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[-/5 Points]DETAILSSERPSE10 24.3.0P.012.MY NOTESASK YOUR TEACHERGiven two particles with Q = 3.10-UC charges as shown in the figure below and a particle with charge...

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[-/5 Points]DETAILSSERPSE10 24.3.0P.012.MY NOTESASK YOUR TEACHERGiven two particles with Q = 3.10-UC charges as shown in the figure below and a particle with charge q = 1.29 x 10-18 C at the origin: (Note: Assume reference level of potential V = 0 atr = <.)SSA0.800(a) What is the net force (in N) exerted by the two 3.10-C charges on the charge q? (Enter the magnitude:)What is the electric field (in N/C) at the origin due to the two 3.10-HC particles? (Enter the magnitude:) N/CWhat is the elec

[-/5 Points] DETAILS SERPSE10 24.3.0P.012. MY NOTES ASK YOUR TEACHER Given two particles with Q = 3.10-UC charges as shown in the figure below and a particle with charge q = 1.29 x 10-18 C at the origin: (Note: Assume reference level of potential V = 0 atr = <.) SSA 0.800 (a) What is the net force (in N) exerted by the two 3.10-C charges on the charge q? (Enter the magnitude:) What is the electric field (in N/C) at the origin due to the two 3.10-HC particles? (Enter the magnitude:) N/C What is the electrical potential (in kV) at the origin due to the two 3.10- HC particles? kV What If? What would be the change in electric potential energy (in J) of the system if the charge q were moved a distance d = 0.400 m closer to either of the 3.10-1C particles?



Answers

Force between charged particles: $F=k \frac{Q_{1} Q_{2}}{d^{2}}$ The force between two charged particles is given by the formula shown, where $F$ is the force (in joules -5 ), $Q_{1}$ and $Q_{2}$ represent the electrical charge on each particle (in coulombs - $\mathrm{C}$ ), and $d$ is the distance between them (in meters). If the particles have a like charge, the force is repulsive; if the charges are unlike, the force is attractive. (a) Write the variation equation in words. (b) Solve for $k$ and use the formula to find the electrical constant $k,$ given $F=0.36 \mathrm{J}, Q_{1}=2 \times 10^{-6} \mathrm{C}$ $Q_{2}=4 \times 10^{-6} \mathrm{C},$ and $d=0.2 \mathrm{m} .$ Express the result in scientific notation.

Here, we're going to demonstrate the difference between electric force and field. So we'll do that by putting a couple charges down on the left will be a positive charge. Q. one. And it will happy Amount of three Nano columns and a little bit of distance away will put a negative charge Which will have an amount of -12 NATO columns. And we will separate them by three cm from center to center. Yes, to find the electric force. So these two objects will interact via electric force known as columns law. We'll see. Coolum can also mean the electric field, but we find that for us by taking the product of the absolute value of the two charges times the electrical constant K. Divided by their distance squared in S. I. Units that K. Is nine times 10 to the ninth. And we want to put in the charges with there S. I. Units of columns. Yeah. And we would put the distance in with meters. And what we wind up with is a force that's pretty small. Yeah. Yeah. And how does that force act basically? It acts along the line joining the two charges. It's an attractive force that would of course pull them together because they are opposites and they will attract. But what about the electric field? So reminder that an electric field is not an interaction per se. It's an imaginary interaction. So each of those charges has its own electric field. The definition says, take a positive charge, and the electric field emanates outwards from that positive charge. And for a negative charge, the electric field goes inwards towards that electric field. And I'm trying to show that the electric field is a little bit stronger around that negative charge. So, ordinarily these will be joined by lines of force because there are two of them. And you would find the electric field everywhere in the region of space by adding The two electric fields um everywhere in space. And that that's kind of a pain. But we can imagine what those lines of force look like all through space. They started the positive charge emanate outwards and they get drawn in to the negative charge. And so you'll have lines of force that kind of look like this, get rapidly drawn in, rapidly drawn in, et cetera. Um so that's the bottom line is you have to add the two electric fields. And what we see is that right smack dab in the middle. So I'll draw line right 1.5 cm away from each charge. We have a very strong electric field. It doesn't just vanish, but we could find that electric field by adding The magnitudes of the two fields. So we'll call that the midpoint and why are we adding the magnitudes? Because both electric fields point to the right at that location. Um and what really imagining with the electric field is what would happen to a positive point charge if you put it right smack dab there. Well, it would be repelled by the positive guy, attracted by the negative guy. And there would be a net total uh, force to the right. And that's what you want to imagine with the electric field. Now, the electric field, you find the same way with columns law for point charges is you take the absolute value of the charge, multiply it by the electrical constant and divide by the distance squared. So we're going to do that. E one is going to be nine Times 10 to the 9th. Uh huh. Types of three Nana columns over the 1.5 centimeters squared. To put that two m. And we'll do the same with the second charge. And we'll wind up with the S. I. Units for electric field. And that comes out to be about six Times 10 to the 5th newtons per column. And again the direction is to the right. Um And as one last example, what if we put eight point positive charge right in the center? Let's make it plus to nana columns. Well now there is an interaction. Um of course that charges interacting with both the positive and the negative off to the side. But now we can imagine that it is feeling the interaction with the electric field and the force on that charge is simply Q. Not an absolute value times the electric field at that midpoint. Um So we're going to have to actually take into account the charge of that particle, which is to antique columns. And our electric field is six times 10 to the 5th newtons per cool. Um, so our units will be newton's and direction will be to the right, Okay, that would be 1.2 Times 10 to the -3 Newtons.

As an example of working with columns law, electric force, an electric field, we'll take a look at two charges that are sitting in space next to each other. one is positive and the other is negative. And let's just give those some values uh minus 12 anna columns on the right And a positive three and columns on the left separated by 30 centimeters. Okay, So columns law gives us two formulas for both the force of attraction between two charges, and that should be written with an absolute value around those two charges. Why? Because this equation gives you the magnitude of the interaction. The direction comes from basically knowing what signs of charges there are and whether situated in space. So the force always acts on a line joining the bear and it will Act one way if they are attracting each other, it will act opposite that on both charges if they repel. Um in the electric field has a similar type of relationship. Again, there's A. K. And the absolute magnitude of a. Q. Over by divided by the distance squared. Um but notice that the electric field assumes just one charge sitting there in space creating a field, whereas the force measures in interaction. So let's take a look at this example. Um the two charges will obviously feel an attraction and that will be a Newton's third law pair. So those are equal and opposite each other. Um And hence you don't want to be putting signs on your columns law calculation. But anyway, uh we can calculate the magnitude of that interactive force. The electrical constant is nine times 10 to the 9th in S. I. Units. And remember we want the absolute magnitudes of those charges. So we're going to put them in both with positive signs. And really it makes no sense to have a negative magnitude. Only astronomers make that happen. Um Yeah, but a vector magnitude is not a negative number. So let's see what we get in terms of Newton's, we've used everything in S. I. units. So this is 3.6 Times 10 to the -6. Um Newton's okay. The next situation we're going to think about is the electric field to add a point right in between those two charges. So we have our negative and positive charges and we are going to look at a point, we'll call that point P. So one 15 cm from both 15 centimeters and 15 centimeters. Now, the electric field is calculated at all points in space by taking however many point charges you have here there too, and adding up their individual electric fields. Um So electric field is a vector and it usually tells you um the force a positive charge would experience if it were placed at a point here we're talking. Look looking at point P. And in order to get the vector right, what you really want to do is imagine the vectors at that point that are created by your two charges. So in our situation, electric field always emanates outwards from a positive charge. And we could draw a bunch of lines coming out of that charge. But what we're interested in is that vector that's on the side with P similarly, electric field is going into um a negative charge. Sorry, that should be minus. And again, we could draw all sorts of factors going into that charge, but what we see is is at point P. Both fields point to the right. So our electric field is going to be the sum of their magnitudes will give us the magnitude of that electric field and the direction will be pointing to the right And again, to find those magnitudes, Remember, we do not want to put in a sign of a charge, we put in a magnitude of charge because if you don't do that, you will wind up subtracting something when you meant to add it and that would be bad. Yeah, and r E minus. We do the same thing. Of course. This is even more critical that we put in a absolute value of the charge. And if all come out, if all quantities are put in S. I. Units, the unit should be newtons per Coolum. Okay, and that works out to be 60 newtons per column. He plus plus c minus turns out to be 60 Newtons per Coolum. And the direction is to the right. And I'll tell you if you put in the wrong sign of charge in this this relationship, you'll think that the positive charge cannot create a electric field to the left and vice versa. The negative charge can't create an electric field to the right. And we see that that's definitely not true that the electric field points of different directions in space around each charge. Um So you have to be careful with that, and finally, if we put another charge into the situation, um So let's put a charge of 2-plus 2. Yeah, the columns there, right smack dab in the middle. We could work out the force on it based on the forest from the two charges nearby, but it is much easier to say F equals Q three times the electric field. Um Where e is the total electric field experienced by that charge at that point, which is 60 Newton columns practical. Um To the right standard minus nine. And that will be to the right as well. So, a thing to remember is that the electric field gives the sense of the force that would be directed on a positive charge. So this would be 1.2 Types 10 of -7 Newtons, and to the right.

In this problem we have on arrangement of a square. Um, where there's charges on two of the corners of the square, so the orientation is not unique. But for this point, I'm gonna choose it to where Q one is here. Cute too is here. Um And then they tell us that point A is in the middle, and point B is on the corner closest to cute, too. So that would, for my picture mean that b is up here. Now they tell us that each side of this square is three centimeters, so that means that 0.3 meters and then they give us some of these values. So Q one is equal to plus to micro Kula mes que tu is equal to minus two micro columns. Okay, so now what we need to do is we need to figure out some of these lengths here because we're gonna have to know these lengths in order to calculate Ah, the potentials. So first I'm going to do this triangle that I have drawn here. So this triangle is simply 0.3 0.3 and then a will be in the middle. So whatever we find the high pot needs to be the length from Kyu won A will just be half of that. So we have the high pot. News is equal to the square rope square root of two times 0.3 squared. And there's a two here because both sides of the same for this tells us that the high part news is zero 0.4 24 meters. Huh? And so now are our radius, Um, from one to a which will also be the same as the distance from que tu es. Since we have cemetery here is equal to half of this number, which is simply 0.21 meters. Okay, so now that we have this in part A, the book is asking us to find the voltage. Remember the voltage for a point charge. It's simply Q over four pi epsilon, not R. And then you must add up. The voltage is from every charge. So if we're trying to find the voltage at point A, we need to find the voltage at a duda que one plus a voltage at a dude. A cute too. So we can do that. The voltage et ai will be equal to Q one over four pi absolutely not are one plus que tu over four pi Absolutely not, aren't too. But since our one is equal to our two, this is simply equal to Q one plus que tu over four pi epsilon Not our one, or are too. And then we see that kyu won and que tu actually have the same magnitude but opposite signs. So when you add them together here, you're gonna get the voltage at point A is simply zero votes, okay, and that's it for part A. So now if we move on to part B, the book is asking us to find the voltage up at the top of our square at point B. So just to remind you this is point B, this is Q one. This is cute too. Okay. And we already found that this length here, okay, which is our one in this now is equal to 0.424 meters. So now we just do the same thing we did in part a. The voltage apart at location B is equal to Kyu won over four Pi epsilon not are one plus que tu over four pi absolute not are too. But now our wanted our two are different values are too is just a side of this square the distance between que tu and B So this is simply 0.3 meters And now you can simply plug in these values being very careful that Kyu won, um, and cute to our micro Coombs He must convert into cool ums and you'll find that the value is negative 1.75 times 10 to the fifth votes. Okay. And the reason this is negative is because que tu was actually a negative charge. You get a negative answer when you plug in it Now in part, See, we're gonna have a point a point Charge Q three equal to negative five times 10 to the negative six cool ums. And it's gonna travel from point A here, up toe point B. Okay. And we want to figure out how much total work is done as that point charges moving from A to B. So we can simply remember that work is equal to negative. Tell to you so this will be equal to negative. Ah, you final. Which is you be minus initial, which is Yuet huh? So this will simply be equal to you, eh? Minus you be. And now we just remember that you is equal to Q not V. Where cute not is the charge that we're considering moving or changing potential energy. So this will simply be equal to Q three, since that's what's moving times V A minus V b. Okay, so now we know he a is zero. So this whole thing goes to zero here, and what we're left with is simply the work from A to B is equal to minus Q two times the voltage at point B. Um, the voltage is negative. Thank you to is negative. So when we plug in these values, we find that the work from A to B is simply sorry. This is a Q three is simply minus 0.877 Jules, and this work is negative, which is, as we would expect, because a negative charge Charge three wants to move from low to high potential, and V B is at a higher potential. So it's trying to move to that point

Electrical potential energy for a pair is given by K. You want You Do Over are which in this particular case, will come out. Toby nine times tend to their nine for gay. Q. One is five. Mana could, um 5 10 17 89 cause, um times Q To ease negative three time stand to the negatives. Nine. Whose, um divided by the distance between them, which is 35 centimeter, 0.35 major so calculating that it comes out to be 103 vote. So this potential is positive. So that means. I mean, there was a native over here, so there's native over here, so that becomes positive. Which means that there at attractive in nature, with their attractive in nature, what is the electric potential at the midway between the two charges? So which means at Midway means that if we have particles one over here and particles to over here, that used to be my r equals 0.35 Now I have to use our prime equals r over to so using that we will have potential with BK Do you want over our prime? Thus you do over are trying. So that will be gay. You want to ask you to over our prime, We use our over to. So that is two times nine times 10 to the nine. That's too. Okay, put the are in 0.35 And we were that you want to ask. You do which is 583. So there's just simplify it better the two times 10 to the natives nine. So solving for these we get the and I need you to be que times nine times to divided by 0.35 1028 Oh, that this just be doing them was 35 centimeter. So clearly I wrote the values wrong. So if we have this, then it becomes like this r equals 1/3 fold that is, at the midway this number I copied it wrong from whatever I was working on. So this number comes out to be negative. 3.85 times 10 to the negative seven. Yours


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