5

Two identical positive point charges; +Q. are placed at two of the vertices of an equilateral triangle of side length a_ Derive the expression for the magnitude of ...

Question

Two identical positive point charges; +Q. are placed at two of the vertices of an equilateral triangle of side length a_ Derive the expression for the magnitude of the electric field at the third vertex of the trangle Your answer can only contain the symbols Q a, and ke (the Coulomb constant) and numerical [actor:

Two identical positive point charges; +Q. are placed at two of the vertices of an equilateral triangle of side length a_ Derive the expression for the magnitude of the electric field at the third vertex of the trangle Your answer can only contain the symbols Q a, and ke (the Coulomb constant) and numerical [actor:



Answers

Equal charges $q$ are placed at the vertices $A$ and $B$ of an equilateral triangle $\mathrm{ABC}$ of side $\mathrm{a}$. The magnitude of electric field at the point $c$ is $\ldots \ldots \ldots$ (A) $\left(\mathrm{Kq} / \mathrm{a}^{2}\right)$ (B) $\left.(\sqrt{3} \mathrm{Kq}) / \mathrm{a}^{2}\right)$ (C) $\left.(\sqrt{2} \mathrm{Kq}) / \mathrm{a}^{2}\right)$ (D) $\left[\mathrm{q} /\left(2 \pi \mathrm{t} \varepsilon_{0} \mathrm{a}^{2}\right)\right]$

16.16. So we have an equilateral triangle, uh, lists equal charges positioned around its center. So we want to show first that the electric field in the center is equal to zero. Then we want to find an expression for the potential with the center of the triangle. Then we want to explain why the potential is not zero, but the electric field is. So the reason that the electric field zero here is because of the symmetry of of this, um, set up at the the center point, we're going to have an electric field directly downward an electric field like this. That's, uh, and declined at 30 degrees and another one like this that's also inclined at 30 degrees from the horizontal. So the x component of the electric field there if we say these all have a magnitude of e one or something. So for X, we have just these two. So we're gonna have e one co sign of 30 degrees minus because this isn't the negative extraction. He won co sign of 30 degrees. And so obviously those cancel out than in the wider action we're going to have, um, two times e one times sign of 30 degrees, just the white components of these two. And then from that we're going to subtract anyone because this is pointing straight down. And, you know, the sign of 30 degrees is 1/2 and so it was also canceled. So no electric field at the center now to find the potential. This is just the sum of the potentials from each one of these at this distance away. And now this only depends on the distance away and not on the direction of anything. Some over I take you, I over our I. But all of the cues and distances are the same. And so this is going to be three. Take you over a now The reason for this being non zero here, first of all, these don't contradict each other because recalled it, the electric field is given by the derivative or grady in of the potential. So the potential is going to have a maximum or a minimum here, so the radiant will still be zero and the electric field will be zero. But this has to be non zero. If we say the potential is zero out an infinity because We have to do some amount of work to bring a test charge to this location, and the potential tells us how much work a test charge has to have done on it to bring it to a particular location. So you've been in a region or a point where the electric field zero. If it still took some work to get there from infinitely far away, the potential is going to be non zero.

High. In the given problem there is an equal lateral trying the Yeah this is the equilateral triangle which is having two equal in magnitude charges. The two charges which are having same magnitude As 12 nano colon side of this, equilateral triangle Is given as a sequel to 2.0 m. I suppose it is named as a B and C. In the first part of the problem. Both of these charged particles which are put at the and see are positive. So if both the charged particles are positive, then electric field due to both of them at the vertex that will be going away because electrical goes away from that positive charge. So here this is electrical due to be and here this is electric field due to see. And as the site of the equal little triangles are same is the distance of this observation point from both of these charges will be seen each and the charges are also same in Madrid your hands. We can say these two electric fields. This is E. D. And this is E. A. They will be same in magnitude And as each angle of equilateral triangle is 60°. So these two electric fields will be having an envelope 16 degree between them teed up Physical to 60°. Now to find the net electric field here at this point. First of all we find individual electric fields. E. A. Is equal to E. B. Which will be given by K into Q. By dispense a square. No, the net electric field will be given by triangle law of tradition, which is the square root of kick you buy He squared him the square less again square off this cake. You buy a sweat Plus two times of the product of these two. So it will also become cake. You buy a square Having a square and cause sign off angle between them, which is 60° and cause 60 degree will come out to be half means this will be one by two and this one by two will cancel this too. So finally in the square root it will be three times off key. You buy a square squared So it will come out to be kicked. You route three by a square. So now plugging in unknown values for K. Disease. 9-10 days to the power nine newton. Meet every square park along the square for charge Q. This is 12 Nano Kula means this is 12 into 10 for -9 column, divided by a square, which is two 10 m squared. So cancelling this square with colon, cancelling this meter square with this meter square and 10 days per plus nine with the Spanish for -9. Finally this net electric field here will be given by 46 .8 Newton purple. Um or approximately this is E. is equal to 47 new 10 per column, which is an answer for the first part of this given problem. Now, in the second part, it is said that if both, if one of the charges positive and another is negative, then if we redraw the figure, the diagram, this was the equilateral triangle, suppose this one is positive and this one is negative. Now This is the this is see this is observation .8. Now, as electrical goes away from the positive charge, so direction of E. B remains same as going away, but electrical approaches the negative charge, so the direction of E. C. Will be down this time. So here this is A. B. But this is easy and they will be having same magnitude as the site of the triangle and the charge, the magnitude of charge Q. Each and aside, all the things are seen. So the magnitude of E. B will still be equal to E. C. And that was equal to. And here we have made a mistake here. This was not A. This was easy. So similarly here also, there should be represented as E. C. Every year. So now the mad radio of these two again will be same, which we have found to be K into cuba A square in the previous part. And this time the angle between them will be 120°. So then we will use The value of course 120 degree. This time it will come out to be minus half. So for the net electric fields again using triangle of tradition with the direction of that electric field. And it will be given by E. B. Square which is key. You will buy a square. Having a whole square plus E. C. Square means again this is key you bye is square having a whole square. And now this will be plus two E. V. Into easy into cost eight. Acosta is minus south, which will cancel that too. So remaining will be G. Q by a square whole square. So we will cancel these two. So the remaining will be just kick you buy a square. So plugging in all the known values here. Now For K. This is 9 to 10 days. 2.9 newton meters square parque colon square. Again, four charged. This is 12 nano column or 12 into 10 hr minus nine colon Divided by the distance. The square of distance which is 2.0 m is Squire. So finally this time the net electrical comes out to be 27 newton per column, which is the answer for the second part of the problem. Thank you

Kenya has asked us to find a charge. Okay? Force on the charge any one of the charges. Okay. When three point charges are located at the waters of an equilateral triangle. So we need to find out force due to any two charges on anyone chart. So here we will find out the force on our charge. Looking at point C. Okay, so how will we do that? Let's see. Initially we need to see force on C. Due to uh this will be called okay, Q one Q two by our squad. Okay. Now the question has told us that that length of that triangle sides is a. So okay here the Point charges out there. So let's give them charges of 1.6 and 210 to the power -19 each. So I think he wanted to do the same. This will be square. And then here we will have a square. Okay, this is f let's say C. And I'll keep your Okay, this is one. Now let's come down. Okay? So here we will have helped due to force you to be Sorry. This will be quite a force on C. Who knew too. For sun. See due to be okay, So this would be equal to again K. And then you will have the same charges and then so square. And then you will have is squared. Ok? Now, since these forces if you look at these horses, one force will act in this direction and then one force will act in this direction. Okay? So there is an angle between these forces. What will be the angle of this between the two forces? This will be equal to 60°.. How do we know that? Since this is an equilateral triangle? So this is 60, this is 16 months. This will also be equal to 60°.. Okay. And in geometry we have learned that the vertical opposite angles are equal to each other, so this is 60. So this will also be equal to 60 degrees. Okay. And since forces a vector quantity, so here we will have four sufferers are returned to be equal to on the route of fc square. And then you have a f. See we square here with the lab to Yeah, CAFCB. And then yeah, we will have cause of 60°.. So this would be equal to Okay, squared 1.6. 10 to the ball -19 square. Here we have a square. Okay, the same thing for this one. So this will be K square. And then sorry, this will be equal to four and this is also equal to four. Okay, The powers the powers on both. These will be equal to force. Why? I'll just tell you in a minute. Okay, serious. They are already in square. And since the forces have been squared to the power so this Okay, but take the square power and this will take the power of four. So the Sylvia again here equal to 1.6, turned to the bomb minus 19 to the power four. And then here we will have here to the power four and then for two. Well right to here and since F A and F B are equal. So for each square there again multiply and we'll get a K square 1.6. Turn to the pound minus 19 to the power of four. Yeah, we're gonna have air to the powerful and then here we will have a cost of 60°.. This will be equal to two. K square 1.6. 10 to the power minus 19 to the power four. Here to the power four plus cost 60 is equal to half, so half. And then two cancels out. So here will have K square 1.6 10 to the power minus 19 to the power four. And then to the power four. Okay, so this will be equal to Route three times. We'll have K outside. We'll have 1.6 10 to the power minus 19 square outside. And then here we'll have a square outside since we don't know the value of A. So we destroyed it here, Route three by a the value of K is 9 to 10 to the power nine And then this is 1.6 to 10 to the power in this 19 square. Okay, so the net value of this equal comes out to be equal to. This will look something like this. 3.99 and 210 to the power -28 divided by eight square. Okay, this air will be the length of the triangle it's side. This will look something like this. Okay, this will be done. Net. There's a return force acting on the charge at any one of the water says. Okay, here I have also multiplied the value of Route three, but along with this numeric. So okay, and the value of Route three magnitude, Route 34 is equal to 1.732 But if you want to apply this with this one, you will get this value. This is there is a return to value acting on the charge at any of the water says, okay thank you.

Hello. You in this problem first have to find an expression for the electric field fell by a charge located on the Y axis, as I've shown here on the left. In the diagram generated by three charges. Cuban Q two and Q three that are located on the edges off adversary on the courtesies, often equilateral triangle off size or off size length. A silent bait. And so also given that Q one is equal to Q two, is equal to Q three. Um, is equal to queue. Okay, so the question is, what is the electric field at this point? So to find that we know that it's going to be the vector sum off the different contributions generated by the first charge. Second charge and the third charge. And so that's what I've written over here. And we also know that in general, between two charged object, the electric field is equal to, um de this inverse square law off, or the school um field that is generated by a charge. And so the direction of that vector is going to be pointing away or two words the charge generated in the field. And so that's what I've They noted. Here is this r and R hat. Okay, so first we're gonna have to find the different contributions from the three different charges. So first, what is it? The contribution from the charge you three or this thing this discharged is already lying on the Y axis. Well, to find that, we just have to figure out what the distance off discharge toothy, um, to the object on the on the y axis is or what? The distance of the test charges. And so to do that, we have to find out what this height is. That is the distance between the origin and the location off charge three. So for that we have that the ECA literal triangle breaks up into this right angle triangle over here. Where we have that one side is h theater side hypotenuse is a We have that. The other side is a over to meaning that a squared is equal to a square plus eight or two old squared by by pipe. Diverse is through. And so just rearranging that we find that H is equal to a over two times Route three. Okay. And so the distance between the object and are the test charge and charge. Q. Three is going to be the difference between the y coordinate off test charge and this height. And so that's what I've denoted as our three over here. So are three is with y minus a over 233 where this combination is just h okay. And so the electric field generated by this charge at the location of test charge is gonna be K Times Q. Because we know that the Q three has a magnitude of charge of plus Q. And the distance between that object are three on the test. Charges are three, and so that squared is why minus a over to Route three squared. And this is going to be pointing in the white hat direction. Yeah, so this is This is the first part. So now that's check. This is E three. Next, we have to find e two, which is coming from charge Q two. Now cue to have taken to be the charge lying on the positive part of the X axis. And so we have this triangle, right? So the distance between theme, the Charge Q two and the test charge is going to be our too. So I'm gonna call that are too like a have over here, and that's going to be given by the hyper two news at the right off the right handle. Right hand the triangle where the one side is why the coordinates of the test charge and the other side is the distance from the origin to charge to which is a over to keeping by to set up. And so are two is going to be equal to the square root of y squared plus over to all squared, which is just the square toe y squared, plus a squared over four. Now, this tells us what the magnitude off E two is going to be at the location off off. The test charge is going to be cake you over y squared, plus a squared over four. But we're not done. We actually have to find what the different components off that electric fields are. So in order to do that, since we've chosen are right handed Cartesian coordinate system. We can find what the X and y components off these different, uh, off this electric field are. And so we say that there is some angle that I have already denoted. Here is data in the diagram. And so that angle is going to be the same as Theano Eagle away from the UAE, access towards the negative or into the into the second quadrant off the Korean system. And so resolving that towards the X and Y axes, we find that e two y is going to be the magnitude of E two types of co sign of data and the and and and the victory to X is going to be minus the magnitude of two times of science data. Now there's a minus. Sign here because notice that this is going to be pointing towards the negative, uh, part of the off the X axis. So what is co Cynthia and what is science data? Well, given that we have this triangle over here, we see that Coast data is going to be why, over our two. But are too we forgot to be square with the Y squared, plus a squared over four. And so coast data is just why over the square root of y squared, plus a squared over four Similarly science data is going to be a war to over or two. And that works have to be, um, a or two over the square with the white square places where over four. Okay. And so now, putting these back into the expressions for, uh, for E two y Andy two x, we find that e two y it's going to be K temps que, uh, over the y y squared, plus a squared over four to the power of 3/2 in total pointing in the UAE head direction. And the E two X is going to be minus k Q A over to over y squared, plus a squared or force it to three halves, pointing in the ex head direction. Now this is the positive x x had direction, and I've taken the where the e have already taken care of the minus sign by putting the minus sign over here. But really, this is just saying that the contribution from discharge is pointing towards the negative x had direction or the words the negative part of the coordinate axis. Okay, Now, similarly, we have the same pretty much the same set up from from the first charge, you one but they're the X component is going to be pointing in a different direction while the Y had direction is going to be the same. So e one, Why is the same as e to? Why Waas? And so that's K times Q Y over y scripless eighth grade or for 2 to 3 halves. But e one X is now plus, so it's plus K Q A over to over y scrap, plus a squared or +43 halves. Proversity positive ex head direction. Okay, so these air the other parts off the puzzle. So putting everything together you could say that you have these fields for the coming from from YouTube. Thes would be the fields coming from from E one or from from Q one and this field over here. E three is the field that's generated by you three of the location off the test charge, and now we have to add up all of them. So we add up the first checked equation, this guy, this second squared acquisition and this third box decoration, and that's the vector sum off the fields and so we can do this component by component. If we add up the components of the that are pointing in the X direction. We find that they're off opposite sign but equal magnitude. And so the X component completely cancels. And for the white component, we can write it as K. Thanks Q times, uh, old brand brackets one over. Why, minus a over to Route three all squared. This is the contribution coming from Q three plus twice the contribution off y over y scrap, plus a squared or four to the three halves. Okay, so this that's what we take. And then we just simplified us to the last form that I'm going to show you here by using some of the facts like that. Four to the three halves is equal to eight. And so we find it. Eve. Why? So the white component that that is generated in the location of the charge is going to be four times G Q times in all in brackets 1/2 Y minus a freak cube or route three old school bracket or bracketed squared plus four y over four y squared, plus a squared all brackets, uh, to the power 3/2. And this is all pointing in the white head direction. So this is what you find the field to be. And so this is part eight and in part B were asked to find what happens if we're sufficiently far away from this thesis charge configuration that we can take. Why? To be much greater than a And So in that case, what happens is that these arguments or these things distance functions that you have, they're going to simplify. And so we're just gonna go to first order in our approximation here. Otherwise, you would capture effects of die poles and so on so forth, higher prior moments. So you would have So you're you're gonna have that. The two y minus eight times every 31 over squared is going to be approximately 1/4 y squared through this approximation and then the other function for y squared, plus a squared to the minus three over to its approximately 1/4 y squared, all bracketed and raised to the power through or to which works out to be, um, won over eight y cute. And so, putting those expressions into the results from part eight, we find that ive is approximately equal to four times get you. That was 1/1 over four y squared, plus four y over eight y cute. Now here we have that the Y in the top cancels one of the wives and the bottom. So there's a square here and we can take out 1/4 y squared. So then the force canceled and we're left with three times K Q over y squared, which is exactly the electric field that you would expect from a one from a single object of charge. Three times, Hugh.


Similar Solved Questions

5 answers
Wenment TVIDIOcodMhkh0' tJ=EannePrtScnhan
Wenment TVIDI Ocod Mhkh0' t J= Eanne PrtScn han...
5 answers
H,OH3C _ CH3 OHCH;CHNH;H,otHo-
H,O H3C _ CH3 OH CH;CHNH; H,ot Ho-...
5 answers
Yoc eaeh Pvnttion dele(mine hke#ker 0 ;s One _ +o- Onc 8 detr(mne whttht( f is Onto 4) F:k_Ry F(xy) X+y Dar?_ 7e by f(y)e (xtyX-Y,y) dAALARUAKep1A cecZ b(e C-{A € P(z)|a {S finitc} Gnd #(A) is H Svm of sl ekmnts of4
Yoc eaeh Pvnttion dele(mine hke#ker 0 ;s One _ +o- Onc 8 detr(mne whttht( f is Onto 4) F:k_Ry F(xy) X+y Dar?_ 7e by f(y)e (xtyX-Y,y) dAALARUAKep1A cecZ b(e C-{A € P(z)|a {S finitc} Gnd #(A) is H Svm of sl ekmnts of4...
5 answers
Find all numbers € that satisfy the conclusion of the Mean Value Theorem for the following function and interval. Enter the values in increasing order and enter N in any blanks you don't need to use: f(e) = 24x? + l62 + 10, [-1,1]NNNSubmit answer
Find all numbers € that satisfy the conclusion of the Mean Value Theorem for the following function and interval. Enter the values in increasing order and enter N in any blanks you don't need to use: f(e) = 24x? + l62 + 10, [-1,1] N N N Submit answer...
5 answers
What is an intermediate within a reaction mechanism?
What is an intermediate within a reaction mechanism?...
5 answers
A quasi-static process begins and ends at the same temperature. Is the process necessarily isothermal?
A quasi-static process begins and ends at the same temperature. Is the process necessarily isothermal?...
5 answers
Respond with true or false to each of the following assertions. Be prepared to justify your answer. Normally, this means that you should supply a reason if you answer true and provide a counterexample if you answer false.The cotangent is an odd function.
Respond with true or false to each of the following assertions. Be prepared to justify your answer. Normally, this means that you should supply a reason if you answer true and provide a counterexample if you answer false. The cotangent is an odd function....
5 answers
Givcn thc following graph of cither sine or cosine: Answer the following questions, using the same choices for each question314510/4For the function f (x) =Asin( Bx C)+D o f(x) = Acos( Bx-C)+D.4. Find A_5 . Find D.What is the period of this funetion?
Givcn thc following graph of cither sine or cosine: Answer the following questions, using the same choices for each question 314 510/4 For the function f (x) =Asin( Bx C)+D o f(x) = Acos( Bx-C)+D. 4. Find A_ 5 . Find D. What is the period of this funetion?...
5 answers
Find the vertex, focus, and directrix of the parabola. Then sketch the parabola.$$y^{2}=3 x$$
Find the vertex, focus, and directrix of the parabola. Then sketch the parabola. $$y^{2}=3 x$$...
5 answers
When a system is at equilibrium, the total force applied to th system is:
When a system is at equilibrium, the total force applied to th system is:...
5 answers
2.50 kgEs0.70, Uk 0.3080 N6,00 kg5 4 3.20-kg block is given an initial velocity 0 12,0 m/s upan inclined plane that makes an angle of 30 | with the horizontal. After it has traveled 6.00 m along thd plane, its upward speed is 2.40 m/s. Find (a) the coeffi | cient of friction between the plane and the block; (b) tha maximum distance of the block from its starting point] (c) the speed ofthe block when it has returned toits starting point
2.50 kg Es 0.70, Uk 0.30 80 N 6,00 kg 5 4 3.20-kg block is given an initial velocity 0 12,0 m/s upan inclined plane that makes an angle of 30 | with the horizontal. After it has traveled 6.00 m along thd plane, its upward speed is 2.40 m/s. Find (a) the coeffi | cient of friction between the plane a...
5 answers
Points) Find the general solution to the system of equations 60x 22y + 9e165x 61y.Both of your functions must be correct to receive creditx(t)y(t)
points) Find the general solution to the system of equations 60x 22y + 9e 165x 61y. Both of your functions must be correct to receive credit x(t) y(t)...
5 answers
Which compound is the most acidic? Note that the acidic hydrogen is underlined.OHOHMeoCompound on the left Is more acidic Compound on the [ght Is more acidic
Which compound is the most acidic? Note that the acidic hydrogen is underlined. OH OH Meo Compound on the left Is more acidic Compound on the [ght Is more acidic...
5 answers
Review Constarbicycle wheel has an initial angular velocity of 30 radIf its angular acceleration constant and equal 0 200 rad/s? what is its angular velocity att = 2.50 5? (Assume the acceleration and velocity have the same direction)Fxpress your answer in radiang per secondAEdradSubmitRequest AnswerPar BThrough what angle has the wheel turned between Expross Vour answvor radiansand2.50 $?Azq4e =rad
Review Constar bicycle wheel has an initial angular velocity of 30 rad If its angular acceleration constant and equal 0 200 rad/s? what is its angular velocity att = 2.50 5? (Assume the acceleration and velocity have the same direction) Fxpress your answer in radiang per second AEd rad Submit Reques...

-- 0.042302--