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The force exerted by an electric charge at the origin on a charged particle at a point $(x, y, z)$ with position vector $\mathbf{r}=\langle x, y, z\rangle$ is $\mathbf{F}(\mathbf{r})=K \mathbf{r} /|\mathbf{r}|^{3}$ where $K$ is a constant.

(See Example 5 in Section $13.1 .$ ) Find the work done as the particle moves along a straight line from (2,0,0) to (2,1,5)

Work done $=K\left[\frac{1}{2}-\frac{1}{\sqrt{30}}\right]$

Vector Calculus

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so this question were given the force acting on an object or a particle. And we're given our s o the forces given by this equation re breath of ours. He quoted k r divided by the magnitude of our cube. And we're given that ours the position Victor. It's a function of X, y and Z and X, y and Z and Turner functions off position. I'm sorry time and then we're given that the path is from 20022 and flat. So it's a straight line segment and then we're of course what? We don't want me procedure is we parameter unsatisfying segment. So first of all, we're going to determine the vector that's parallel to this line segment. So we just subtract x minus x one y tu minus what I wanted to see to my New Zealand. So two lines 201 minus 0050 is five. So we get this vector right here and two para mature eyes X y and Z X is just x one plus e t. It is, uh, two plus zero times to you, which is to why is why one plus BT, which is zero plus one t or just won t and C is C one plus C t, which is zero plus five tier. Just five teeth and again key ranges from 0 to 1. Um, so if you plug in one you get your best are people again? Zero You get your initial point, which is true. 00 If you plug in 1 40 you're going to get 215 which is your friend. All right. Now are is just, um eso another way to right So we can write our t s to t and 50 and that our prime is just a derivative of that with respect ity there a video to his dear their derivative of tea with respected to use one too derivative five with respect to tease Sorry. The derivative of five Cuban respective teas just by. And then here's one extra things and serve the force formula or equation contains the magnitude of our became determined the magnitude of our easily. It's just square root of two squared plus t square was far T square, which is just the square root off four plus 20 sixties where all right, you know, since we're gonna find the work. We know that the work is the, uh the integral along the curve of f d r. And a foreigner breaks in terms of t It's just integral from 0 to 1 of efta are prime. All right. So again, we know that f is just Ah, okay, okay. Divided by the magnitude off our cubes of our f is right over here. It's just just K. Yeah. So this is so this should be in our dot are crying, but t t t now. Okay, it is a constant, all right? And so we already determined the magnitude, um, more just just a square root of four plus 2060 square. We're gonna puke that. And then our position vector are is just ah to t and five t and are derivative of the position. Vector 01 fun, Dickie. And now this is just a regular dog broader. This just two times zero plus t times one plus five times. All right, so that we get this right over here. So this is 0 to 2 times 00 t plus 25 2016. So finally, get this integral right here. All right now, this integral? Um, we're gonna use the, uh, you substitution. So we're gonna sit. Are you equal to four plus 2060 square. And then if we take the drivable that we get you is 52? Um, t t t And now, um, since so this this might look a bit strange, but what we're gonna do is we can write 52 is 26 times too. So if we divide to by both sides were going to get do you buy? Do you buy? You divided by two is equal to 2016. Now, why did we break this 52 into two attempts? 26 because of you. Notice. Right here. We have a 2060 t so that we want to write everything in terms of you. So this is just to you. And here we have you to the power off. Three. Perhaps this k we can pull to the outside since it's constant. And now what we have to do is change these limits of integration, which are terms of tea into you. Well, you is just four plus 2060 sweat. So now we change the limits of integration and This is our king. So this is the U divided by two of sores 2060 VTs Do you divided by two? And we can also we can pull the chaos ever constant and we can pull this 1/2 as constant. That's why on the outside we have came divided like All right, so now we change our limit points. So we get k over to into growth, um, for to 30 of u to the power of negative 1/2 to you because three three, divided by two is just three divided by two is just 1/2. And if we want to raise it from the denominator to the numerator, so d to the power of three. Over to Mrs Sorry U to the power of three over to just U to the power of point. So no, sorry. 1.5. So it's 1.5. Then if we raise this to the numerator, just becomes negative one. All right, now we integrate this Normally, we just add once it becomes negative one negative 1.5 plus one, which is negative 0.5 and then we divide by negative 0.5 which is just negative too. So here we have to. So here we got two and two, they cancel and there's a negative which we can pull to the outside. So we get negative K and then now you to the power of negative 1/2. This is just one divided by spirit of you. And we know that are used, that the value goes from Barre before 2 30 You know, we have this spring here, and this there is a minus on the outside. We can distribute it in. So we just get so all we have to do is switch. It's when we get there. K is equal to, uh, sorry. The integral is equal to K times 1/2 minus one, divided by the square root of third. And that's the value of the work.