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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 $ \textbf{r} = \langle x, y, z \rangle $ is $ \textbf{F(r)} = K \textbf{r} / | \textbf{r} |^3 $ where $ K $ is a constant. (See Example 16.1.5) 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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Johns Hopkins University

University of Michigan - Ann Arbor

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Idaho State University

So here, let me explain the new concept. Yes, I know that this kind of work down by a force problem, you can do it by computing's F that they are which you first, right out the F then Asai right here and you feel are prime. But you can easily get by the para militarization off the off the kerf, which is given by a settlement. So he's use a standard perma authorization. You can come to the dock for the off them and for X y Z were replaced by two tea and five and maybe a few steps of U substitution will gives us answer. However, there's a much easier for for this problem. However, your force field can be returned a potential the radiant off some potential function Isa grating off some scaler function. Not all force field can be read like that is a very special kind of force field. In this case, you can easily verified that the radiant off this scaler function is this force field, or is this battlefield that will just tell you that it's the work down? This simply is a different sobs officer potential function. So hey, embark on borrowing the concept from the next section. But I think it worth mentioning it now because if you go through this long process, which I did a lot of times before, it will be unnecessary for for a problem like this, so f off to one. So answer Well, simply beauties and nice things about this is it doesn't have to be a straight line. You can go through any passes. Assume was your sign. Point a and point. It's Asem's. That energy is always like this and that Let me emphasize that there is on ly. It only works for for the force field that can be written as a potential. If it is not the case, you cannot apply this result So this will be negative K over twenty five plus four plus one thirty minus that k over too, which is OK one over to minus one over square root of thirty. And if you want, you can afford to simplify that