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Question 13ptsFino te couationcftneline tnat Paresthrouan tnepoint Ru(s 2,1) inthe directionofthe vector 3ly# the equaticnis Flt)=(),st) =() >Simplityeja Expresl...

Question

Question 13ptsFino te couationcftneline tnat Paresthrouan tnepoint Ru(s 2,1) inthe directionofthe vector 3ly# the equaticnis Flt)=(),st) =() >Simplityeja Expreslcn foryourvarjblecin Eindudeany ISCE}VnoeceeoneTnbol VaWnaioumetQuestion 23 ptsFindithn eqJationoftheling that Pazse; throuphpoints Prl, 4,16)Q(21,15,8} PQ to firdthe equation FW) I() Mt),#(0) >Usethepoint inotnevecctSimolity ejdExpresicn andcn Eindudeany ISCE} foryouryargbleCeicaN/ SmnboliMAKE Nure vou eQuestion 33ptsFindthe gcJat

Question 1 3pts Fino te couationcftneline tnat Paresthrouan tnepoint Ru(s 2,1) inthe directionofthe vector 3ly # the equaticnis Flt) =(),st) =() > Simplityeja Expreslcn foryourvarjble cin Eindudeany ISCE} VnoeceeoneTnbol VaWnaioumet Question 2 3 pts Findithn eqJationoftheling that Pazse; throuph points Prl, 4,16) Q(21,15,8} PQ to firdthe equation FW) I() Mt),#(0) > Usethepoint inotnevecct Simolity ejdExpresicn andcn Eindudeany ISCE} foryouryargble CeicaN/ SmnboliMAKE Nure vou e Question 3 3pts Findthe gcJationofthe Iinc tnat pazaithroumntne point I0, 0 0 andipinlcitote Iinc L9 > Simplit eja Epresicn undezntindudeany IecES foryourvarble unnecessary symbols Make sure Yu et Question 4 3 pts Find tha Aquationcttreling tnrolah (yt ta#- Famandiculirto cozh Fti0=X ULS r(t) . 9(). 44) > simplmejm @prezicn andcintinciudeanyeico_ DrUnncc foryour arabla 3Mbolz zurc YouIet



Answers

$$ \begin{aligned} &\text { Let } P \text { be a point, at distace } r>>l \text { and at an angle to } \theta \text { the vector } l \text { (Fig.). }\\ &\text { Thus } \vec{E} \text { at } P=\frac{\lambda}{2 \pi \varepsilon_{0}} \frac{\vec{r}+\frac{\vec{l}}{2}}{\left|\vec{r}+\frac{\vec{l}}{2}\right|^{2}}-\frac{\lambda}{2 \pi \varepsilon_{0}} \frac{\vec{r}-\frac{\vec{l}}{2}}{\left|\vec{r}-\frac{\vec{l}}{2}\right|^{2}} \end{aligned} $$ $$ \text { Also, } \quad \begin{aligned} &\left.\varphi=\frac{\lambda}{2 \pi \varepsilon_{0}} \ln |\vec{r}+\vec{l} / 2|-\frac{\lambda}{2 \pi \varepsilon_{0}} \ln \mid \vec{r}-\vec{l} / 2\right] \\ =& \frac{\lambda}{4 \pi \varepsilon_{0}} \ln \frac{r^{2}+r l \cos \theta+l^{2} / 4}{r^{2}-r l \cos \theta+l^{2} / 4}=\frac{\lambda l \cos \theta}{2 \pi \varepsilon_{0} r}, r>>l \end{aligned} $$

Let P. Is the point. Yeah. At our distance. Uh huh. Which is much greater than an anger at an angle theater with vector elders. Sorry, L electric field at P is given by Land up on two by absolutely not. Are vector plus and by two vector divided by its square off magnitude plus land up on two pi absolutely not Are better -L by two better divided by its magnitude. So it could be written it's yeah. Clammed up on to buy absolutely not. Are vector plus and why to make her upon our square plus any square by four plus hard and course of theater plus land up on two pi. Absolutely not. Are we clear minus and why to be a terrible our square place and square by four minus R. L. Castrato. So after simplification you will get an electric field to be Land up on two by absolutely not. And upon artist squared minus to end are vector upon argue course of treatment. Hence it can be written it's even after physical too lambda upon to buy. Absolutely not artist square hardest square for our is greater than no potential activity will be Land up on two by absolutely not none of are vector plus Held by two factor minus lambda. Up onto my absolute note none of are vector minus L. Y. Director. So finally it can be written us lambda upon to buy absolutely not a lot of R squared plus are in because of the surplus any square wait for upon our square minus are in because of the time Plus early square by four That is lambda and cause of twitter upon by absolute not art fun or is much greater than that. So thanks for watching it.

Given that P one is equal to 57 negative one and P two is a multitude not making, too. We want to describe the vector p one p too, using I, J and K. All right. So first. Well, let's just kind of remember what people want to p two means bowl. That means if we take the point p one and we add the vector, he one p two to it. We should land on p two. So since we know what he wanted, Pete, you are Let's just go ahead. It's attractive. People in over get P one. P, too, is equal to P two minus p. What now? Let's go ahead and plug in P one too soapy, too. And I'm gonna write these Justin vectors as opposed to points so to nine Negative Too minus. And then p one is 57 minus one so we could go ahead and subtract. Each of these component wise will do to minus five, so that's going to give us minus three and then pulls a tract. Our why components of nine minus seven gives, too, and then, lastly will do negative to minus negative one. So that's just going to be negative. All right, So now we have this here using the bracket notation, And now what we want to do is rewrite it. Using I j. K. So remember, the X component is our eye component. The wide components are J. And the Z component is R k component. So this is just gonna be negative. Three I plus two j minus k. So this here is that vector written using the I J K notation for a vector.

Hello and welcome. We're looking at Chapter 12 Section two. Problem. 18 were asked to take these two points p one and p two and find the vector P one p two. We need to find that vector and standard unit vector form to recall the standard unit vectors. There's three of them. Oh, I that is just the X component. So it's a one on the ex con zeros everywhere else and there's J that's zero on the X and then the one on the why. So that Jay is just a UAE component. So that's all you need for two D. But this is obviously a three d problem. So we also need kay Que is just the Z component, the third component. So these are all unit vectors. If you calculated the magnitude of each of them, they do one. So it's a standard unit vectors. There's three of them for three D for the three d case. So before we can use the standard unit vectors, we need to find the component form of the vector formed from these two points. So the way we do that is we do terminal minus initial, So the terminal point is the second point to end point P two in this case in the initial point of P one eso for each component 1st 2nd 3rd or X y z. If that's what you prefer, we do the terminal point minus the initial point, so this would be negative. Three. Is that common there? Negative. Three minus one. That's That's the terminal. My estate national. The first component the X component that would be terminal nice initial zero minus two in five. My zero simplifying that we get negative. Four Negative, too. Five. And now, when converting a component form into standard unit vector form, we just have to look at the X component y component Z component s o negative for is the ex components. That's what's gonna be multiplied by the eye. That's the X component Standard unit vector. Just be negative for hi. If you actually expanded eye out and multiplied it by negative for you get the vector negative 400 So it's just isolating the ex complaint there when we combine it with minus two j the y component and plus five. Okay, the Z component. If you actually expanded this out, did the scaler multiplication and combine them all. You'd get right back to the component for So this is just a useful form. Standard unit, vector form of the vector P one P, too. So what we did, we were given two points. We found the component form of the vector formed from those two points. And then we very quickly converted it into standard unit Vector form just by looking at the X component. The Y component busy component and assigning it to its correct, are multiplying it by its correct standard unit vector. That's exactly what we're asked to do.

To find which of these are, ah equal among P. Q. R s and to you. First, let's rewrite P Q R S and T u in terms of vectors, So P Q. That's why new vehicle, too. So we'll take Q one mine execute two or q one minus P one. So it's gonna be a negative one minus negative three for the first component. And then the second component is going to be two minus negative one. So to minus negative one here. So this is going to be equal to now. Negative one minus. Think it'd three is gonna be the same as negative one plus three. So that's going to be equal to two. And then next, the next component is going to be so two plus one here. So that's going to be three for the second component right now. Let's do the same for rs rs. So we start from our go to s, so we're gonna be three in minus one here, So three in minus one and then comma five minus two. So that five minus two here, that's when I'm people, too. So this first part is to here and in the second part five minus two is gonna be three. So that's going to be three here. Next t you. So we start at t go to you, so we're gonna do U minus t. So six minus four here. First six minus four. I mean comma and then four minus two four minus two is gonna be able to and then here six minus forests to so you have to and then four minus two is also too. So we're gonna have a two here as well. Now we can compare the three vectors, so we see that to to It's not going to be the same as 23 and 23 So these two are equal, so p Q is equal to rs, but not equal to t you.


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