" and "B" and it's still the same amount of force. Nothing changes, because it's just multiplication either way. -Remember that this is Newton's 3rd Law -- it was inevitable that it should work out this way

What are conductors?

-are materials that are very willing to allow the free flow of electric charge on their surfaces these are things like metals like iron copper aluminum -not every conductor is a metal but almost every metal is a conductor

What is the relationship between v and e

-electric potential and electric field -the electric field (vector E) is the negative gradient of potential (V) -forces and fields potentials and energies are inextricably linked because electric field measures the rate of descent of potential

What do equipotential curves show?

-equipotential curves are intended to show you exactly what they sound like equal potential all the way around -if you use the voltmeter and you dragged it along those curves you noticed that every point in space on one of those curves is at the exact same potential (like elevation) -electric potential is the electric analog of gravitational height in volts

What are the 3 rules to interpreting/drawing electric field lines?

-first rule: is that they always show you the direction a hypothetical positive charge would be pushed or pulled (imagine placing like a little test charge somewhere in the neighborhood of the visible charges and their field lines should show you where the test charge would be pushed if it's positive) -second rule of electric field lines is that the more densely packed they are in a certain region the stronger the field (they have no definite length so we had to come up with some other way to illustrate places where the field is strong) (whenever field lines are farther apart or whether field lines are non-existent the field is either very weak or in some cases it might even be zero) -third rule of drawing electric fields with field line diagrams is they can never cross each other there may be places where you don't see any electric field lines like right in between the two charges (it's not possible to have multiple electric field lines at one point so they can't cross each other)

What is charge conservation?

-it means that you can't create or destroy charge -electric charge is always carried by individual electric charge particles and they don't disappear therefore electric charge doesn't disappear you can move particles from one place to another you can take an electron and move it over here you can take a dust particle that has a lot of charge on it and move it over there and that's a way to move charge around but charge doesn't disappear -although they may move from one place to another onto one object off of another they will not vanish

What is the correct energy sequence of this Ca^2+ ion when released from rest at Point E. -100V -50V 0V +50V +100V (E)

-the charge falls to the left, speeding up along the way. -The charge starts with no kinetic energy, but lots of potential energy (PE=q⋅V) -As it "falls," it trades potential energy for kinetic energy. -Notice that the total energy (KE+PE) stays constant during the entire trip

What are the most basic carriers of electric charge

-the most basic carriers of electric charge in the universe are protons and electrons. -that a proton has plus one e (+1e) charge or an electron has minus one e (-1e) (These are just labels that we give to those charges that's not something measured in coulombs

electric field vectors measure what? How?

-they measure how steeply the potential is declining -the steeper the potential declines from one spot to the next the stronger the electric field points in that direction -the electric field in this region is strongest anywhere the potential is changing most rapidly (biggest arrows) -where the descent is quite shallow you find very weak electric field vectors so e doesn't just point toward lower potential and it doesn't just point the fastest way downhill it also measures how strongly the slope descends and the bigger that slope the steeper that descent the stronger the electric field is in that region

A negative test charge Q = -0.6 C was moved from point A to point B in a uniform electric field E = 5 N/C. The test charge is at rest before and after the move. The distance between A and B is 0.5 m and the line connecting A and B is perpendicular to the electric field. How much work was done by the net external force while moving the test charge from A to B?

0 J

Which of the following is equivalent to 1 Volt? 1 Joule/Coulomb 1 Joule⋅Coulomb 1 Newton/Coulomb 1 Coulomb/Joule

1 Joule/Coulomb

What are the features that are always present when putting excess charge conductors in electrostatic equilibrium?

1. If you put excess positive (or negative) charge on the conductor (any shape), the positives (+) evenly/as far away as possible, distribute themselves around the outermost surface 2. this causes all the electric field inside the conductor cancels everywhere you just saw that there will be zero electric field strength inside a conductor when it's in electrostatic equilibrium 3. Electric field lines point perpendicularly outwards in all directions from the conducting object

What is the most important things when it comes to electric field?

1. they are the tool that other electric charges use to push and pull on each other they are the answer to the question - how do electric charges push and pull on others at a distance? 2. electric charges can be pushed and pulled by electric fields 3. electric charges create their own electric fields to push and pull on each other

Electric field vectors always point in the following (self-consistent) ways:

1.In the direction a positive charge would be pushed or pulled 2.Toward lower potential 3.Perpendicular to equipotential curves ±They also show the steepness of the potential. ±In places the potential is changing fastest, and where equipotential curves are closest together, the E ⃗ vectors should be bigger.

Solve the equation 3a^2+6=18 for the symbol a?

3a^2+6=18 3a^2=18-6 =12 a^2= (12/3) =4 a=±√4=(± 2)

What is an electric field and is it a force?

An electric field despite the way Michael Faraday called it is not itself a force any more than a 'rope' or a 'pole' is a force. obviously a rope is not a force. it's electric fields that stretch out into space and touch other charges even at great distances and they ultimately provide the push or pull even though they themselves are not the force itself they cause the force at a distance

Any individual electric charge creates its own electric field __________. If a Coulomb Force is to be felt, _______.

Any individual electric charge creates its own electric field "even if no other charge is around" If a Coulomb Force is to be felt, "At least two charges must be present" How can you have a force when there's nothing else to push/pull against?

Electric charges clearly move on the metal wall as the balloon comes closer. Why does the excess charge on the balloon not move around on its surface as it approaches the wall?

Because those charges can't really move at all. The balloon is an insulator. Nothing happens to the electrons on the balloon. Why not? Because the balloon is made of an insulating material like rubber and/or latex. Insulators do not permit the free flow/transfer of charge (at least, not without a lot of effort/physical contact).

Explain how/why Coulomb's Law correctly describes the following prediction:Electric charges that are farther apart should feel weaker forces due to each other.

Coulomb's Law:(F_E=k_E |q_1 |⋅|q_2 |/d^2 ) -If the electric charges are farther apart, then d is a larger number -Since it's in the denominator (and especially since it's being squared in the denominator), it makes the whole fraction k_E |q_1 ||q_2 |/d^2 much smaller overall

Why do you think UK hospitals banned employees from wearing Crocs™ in 2007?

Crocs are excellent INSULATORS. Most likely, shuffling your feet on the floor while wearing them scrapes excess electrons off of the Crocs and leaves your shoes (and your entire body) with a slightly net POSITIVE charge.

horizontal and vertical components of an electric field vector equation

E_x = -(delta V/delta x) E_y = -(delta V/delta y) EX: E(x)= - (change in V) / (Change in X) =50/3.5 =14 (1 N/C is equivalent to 1 V/m, the units of the quotient you just calculated. They are, in fact, identical units.) show us the rate of descent of electric potential from 1 location to the next (looks like slope change in v over change in position)

Electric Potential Created by Individual Charges equation? -As you get closer to a single electric charge from any direction (d↓), the potential should become _________. -Directly on top of the charge (d=0 m), the potential should look like _________________.

Electric Potential Created by Individual Charges equation? V_charge=KE⋅(q/d) -As you get closer to a single electric charge from any direction (d↓), the potential should become taller (or deeper). -Directly on top of the charge (d=0 m), the potential should look like an infinitely tall hill (q>0) or an infinitely deep valley (q<0)

What is an electric charge?

Electric charge is a fundamental property of matter and it can't really be described in terms of anything else things like mass shape volume density those are all great properties of matter but knowing any one of them doesn't tell you anything about how much electric charge something has electric charge is just a different dimension of understanding matter. Some things have electric charge some things don't some things have a lot of electric charge some things have very little. If objects have electric charge then they can be subjected to electric forces.

Electric fields are created by ___________ but electric fields only exist in regions where the _________

Electric fields are created by "individual electric charges" but electric fields only exist in regions where the "electric potential is changing" after all if the electric field is pointing downhill there has to be a hill if electric field only shows you the direction toward lower potential then there must be a change in potential wherever you see them if you have a flat potential there can't be an electric field there because there's no direction downhill to begin with

What do electric fields cause? What are the units of an electric field strength E →?

Electric fields cause Coulomb Forces on electric charges, F → E = q ⋅ E →. Unit: (Newton/Coulomb) In fact, it might even help you to think of electric field as the strength of the "force per charge," so that the bigger the charge that's placed in that field, the bigger the force on it.

Electric fields exist anywhere that _________.

Electric fields exist anywhere that electric potential changes.

Electric potential energy exists and can be converted into _____________ by __________.

Electric potential energy exists and can be converted into kinetic energy by falling individual charges create their own electric potentials

Atoms (and therefore molecules) are comprised of electrically charged and uncharged pieces of matter. When a transfer of electric charge occurs (i.e., a spark, a shock, etc.), it's because actual pieces of electrically charged matter leave one location and migrate somewhere else. This means that some part(s) of some atom(s) are transported during the event. Which of the following subatomic particles is most likely responsible for this "hopping" from one object to another during a charge transfer?

Electrons are most likely (by far) to be transferred from one object to another, because they're not tightly bound in the core of the atom. They orbit around the nucleus, and although they're not "free," it's a lot easier to coax them away from the nucleus than it is to rip the nucleus apart and unleash free protons.

If you bring the electrically charged balloon closer to the metal wall, what happens to the electric charges that make up the atoms of the wall?

Even BEFORE contact between the balloon and the wall occurs, the balloon pushes NEGATIVE charges AWAY on the wall. Even before contact occurs, the electrons on the metal wall begin to move away from the negatively charged balloon. This is called "charging by induction" because we induced (meaning "indirectly") a net positive charge on a certain spot on the metal wall (temporarily). Why is this possible? Because the wall is a conductor, as it's made of metal. If it weren't a conducting surface, we'd never be able to coerce electrons on it to move without vigorous scrubbing such as with the shirt.

What is equation to know how strong force is that electric charges exert on each other?

F(E)= KE (|q1| ⋅ |q2| / d^2)

What does an individual electric charge create? Equation?

F(E)=Q ⋅E individual electric charges (q) they create their own electric field (even if nothing else is present). if you have a new charge some other charge (Q) that you put in its neighborhood then you can have a coulomb force on that second charge. (IF they were the same, they would repel) remember that as soon as you know the electric field somewhere you automatically know how strong a coulomb force would be on any new charge that you put there so if you understand the q's electric field everywhere in space then you automatically know how strong the force would be on any other Q charge that you put in this neighborhood it's always going to be the new charge times whatever the electric field strength is there and again this is not different from the coulomb force you already learned it's just a new way to write the same thing

Describe what positive and negative vectors do to the electric fields and charged objects equation? What is the equation?

F(E)=q⋅ E (F(E) and E vectors) OR F(E)=-Q ⋅E if q is a positive charge then it's a positive scalar and when you multiply that by e they point in the same direction the force it points in the same direction as e because q is a positive charge so a positive charge would get pushed in the same direction that the electric field points however just like we saw on the previous slide if you instead have a negative charge then the coulomb force looks like minus q times e and you're back in the situation we just saw in which you're multiplying a negative scalar minus q a vector e and so the result that you get from that is a force that points backwards the electric force the coulomb force felt by a negative charge is pointed in the opposite direction as the electric field and it's all because q is a negative scalar so negative charges always feel coulomb forces in the opposite direction as positive charges whichever way a positive charge would be pushed a negative charge gets pushed the other way if the electric field is pointing somewhere and it pushes a positive charge down into the right you can bet that a negative charge would be pushed up and to the left and it's all because the sign of the charge is different

The magnitude ("strength") of the electric force on each charge is given by?

F_E=k_E (|Q|∙|q|)/(r)^2 -A major feature of Coulomb's Law (the way that electric charges exert forces on each other) is that if you separate them TWICE as far apart, they push/pull only one QUARTER as hard as before. -If you separate them TEN times as far apart, they push/pull on each other only one HUNDREDTH as strong as before.

Identify the directions of all forces acting on the Na^+ ion -------------> F_cl- Na+ -------------> (electric field) ------------->

F_cl- <-- Na+ --> q∙E -There are two forces acting on this ion. -It feels a Coulomb attraction to the Chlorine ion to its left, but it also feels a Coulomb force pointing to the right due to the electric field. -The electric field provides a Coulomb force in the same direction as the field vectors because the charge is positive.

Which location would a potassium ion (+1e) have the highest electric potential energy?

For a positive charge, the highest electric potential energy (in Joules or calories) will be at whatever location has the highest electric potential V (in Volts).

Which location would a potassium ion (+1e) have the lowest electric potential energy?

For a positive charge, the lowest electric potential energy (in Joules or calories) will be at whatever location has the lowest electric l potential V (in Volts).

In terms of potential energy, charges don't fall downhill, then fall to ________. Measure in what unit? High elevation= ?

In terms of potential energy, charges don't fall downhill, then fall to "lower electrical elevation and this gives them kinetic energy". Measure in what unit? Volts High elevation= high electric potential

On the left side, equipotentials are closer together, which means the potential is _________, so the field is ________

On the left side, equipotentials are closer together, which means the potential is steep, so the field is strong

Suppose the electric potential at Point E is V_E = + 10 Volts. How much electric potential energy (in Joules) would a 3 Coulomb charge have if placed there?

PE (E) = (q)(V) = (3)(10) =30 Electric potential is the cause of electric potential energy for individual charges, PE = q ⋅ V.

In terms of potential energy, if the charge speeds up when it falls _______? why?

Positive: if you speed up when you fall then your change in kinetic energy must be positive because you sped up and that also means that your change in potential energy must be negative you lost elevation Negative: if it's a negative charge the product of q times delta v has got to overall be a negative since nobody gains potential energy by falling the change in potential energy must be a negative quantity because everyone loses potential energy when they fall so nature says okay if the charge is already negative then the only way we resolve this is if your change in potential your change in elevation electrically is positive so what does that mean your change in potential your electric elevation goes up if you're a negative charge if you fall you fall towards higher potential what does that mean well it means quite literally what it says negative charges fall towards higher potential electrically that's like falling up they still lose potential energy just like positive charges do if you will allow them to fall and pick up speed everyone agrees that you must lose potential energy when you fall but now we have the problem that negative charges disagree about which direction they want to go they fall backwards whichever way a positive charge would lose potential energy negative charges lose potential energy by falling the other way now do you remember that electric field lines always point in the direction that a positive charge feels a force that's because electric field lines always point toward lower electric potential you might call that electrically downhill electric field lines show you which way is downhill electric field lines point toward lower potential and that's exactly why negative charges feel a force in the opposite direction of electric field lines they naturally fall electrically uphill they fall toward higher potential that's how they lose potential energy

What features can an electric field line diagram show us?

Regions of relative strength and weakness of the electric field in the space surrounding electric charges (Places where the field lines are closer together indicate stronger electric field strength than places where the field lines are farther apart.) The direction that a new (hypothetical) electric charge would be pushed or pulled at a certain location (Field lines can show us this. The actual electric field vector at any location is always tangent to the field line that runs through it. And we know that electric field always shows the direction a positive charge would feel a force (note: negative charges get pushed in the opposite direction), so we can always determine the direction that any charge would be pushed or pulled at any location, as long as we can see the field lines everywhere.) The sign of an electric charge creating the electric field (Electric field lines always show the direction that positive charge would feel a force. For example, if you see field lines pointing toward a certain charge (or region of space), then that must be a location that a positive charge would be pulled toward. This is a clue that the electric charge making the electric field in that region might be negative. The opposite is true if the field lines seem to be pointing away from a certain region or charge.)

Give the yellow balloon excess electric charge by scrubbing it on the shirt. What does the yellow balloon do next? Why does this occur?

The balloon picks up excess NEGATIVE charge, leaving the shirt with an excess POSITIVE charge. The balloon then accelerates TOWARD the shirt because the OPPOSITE sign charges apply an attractive electric force to each other. The balloon accelerates toward the shirt. This happens because the shirt and the balloon each now have excess charge, and notably, of opposite signs. This causes an electric force of attraction between the two.

Place a single positive charge (red) or a single negative charge (blue) in space and observe the electric field (white arrows) it creates. To see an individual electric field vector somewhere in space, place the sensor (yellow dot) at a certain location. The electric field?

The electric field created by a bigger charge is stronger everywhere compared to the electric field of a smaller charge The electric field always seems to point in whatever direction a positive charge would feel a Coulomb Force.

How do electric charges push and pull on each other? What factors determine? This is called the "Coulomb" Force (sometimes simply known as the "Electric" Force).

The electric force between two electric charges should either (1.) pull directly towards or (2.) push directly away from each charge. (The Coulomb Force causes electric charges to directly attract or directly repel. Those are the only directions it can point between any pair of electric charges.) If Object A has a very big electric charge, then it exerts a very big Coulomb Force on charged Object B. (This is reasonable. If one of these objects has a big charge, it probably exerts a big influence on the other object.) If two electric charges are farther apart, they apply weaker Coulomb Forces to each other. (Separate them farther, their influence on each other goes down. This is a reasonable expectation. After all, if two charges are separated on opposite sides of the universe, should we expect them to have much effect on each other? Probably not. But if you bring them closer together, they will definitely push/pull harder on each other.) The Coulomb Force between two electric charges should be attractive if the charges are different signs, and it should be repulsive if the charges are the same sign.

What is the equation that electric fields apply to charges objects?

The same Coulomb Force is actually a consequence of the existence of electric fields: F(E)=q ⋅ E (F(E) and E vectors) As soon as you know the electric field (E) at a certain location of space, you instantly know how much force (and in which DIRECTION) and charge would feel if it were placed there. q times e knowing the electric field is a shortcut to know how much force any charge would feel if you were to put it there just swap out the q for a different q and you'll know a new force on a new charge electric charge can be pushed by an electric force we now understand that to be caused by the field that exists in that electric charges neighborhood

What are 2 ways to write equation about the strength of the Coulomb Force between two electric charges

The strength of the Coulomb Force between two electric charges q and Q looks like F_E = k E ⋅ | q | | Q | /d^2. But we now know a new way to write and interpret this same force, F_E = | Q | ⋅ E, as if the force that Q feels is being caused by E, the electric field created by q. E_charge=k_E⋅ |q| / d^2 is the strength of an electric field created by any single electric charge

A sodium ion is placed in a uniform electric field. The word "uniform" means that this electric field is "the same everywhere [in strength and direction]." The electric field has a strength E = 350 The electric force felt by this sodium ion will be?

The strength of the Coulomb Force on this charge will be F_E = |q| ⋅ E = ( 1.6 × 10 − 19 C ) ⋅ ( 350 N C ) = 5.6 × 10 − 17 N F_E = | q | ⋅ E = ( 1.6 × 10 − 19 C ) ⋅ ( 350 N C ) = 5.6 × 10 − 17 N (1 e ≈ 1.6 x 10-19 C). Pointing in the direction of electric field because positive charge. (negative sign on the charge flips the direction of the force to be opposite that of the electric field that's causing it.)

Coulomb's Law states that the Electric ("Coulomb") Force between two electrically charged objects ( F E = k E ⋅ | q 1 | ⋅ | q 2 | d 2 ) depends on the product of the two charges (q 1 and q 2) divided by the square of the distance d separating those charges. If two electric charges are separated THREE times as far apart as they were before (i.e., d → 3 ⋅ d), how does the Coulomb Force that these two charges exert on each other change?

They push/pull on each other with a force only 1/9 as strong as before. If the original Coulomb Force between these charges looked like: F E O r i g i n a l = k E ⋅ | q 1 | ⋅ | q 2 | d 2 When they're separated farther apart, so that d → 3 ⋅ d, this new Coulomb Force becomes: F E N e w = k E ⋅ | q 1 | ⋅ | q 2 | ( 3 ⋅ d ) 2 = k E ⋅ | q 1 | ⋅ | q 2 | ( 3 2 ⋅ d 2 ) = k E ⋅ | q 1 | ⋅ | q 2 | ( 9 ⋅ d 2 ) Notice that all the symbols in this final answer look exactly the same as the original Coulomb Force, but now we have a factor of 1/9 on the entire fraction (i.e., the whole fraction is being divided by 9). So it's now 9x weaker than it used to be.

How much potential energy does the 2q charge have while placed there? q(+) ---a--- 3q(+) ---a--- 2q(+)

Total Electric Potential Created by Multiple Charges -Because an electric charge exists at this location, it possesses potential energy due to the electric potential hill that it's sitting on: PE=(2q)⋅V_total= ▭(2q⋅(k_E3q/a+k_Eq/2a) )

What is the total electric potential at the point in space shown at the far right? q(+) ---a--- 3q(+) ---a--- (⋅)

Total Electric Potential Created by Multiple Charges -The total electric potential at the location of the far-right charge is: V_total=V_3q+V_q=k_E3q/a+k_Eq/2a

How much potential energy does the far right charge (+2q) have while placed at its current location? (a is distance) (q+) ---a--- (3q+) ---a--- (2q+)

Total Electric Potential Created by Multiple Charges: The total electric potential at the location of the far-right charge is: V_Total=V_3q+V_q=k_E 3q/a+k_E q/2a -Because an electric charge exists at this location, it possesses potential energy due to the electric potential hill that it's sitting on: PE=(2q)⋅V_total =(2q⋅(k_E 3q/a+k_E q/2a) )

Equation for electric potential created by single electric charge?

V_charge=K_E ⋅ (q/d) (no absolute value)

How can coulomb force between two electric charges can be written more generally

as a consequence of the existence of an electric field: it's as simple as q times e. e could be made by an individual single electric charge or it could just be an electric field that just happens to be present in the room it might not be made by just one electric charge but no matter what electric charges will feel a coulomb force q times e if there is an electric field present in their region

electric field vectors always point _____________ Know that electric fields only exist in _____________________ Know that electric field measures the __________________ equipotential curves show lines of ____________________

electric field vectors always point most efficiently toward lower potential Know that electric fields only exist in regions where the electric potential changes Know that electric field measures the rate at which electric potential decreases per meter. equipotential curves show lines of constant electric potential

What did Michael Faraday do?

he gave us an answer to the question of how do electric charges reach out into space and apply forces to each other at a distance and he imagined what he called invisible lines of force (electric field) learn that electric forces are actually caused by electric fields thanks to michael faraday

Describe what vector 3 ⋅ v1 then will look like? What about v1 ⋅ -2?

if it looks like 3 ⋅ v1 then it's going to look exactly like v1 except it's going to be three times as big but it'll point the same direction however if the scalar is a negative number such as say -2 ⋅ v1 it doesn't just change the size of this new vector that you create it also flips the direction backwards

How can electric fields and electric charges coexist?

in general there are two ways that electric fields and electric charges coexist: 1. one way is electric fields apply forces to anything that has electric charge if you have electric charge an electric field can push you or pull you 2. the other way is that individual charges create their own electric fields so if you have charge you can be pushed by a field and if you have charge you create a field to push other things but you only get into this game if you have electric charge that's the cost of admission if you have electric charge you can be pushed and pulled by fields and you can use fields to push and pull other things that have charge electric fields apply forces to charged objects electric charge can be pushed by an electric force we now understand that to be caused by the field that exists in that electric charges neighborhood

what does electric charge do?

it allows you to feel an electric force if you have it you can feel it if you don't have any electric charge then you cannot feel electric force and it doesn't matter how strong an electric force is it cannot be applied to something that doesn't have any charge

What is electrostatic equilibrium?

means charges are not moving electric charges are static they're in equilibrium meaning all the forces on them cancel everywhere they could be applying coulomb forces to each other but all the forces have to add up to zero so that nobody goes anywhere

A proton is released from rest at point B, where the potential is 0 V. Afterward, the proton... -100V(A) 0V(B) +100V(C)

moves toward A with an increasing speed.

Objects on hills have ___________ that was put into them to get them up to the top of a hill that you can then turn into ________ by falling down. This is called ____________. Gravity is a force that we call __________ it means that it _______________. Electric charges have a _____________ called _______________.

objects on hills have "(stored) potential energy" that was put into them to get them up to the top of a hill that you can then turn into "motion (kinetic)" by falling down. -objects have energy that's stored in them simply by being high up above the ground because something had to put that energy into them to get them up there and because gravity is a force that we call conservative it means that it gives you all of that energy back in the form of motion when you are allowed to fall. -We called this quantity "gravitational potential energy" because it was energy that you got from having work done on you against gravity to push you up somewhere and then it's stored in you -Electric charges have a potential energy called electric potential energy

What are insulators?

they are whatever material they need to be that prevents you from easily moving charge around on their surface or inside them a good example might be plastic or glass or rubber it's really hard to put excess electric charge on an insulator and it's really hard to take it off (wear special electrical safety gloves)

What are electric field lines?

they're kind of like vectors in the sense that they point in the direction that the electric field points however vectors have a definite length the size of a vector actually means something those are vectors what you see in pink those are electric field vectors and that means something the size the length of those arrows is supposed to indicate the strength of the field by contrast electric field lines have no finite length and they're not just arrows that point in one direction the way vectors are they can be long and curvy whereas for a vector the length of the vector tells you the strength electric field lines don't have a length they really don't terminate or emanate anywhere except at individual electric charges so knowing how long a field line is doesn't mean anything

how do we measure electric charge?

we actually invented a unit and we called it the coulomb and it was named after after Charles-Augustin de coulomb

A charge is released from rest in an electric field. Neglect non-electrical forces. Independently of the sign of the charge, it will always move to a position ____________

where it has lower potential energy.

equipotentials show continuous lines of equal potential, they act like a topographical or elevation map, where is highest and lowest? Co2

±The highest electric potential is usually the positive charge in the center (a tall hill), and the lowest potentials are the two negative charges (two deep wells)

About how strong is the force on the electron due to the proton's electric field at distance 5x10^-11 (in N)? (E is (4×〖10〗^11 N/C))

•Approximately how much force (in N) do they exert on each other at the average orbital distance? -Once we know E, we quickly know the force that E can apply to a nearby charged particle, q: F_E=|q|⋅E -At that average distance away from a proton, we have: F_E=|q_(e^- ) |∙E_(p^+ )=|-〖1.602×10〗^(-19) C|(4×〖10〗^11 N/C) ≈▭(4×〖10〗^(-8) N) -Huge field still produces a tiny force. On the other hand, that's still a massive force to a tiny little electron.

How to know where electric field points?

•Electric charges create their own electric fields •The strength of these fields looks like E_charge=k_E⋅|q|/d^2 •The way to know where these fields point anywhere in space around the charge creating them is pretty simple: -Just ask yourself where a hypothetical positive charge would be pushed or pulled -Because that's the direction of the E-field created by a charge

Explain how to find Total Electric Potential Created by Multiple Charges

•If you can calculate the electric potential some distance away from a single charge, then you can do the same thing when many charges are present in that region. •Because the total electric potential somewhere in space can be "high" or "low," but it's not a vector like electric field - it has no direction •You just add all the hills and valleys and see which one wins, and you're done - just like elevation. Ex: V_A=V_1A+V_2A= k_E q_1/d + k_E q_2/d (If one of the q's is a negative scalar, you don't need to introduce your own minus signs to show this, the charge q will be a negative number already)

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