questions-1-2-and-3-are-short-free-response-questions-that-require-about-13-minutes-to-answer-and--4

Question

Questions 1, 2, and 3 are short free-response questions that require about 13 minutes to answer and are worth 8 points. Questions 4 and 5 are long free-response questions that require about 25 minutes each to answer and are worth 13 points each. Show your work for each part in the space provided after that part.
(DIAGRAM IS NOT AVAILABLE TO COPY)
Two charges, $q_{1}=+3 \mathrm{n} \mathrm{C}$ and $q_{2}=-12 \mathrm{nC}$ , are fixed in place and separated by a distance of 3 $\mathrm{cm} .$
(a) What is the magnitude of the electric force between the two charges? Is the force attractive or repulsive?
(b) At what point between the two charges is the electric field equal to 0?
(c) What is the electric field at the midpoint between the two charges?
(d) What is the magnitude of the electric force on a charge $q_{3}=+2 \mathrm{nC}$ placed in the middle of the two charges?

Linda Winkler · Accepted Answer

as an example of working with columns law, electric force, an electric field, we&#x27;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&#x27;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&#x27;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&#x27;s take a look at this example. Um the two charges will obviously feel an attraction and that will be a Newton&#x27;s third law pair. So those are equal and opposite each other. Um And hence you don&#x27;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&#x27;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&#x27;s see what we get in terms of Newton&#x27;s, we&#x27;ve used everything in S. I. units. So this is 3.6 Times 10 to the -6. Um Newton&#x27;s okay. The next situation we&#x27;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&#x27;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&#x27;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&#x27;re interested in is that vector that&#x27;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&#x27;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&#x27;ll tell you if you put in the wrong sign of charge in this this relationship, you&#x27;ll think that the positive charge cannot create a electric field to the left and vice versa. The negative charge can&#x27;t create an electric field to the right. And we see that that&#x27;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&#x27;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.

Salamat Ali · Answer

a magnitude of force but a major force between two charged particles given in a problem, it&#x27;s given by a fee that is equal to K. Coolum. Constant times. Kyu won Times 22 divided by a distant square are square that is equal to substituting the values for Coolum cost in charge. One in just two we get here or minus 3.6 times 10 to the power minus six. New, too. And since one of the charges negativo there is supposed to their force off the force between these two charges is attractive. Attract you in part B the electric field. The point at which the electric field with due to just you wanting que tu is zero can be found by writing equation for a few one. It&#x27;s okay, it&#x27;s Colin q one divided by. So here we have Q one times. So the electric field for first particle. You wonder why did by on one square plus o que tu que tu divided by are too square eyes equal to zero, then substituting the Valium for kyu won in qd, we can find the relation between our one an hour, which is a two on one. It is equal Thio too country, then using the relation for one and are too Do that is 0.3 we can solve for a are due and substitute the value than with relation we get here is our two is equal to two times over one that is equal to 0.2. So this suggested that the electric field is it Oh, it at, uh 0.1 from the key one and 0.2 from the q one between the two charges. So I see at the midpoint the electric field will be There&#x27;s some off some off the fields off one in you too. So here you have on our square Que, um two divided by our squares is the same for more because off the midpoint, then the resultant electric field we get by substituting value is minus 3.6 times 10 to the power into the power of three Newton but whom you didn&#x27;t. But hula on this is towards the cute, too. So the charge the field lines originate out off the poster charge and get into the negative tracks. So it is. They feel this tour&#x27;s Q two indeed their force experienced by the to micro Michael. Cool. Oh, um, charge is given by two times over 10 to the power minus nine. Coolum went to apply by the electric field, which we found your sentry 0.6 times. 10 to the power into the power tree. When you find these two numbers than the force we get here is a minus seven point two times 10 to the power a minus, minus six Newton.

John-Paul Mann · Answer

in this problem we have on arrangement of a square. Um, where there&#x27;s charges on two of the corners of the square, so the orientation is not unique. But for this point, I&#x27;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&#x27;re gonna have to know these lengths in order to calculate Ah, the potentials. So first I&#x27;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&#x27;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&#x27;s simply Q over four pi epsilon, not R. And then you must add up. The voltage is from every charge. So if we&#x27;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&#x27;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&#x27;re gonna get the voltage at point A is simply zero votes, okay, and that&#x27;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&#x27;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&#x27;re gonna have a point a point Charge Q three equal to negative five times 10 to the negative six cool ums. And it&#x27;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&#x27;re considering moving or changing potential energy. So this will simply be equal to Q three, since that&#x27;s what&#x27;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&#x27;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&#x27;s trying to move to that point

Maria Gabriela Cota Moreira · Answer

in his exercise, we have to charges Q R and can chew place as I show here in Cedar Onda. We know the distance between Key Juan and upon to be and we know the distance between key to in a point B and we know the distance between Wana key to Besides, uh, we know the charge of Kyu Wan. We know the magnitude, but we know it&#x27;s signed. We don&#x27;t know what is the charge off key Chew. And also we know that Q one and Q two together produce electric fields on point B that points downwards in the Y direction. But we don&#x27;t know what ISS its magnitude. I also rolled here of globes constant. Um okay, so question they asked us, Just draw some dry a grams, um, showing all the possible configurations off charges. Um, so Okay, so let&#x27;s start with, uh, if key Juan IHS positive. And if g k two it&#x27;s positive what will be the electorate configuration? What we did configuration off the electric field. So we know that that&#x27;s refute of a Boston charge. Points all ports from the source charge. So in point B, what, we&#x27;ll have an electric fields like this for particle to and, uh, for particle one. Okay, well, and it&#x27;s also positive, So diligent fields points outwards from the particle. Sorry, this should be parallel. Should is like here. And this is a June. So this is the first diagram. The second possible configuration is if key Juan is negative and 52 remains positive. So again, the electric field off to after two points outwards. Oh, sorry. There&#x27;s no hats in here. But now, since K one is negative, instead of pointing outwards, the electric fears repointing inwards towards the source charge. So we&#x27;ll have that you want. It&#x27;s like this. Okay, so that&#x27;s your third configuration. Um, is no Okay, Chew is Mueller. That zero is net is negative and okay. Want It&#x27;s positive. So, like our first diagram If Kiwanis positive gelatin field points outwards from the source charge and we have that you want Well, you pointing like this and now teacher is negative, so the electric field will be pointing towards the source charge. So here along this line, this will be each you and the last possible configuration is if both charges K one and K two are negative so if they want is negative. Uh, tentative field for dessert by discharge on point B. Well, repointing towards the source charge so it won&#x27;t will be like this. And, um if George who is negative? The electric fields. Sorry. Well, repointing, uh, towards the source charge So it to we&#x27;ll be like this. Okay, uh, now, question be asks, is among all those four possibilities what should be the sign off both k one and K two so that we all pertain the configuration I shows here in this first set up. So what should be the sign off k one and K choose so that e the resident electric field is parallel to the minors by direction. So let&#x27;s look at the first worse possible case on which they want to get your both positive. We can see that if we some the two electric fields. Sorry. Um, if we send it to electric fields, the residents one we&#x27;ll re pointing upwards. So here he is, parallel to the positive y direction, which is not what we want. So the first, uh, the first diagram is wrong. For what? We want to find the second diagram. If key one is negative and get responsive. Well, again, you sum up the two vectors. Sorry, Uh, some of the two vectors and refines that the electric field? Nope. Sorry. Uh, the residents electric field. Well, repointing, uh, to the, uh, minus X direction. And this is is not what we want. So this is not the configuration we&#x27;re looking for. The configuration three again. Uh, no, we have okay, too negative and key. Well, positive. So summing the two vectors, we have a rest and field that will be pointing to this direction. It has an x component, so it&#x27;s not what we want. And finally, we have the configuration on which both k one and get you all negative. And for this configuration, we have that the some off wolf electric fields. Kim repointing dar words into minus my direction. So this is a configuration. We want p one and get you both negative. Great. So now question. See, X is to calculate the resident electric field e for our set up. So down here, I already drew, uh, a diagram for our case. So we&#x27;re working with recall the scenario on which key Juan is Millard and zero is negative and okay. Chew is also Lega tive. So first things first, we know that the Resident Electric Field e should be pointing downwards in the Y direction. So should be the son off the y component off the electric field generated by particle one plus the y component off the electric fields generated by particle to this is why so it should be He won. Why? Plus e chu y Okay, So what iss you want? Why plus each white So in when you want. Why that started looking at this. We have that This is the projection on the Y direction off the electric field you want. So if we take this angle here? Nope, sorry. All joy in green suits. Um, he&#x27;s there for us to see if we take this single here. We have that you will. Why? Easy. Go to the sign off the music culture, the mother, the magnitude off the one times the sign off the single And it happens that the single physical data that because we have a two parallel lines here and one like crossing so have this relationship between these two angles. So that&#x27;s not writing. We have that e It is equal to the magnitude off Iwan times sank data plus now for delicate field off Marco too. The y component should be the projection over this angle here on the Y direction and for the same reasons as this angle is equal to theta we had this single is a cult If I so it is the magnitude off a two times the sign off five So it is e to time sign Oh Fi. And we know that well, since they want NK to our boat from charges we can use columns law. So we have that Iwan physical to close constant times the charge the magnitude off the charge off k wan uh, over the distance between Charge one and point B which will call our one squared and for a to the same thing. But with the charge off particle to over the distance are two squared, which is this This is nothing This business here but one thing we must notice that we can&#x27;t yet jumping into calculations and calculate the value off B because we don&#x27;t know the question. Don&#x27;t give, don&#x27;t give us. Give us the value. The magnitude off charge, Cheechoo. We only have the one we know. Arwa. We know, aren&#x27;t you what we have to find Kate? You so too funky to We can go back here to our little diagram and noticed that for the resident field to be pointing downwards the negative y direction This means that the X component off both the one any to must castle with each other so that we have something like this minus one blows teach. You should be equal to zero. So highlight this letter in question here. And similarly, as we did for calculating the magnitude off the residents electric field, we have that, uh, we can rewrite this expression as the magnitude Iwan times call sign all faded. And this is Eagle Passengers to the other side of the equation to eat too. Times the Cassano fi. And with this, we have the relationship between, uh, we can rewrite this SK times Kyu wan over I our own squared musical too. Okay, times Ki chu over our two squared when we can cancel out both grooms constant on both sides of the equation And we can isolate kitchen which is the value we need to calculate what it kept. The question asks us the rest on field and we can isolate this value. And we have that K two is equal to our choose squared. I&#x27;m sorry. I&#x27;m missing, um here cause time. School side of fado on times Go San five. Okay. No, everything is okay. On this is our to, um are too Times key. Wong Kei Wan over R one squared times Call sign data over co sign phi. We know our one. We know our chew we know are one we know Kiev on. This is all given by the question and we can express the co sign off data and consent if I buy quantities. We already know if we noticed that. Uh, Fada is this angle for them from the bigger triangle and that are one Is this distance here? Our chu is this distance here and we have this distance here below is 13 centimeters and we have that co sign off. Data is equal to, um five. I&#x27;m already sir. I&#x27;m already soliciting defender so rewriting. Just as his numbers here are one over this that alcohol d over D and co sign off. I physical chu are two over D Andi, substituting all the values that are given by the exercise we have that K two is equal to, uh, 12 times stand to the minus two squared times three times Send to the minor sex over five times 10 to the minus. Two squared times cosign A fader divided by percent. If I which is just 5/12 and this is equal to well, we know that the charges negative. So all rise with the sign here minus seven points to So I&#x27;m saying two de minus six Coolum And now we have the better off torture and we&#x27;re ready to calculate the residents electric field. So let&#x27;s take this expression here. And so we suit all the values and we have that, uh, the residents electric field physical too. E won&#x27;t time. Citatah Well, we can express I think that is equal. Chew Cynthia, is it go to or chew over d sine phi Is it that you are on over d Same thing Musical sign on. I&#x27;m taking these relations from this over this bigger Tranel and, um submission student values We have that e physical too. Tree time. Stan to demine six times 8.99 Times 10 Denying over five time Stand to demand issue squared times. Um, this is science data. This is a go to 12/13 plus the charge off. Orica too. So seven points two times 10 to the Middlesex time schools constant 8.99. It&#x27;s understand to do nine. I&#x27;m running out of space here, Um, over 12. To distance between a particle to A and point B 12 time send to the months. 2 m time. Sign off data. Sorry. Santa Fi, which is are rolling over D five over, uh, 13. And this is equal Chew, and we find that e is equal 21 0.17. So I&#x27;m Stine to the six Newton Burke Ueland. And with this with so over exercise

Ze-Han Lee · Answer

So in this problem, we have two charges on the Y axis. So this is X. This is why the charge one is ah Q And the charge too. Charge to his negative is too cute. And this is a and this is this is negative to a All right. So in part A, we want thio determine the fools felt by charge one from charge, too. So according to the because these two challenges, they have the same sign, right? So their direction are is repulsive. So the force felt by Q one should be pointing up. And the magnitude is, ah, one over four pi ups or not times Q. Times two q divide by three a squid, right, because they&#x27;re distancing three a. So this gives you why over full pi ups or not? Two. Q. Squared of duel by nine a squid All right, and the direction it&#x27;s pointing up part B. So you want also we won&#x27;t find now the electric field at the origin you want, which is the electric field created by the charge Want It&#x27;s negative because charge one is ah, supposed cures Positive. Okay, then Electra view at the origin created by charge of one&#x27;s pointing down. So we have this negative sign. One over for pie. APS or not, Times Cube over a squid Wie Het Yeah, any two. Which is the electric field created by charge, too, Is what over for pie? Absalon nods Times two Q Divided by two. A squid Whitehead&#x27;s right. So the combined Electra feud is the summation of E one e two and this gives you a negative one more full pie. Abso not times Q over to a squid. My head. Yeah. So the direction is pointing down and the Parsi Ah, we want to know if there is any point on the X axis so that the Nateq Electric field is zero. So that is downstairs? No, because as you can see, that we have Charlie Webb, Charge Q here in the charge took you here either. It&#x27;s ah, on the left on the right. So suppose it is on the right. All right, so the x component of off the electricity is not the euro. It would be pointing to the right it is on left than the combined election field is, uh I mean the X component off the combined Electra food is pointing to the left. So even if it is at the origin as we fought in previous a CZ we found in part B, it is no zero. So we&#x27;ll say that. So there is no any point on X axis so that the net electrified zero. Okay, so it&#x27;s no, and a party won&#x27;t find the location. We&#x27;re the electric few where we won&#x27;t find a location on the Y axis. So that electrocuted zero. So we have. Ah, well over four pi. Absolutely not. Times Q over X square. So here axes that distance from the er from the interesting point to the cute to the challenge one. Okay, so this is Max. So this equal worm over for Opie. Absolute. Not times Q sir, It comes to cute. Divide by three a minus X squared. OK, because the total distance is three a three a. M on its axis this distance okay, and even solve this equation. You have 10 that toe axe equal three square to minus one, eh? And, uh ah. So the coordinates off the other&#x27;s points is ah, 04 minus three times square, two a. All right. And in part E. We want to know the if we put the test charge neck of Q on origin at the origin. So what is the initial acceleration? So we know that because we already found the electro fuel at origin, which is ah, negative when we&#x27;re full pie. Absolute lots of times Q. Over to a squid times my head. Right, So the bulls on the press charges E times negative. Cute. All right, so this gives you went over for pie. Absolute, not Q Q b o Y two a squid, my head, then exoneration equal f over an. All right, so this gives you the initial acceleration, which is, uh, we&#x27;re full pie. Absolute times one over M Q Q over to a squid, my head So I can see that the direct that direction is pointing up. And here is the magnitude all right,

Ze-Han Lee · Answer

in this problem, we have a set up Q one. You, too. And their separation is the 13 centimeters we have also interested. Point over here on P and the distance we don&#x27;t pee in the queue Wines, five centimeters that this had been puta que tu is 12 centimeters. Ah, so for part A, we have four configurations off. There were signs of Q and Q two. So we want to label the Electra feud for different configurations. So suppose we have positive and positive then for the point p. This is pointing up this pointing to this direction for that one is pointing to this direction. So the white component has to be pointing up. And if we have positive and negative, then for the interested point, biggest key wise, positive. Manny&#x27;s point to this direction we&#x27;ll keep people to two is negative. So is pointing to this direction. In this case, the X component is not zero, as can see clearly for third case, negative and positive. Ah, we have this point. So because otherwise negative. So the electric field is pointing to this direction, and a que tu is positive, so it&#x27;s pointing to this direction In that case, X component is not Ah, is No zero as well. And for the last case, if we have ah, negative and negative. So we have two points. We have this interesting point over here because we being skewering cute, they&#x27;re both negative. So the electric it is pointing toward them. In that case, it is possible to set to the X component being equal to zero. And the white components is pointing down, okay. And so for part B, the only possible case, just a negative and negative. Parsi Ah, we want to find out the very of Q two into the very off the Electra feud. So we already know that to the at one peed. Ah, the electric field is pointing down, so there is no any component along the X direction. So in that case, E induced the bite cure one, which is the one equal two negative 1.4 pi absolutes. Ah, we just look at the X component. So you want X Times? Q 21 is our knock. You one? Yeah. You want divide by the separation between the, uh, q and Pompey, which is five centimeters squared. So this is this is a you want because we&#x27;re interesting to the Were you interested in the X component? So we need a time&#x27;s a factor. Five over 13. Okay. And, ah, the X component induced of my e two is Ah ah. So we should not say cute Q one here. We should just utilize the body of Kyu Wan. So right here it&#x27;s simply three micro clubs. And for a to x equal, negative one over for pie. Absolutely not cute too. We just look at the magnitude. Okay, So divide by 12 centimeter squared and times 12 over 13. So this factor gives us the X component off E too. So we have Ah, the one x you see. You want X equal? A to X, right? Ah, right. Embassy. So you want X is pointing to this direction e to exit pointing to this direction. Right. So this one should be positive. So, uh, their relation we will have is ah, the one next plus a two x equals zero. Okay. And you can actually solve for cute, too Equity. Negative. Sound point to Michael corms, and then we can try to figure out to the magnitude of the electric field at one p. So the electric feud with people to e people. Ah, because we in this case, we just look at the white component. So there&#x27;s becomes negative one or four pi ups or not times three micro corms. They will lie five centimeters squared and times 12 over 13. So this gives us the why. Component. Okay. And a minus one over four pi absent a lot times 7.2 micro forms. Do you want my 12 centimeters sweat and times five over 13. So the final value is 1.17 times 10 to the seventh. New temple calm, okay?

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Questions 1, 2, and 3 are short free-response que…

Problem 2 Easy Difficulty

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