for-each-of-the-questions-46-50-two-of-the-suggested-answers-will-be-correct-select-the-two-answer-4

Question

For each of the questions 46-50, two of the suggested answers will be correct. Select the two answers that are best in each case, and then fill in both of the corresponding circles on the answer sheet. 
Two spheres are separated by a distance d. The first sphere has a known positive charge. The second sphere has a charge of a magnitude greater than the magnitude of the charge on the first sphere but unknown sign. Given this information, at which two of the above locations could a positive test charge potentially experience 0 net
electrical force? Select two answers.
(A) A
(B) B
(C) C
(D) D

Vishal Gupta · Accepted Answer

Hi in the given problem here there is a positive charge Cuban and another Unknown charge. You two means its nature is unknown. It may be positive or it may be negative. Then there are three observation points .8 outside Cuban won&#x27;t see exactly in the middle. Want to be towards you. one means close to Cuban and point D. Close to cuba. And here we have to find the point at which charge a third point charge may experience zero electrostatic force. So using the expression for the force experienced by a charge when it is kept in an electric field that is given by the product of charge with the electric field. So for this force to be zero, necessarily electric field at that point should be zero. So actually we have to find the position of zero Electrical intensity due to this distribution of charged particles. A system of two charged particles. So There are two cases, guess one if you do is Positive and it is given that the magnitude of Q two is more than Cuban. This is given here in this problem. So if you do is more than Cuban, then Electric field will be zero at any point between them but not at sea because the magnitude of child vertical is not seen definitely. The electric fields will be opposite in direction at every point, but they will not be equal in magnitude at one C. Because the charges are different magnitude and this we are seeing using the expression for elective fields okay into cuba are square, so for them, middle point the R will be same but as the queue is different, so electrical will be different. So even after having the opposite direction they cannot be same in magnitude. So they will not be cancel each other. So electrically cannot be zero. Now ask you to is more So as the charges more so that the distance should also be more in order to have equal electrically. So as Q2 is more so to stand. So for observation point at which electricity zero, the stance of observation point from You two should be more means that is actually going to be there. The electric field maybe zero. So this is the first option at which dielectric field and hence the electrostatic force. Maybe zero. Now case too, if you do yes, negative in nature, then definitely electric field cannot be zero at any point between the charged particles. It will be zero at any wind outside that charge, which is smaller and magnitude. And such a point is point A. So another option for this given problem is a. Hence we can say here of options E and be are correct.

Ze-Han Lee · Answer

So in this problem we have to charge for years, why is the policy of a Q and the otherwise? We know the magnet. We know that magnitude. It&#x27;s not your thank you supposes Q. Three. But we didn&#x27;t understand the Q. Okay, and then we have four positions. A. This is. See, this is a B and there&#x27;s the D. Okay, so I want to know that. Ah, what what? A two possible cases. If we put that away place and that charge, Ah, to the full location. So that electrophoresis zero. So suppose Q is negative, and this is ah, if we place any charge inside. Ah, like the electric field inside is always pointing to this direction, right? So, well, a, um, for the first, it&#x27;s not real. If q&#x27;s negative, then we should discard the possibility of the B C D right, because any position inside will be pointing to this direction. So the only the only possible choices? A. It&#x27;s possible. And for the seconds now, really, if Q is a positive, then we can no place the charge at 0.8 right, because otherwise the electric field appointment were pointing to this direction. But Ah, bcd their possible. And because cues and larger and cue that smoke you so and that says we should to place the charge closer to the small Q. All right, so it should be. This is so only two possible locations are gonna be okay.

Kristela Garcia · Answer

and this problem. You have two spheres with a charge of four Q and negative to Q, and this produces an electric field in between the two spheres. Now, because this is a positive, then the electric field, due to the charge of the left is going away from it. And then because this is negative, the electric field due to this charge is going towards it. So to find my net electric field, I can add the effects of both of these together now again. Because these are vectors. We&#x27;re actually going to ignore the fact that this is a negative. So our electric field for one charge is cake. You over r squared because they&#x27;re the same distance away from each charge. I can leave it in terms of our and then a K is just a factor. So really, we have kay Q over R squared Puss cake You over r Squared gives me my e total right in the center again. That&#x27;s not a generic equation that only works for this situation. So then I can go back and put in the values I actually have. So I have K four Q R squared and two Q R squared gives me six k q r squared, and now I want to double this effect. So instead of having six, I want to get 12. So we need to do this in terms of the values that were given and for example, part A. If we increase Kyu won by a factor of four. So if we have Kyu won and we quadruple it, I get 16. And for the next one I get I can actually just leave that one as is, and I get 18. So this actually didn&#x27;t quadruple it. So then I can look at the second situation which is increasing. The magnitude of kyu won by a factor of 2.5. So if we get four and we multiply it by 2.5 and we add two, then I get 12 que que over r squared. So we did double our electric fields. Obi is in fact, one of my answer choices. But it does say that we will have two correct answers so we can keep going for part C and asked us to increase the magnitude of cute too. So that means the 1st 1 stays the same which is just four and then our 2nd 1 actually is multiplied by 2.5, which gives us an electric field of nine K Q over our square again. Here you can see we did not double. So see is out and I&#x27;m gonna do de appear at the top, increasing the magnitude of queue to buy factor for so again the 1st 1 stayed the same. So we leave for and then the 2nd 1 is now multiply by four. So this is four times two gives me eight eight plus four gives me 12. So this one also doubled compared to my original, So d is another one of our correct answer choices.

Athiru Pathiraja · Answer

So, given that the radio first years are one with cars on the first years, Q one, the radius of the second spheres are too. So if you will want to part a field on the surface, the electric field on the surface of the first year can be represented by the following equation. So cake you, uh, one over our Warren squared to the potential on the surface of the first fear. She also okay, you won. Those aren&#x27;t one, and it&#x27;s equal to E. R. It&#x27;s a non. If you move on to Part B, none of the stairs two stairs or joining charge flows from one to the other until it potentially are equal to let the new charges the he wants you to, respectively. And the potential in the first fear is be one x equal to K. She learned of the Q two over radio. Sorry, Andi Potential. On the surface of the second fear, B two is equal to K. You, too release especially asks about that in this series of the second. So now if you said both of them equal to each other. So if he said, B one, the long is equal to be to. So therefore, we can get an expression that to one of a Q two equal to our are one over hard to and charges the two charges qihua ask you to equal to charged upon. Therefore, if you saw a cute one gonna be Teoh capital to one times are one over R one plus party and now for charge to as well, just you are. And now it&#x27;s gonna be also, if you won, times are choose. Sure are one, aren&#x27;t you? Yep. And now, if you move on to part C to the potential on the surface off the first year, as he said, is equal to okay, Que Sorry, Q Simple que the first charge two over our And now we can replace key one by the expression from part B. So it&#x27;s okay to on over I want to start. You find the potential On the second surface of this is the greatest surface. Second surface is the same thing, so Q one you cakey to our And now if you replace it with the expression from part B, we&#x27;re going to get that key. Has Q one it over r one plus R two not moving on the party. The electric field on the surface of the first year you arm is equal to We want over our are one. So now we know that this we can replace both to be on the beginning, face to be from apart from apart. I see. So now it&#x27;s just equal to K times if you want. So it&#x27;s just thinking this whole thing down here. So the expressions from party just straight down into electric electric field. But except this time, it&#x27;s multiplied by an ex radius as part of the equation. So now it&#x27;s just our one time and fantasies arms us, aren&#x27;t you economic? You will want to the electric field, too. The one be to over our two equal to que que I forgot. I forgot about Cuba on here because the electrical like to feel on the second on the first surface. Now. No, this is that I could feel the second surface, and that&#x27;s right there to here. So now it&#x27;s once again the same thing. So it&#x27;s are too. Times are loan plus are to in fact deceased whole bracket. That&#x27;s it for a district for this problem. Thank you for listening

Darshan Maheshwari · Answer

let&#x27;s talk about this question were given that the metal sphere release Irwin has a charge of Cuban and we have to take a potential to be zeroed in finance distance from the sphere part a talks about what is the electric field and electric potential. The surface of the sphere. So if this is a sphere having a charge of Q1 and some metal sphere, so definitely everything will be on the surface. That&#x27;s the property of the method. If we talk about the electric potential or the electric field that anywhere on the surface then discharge, actually spear actually behaves like a point charge at a distance of our one from that point. So, we know that if we talk about the electric potential that&#x27;s going to make you and over four pi apps not are one. And this can be written as KQ one over R one. So this is a required potential. Uh, and if we talk about the electric field, that&#x27;s going to be Q 1/4 pi apps, not our one square. So that&#x27;s gonna be candidate and ask a Cuban over our one square. So that is required electric potential. Uh, there&#x27;s required of theirs required electric field and electric potential of the surface. Another sphere, the sphere is now connected by a long term conducting wire to another sphere of videos are too, there&#x27;s several meters from the first fear. So there&#x27;s a reason why they are saying that it is several meters from the first fear, I will talk about it. And uh, for a couple of minutes before the connection is made. The second spheres uncharged. So let&#x27;s, let&#x27;s see how this looks like. So, we have uh, this is the first fear with Cuban and our one and this is the thin cable with which we are connecting another sphere of some unknown charge Q. And radios are too. Now after the electro starting equilibrium has reached what are the total charges on each fear? All right. So, uh, since we are connected with the conducting A wire, it means that the charge will flow from Q1 to the other. Uncharged is fear Q. This is uncharged. I&#x27;m really sorry that there&#x27;s no queue here. Uh, use such that they&#x27;re both potential remains the same. Now, the point is over to Cuba and the potential is definitely due to Cuban and due to the sphere. But since they have said that the this several several meters from the atmosphere, which means that we can ignore the effect of discharge over here and vice versa of discharge over here. So, at this surface, the only potential is created by the charge to one and the same is applicable for others. Fear as well. So, let&#x27;s say the X charge moves from here to here. So, the new charge honor, that&#x27;s gonna be Cuban minus X. And the charge on this is gonna be excellent. That was in charge initially. So this happens or until when their potential becomes same. So, KQ wine, Cuban minus X. Over our one must be equal to K X over are too so Case cancer, this is a proportion that has to across modification here, we have Cuban -6 times R two is going to be equal to X times are one. Uh open up the parentheses here. So we have Q one or two minus R two. X is equal to X. R one. Let&#x27;s bring this one over to the right. So we have Q one or two is equal to X. S. Stick. And as a common factor we have our one plus R two and the final value of X comes out as dividing both sides by R one. R two. We get Cuban are two over R one plus R two. So this is the charge which has flown from this office one to this office to uh we need to find a total charge on each fear. So this is the charge on sphere to and the charge atmosphere one is going to be Cuban minus six. So that&#x27;s going to be equal to introduce some math over here. So Q one minus Q one, R two or R one plus R two if you take the common calcium and simplify this, that&#x27;s going to come out as Cuban are one over R one plus or two. So this is the required charge on fear lot uh in a charge. And now uh party asked about electric potential at the surface of each fear and the electric field at the surface of each fear. So electric potential will be seen for both the sphere because that&#x27;s when the charging charge flow stops. So we can just find the electric potential at anyone, anyone uh sphere. So let&#x27;s talk about sphere two. So that&#x27;s gonna be K. X. Or R. Two. So that&#x27;s gonna be K times X. X. Is uh we already found the X. Data as Q one R two over R one plus hour too. And this denominator are two remains as it is. So these two are cancer. So the potential consort as we of each will be KQ one over R one plus R two. Uh And then we need to find electric potentials electric potential our sphere one, I&#x27;m sorry, the electric field or potential one, that&#x27;s gonna be K Q. The Q. Is the new charge, right? Let&#x27;s call it Q one dash KQ one dash over our one square. And this is the value of Cuban dash. So that&#x27;s going to be equal to okay, uh, times Q one R one or R one plus R two Times one or or one square. So that&#x27;s going to be equal to Q one over are one times are one plus or two. So that is the electric potential. There&#x27;s electric field art sphere one. And if we talk about sphere to uh, if it&#x27;s all the words fear to, that&#x27;s going to be with the same formula. That&#x27;s gonna be K. X. Or R Two square and excess Keio where are two square remains as it is an excess Q one R two Q one R two over R one plus or two. So that&#x27;s going to be equal to K. Cuban Over our two times are one Plus R two. So this is the net electric field. Uh, He, too, thank you.

Ben Nicholson · Answer

24.62. We have two spheres. Uh, one is charged initially with a charge Q. The 2nd 1 is twice as big Radius is This place is big depending on how you measure the bigness of the spirit of its volume. It&#x27;s like eight times as big. But anyway, it&#x27;s initially uncharged. Then we connect them with a thin wire that is too thin to retain any charge. And we want to see, uh, you know, how did the voltage is on each sphere? Compare. And then we want to figure out what fraction of that charge is on, sir. One and is on severe, too. And then find the ratio of their surface charge densities. So a, we have two conductors are taken a single conductor rather because, you know, the sphere is connected with wire to the other sphere. And this is all one big conductor. And so they&#x27;re voltages are equal now for, uh, B and C, the, uh, voltage of sphere one is going to be Q one provided by just use. Coolum is constant. If you won over our one, it&#x27;s going to be equal to be too just take you to over our two, and we further have you one plus que tu I have to add up to Q and that are two is to our one. So we can then sulphuric you want. And you too. Uh and we find, um Q one over Q is equal to 1/3 and cue to over Q is equal to 2/3. So this is B. This is C Hey, now d the surface charge density is, of course, just the, uh, charge divided by the surface area four pi r once weird bye bye or hi R P squared. So then this equals Q one over a cute too. Times are too over are one squared. So then putting in the fact that we know what multiples of each other all of these things are, we find the ratio of the charge densities is too

Problem

For each of the questions 46-50, two of the sugge…

Problem 49 Easy Difficulty

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