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04:27

Kai Chen

(I) A 110-kg tackler moving at 2.5 ms meets head-on (and holds on to) an 82-kg halfback moving at 5.0 m/s. What will be their mutual speed immediately after the collision?

00:56

Donald Albin

I) How much tension must a rope withstand if it is used to accelerate a 1210-kg car horizontally along a frictionless surface at 1.20 m/s$^2$ ?

04:39

Muhammed Shafi

(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?

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next, we'll revisit our example of the charged ring. In this case, we have some ring of wire with the radius of A and a total charge of Q. We want to find the electric field. Before we had made an argument, we broke this ring up into little pieces like this and said that the electric field would have some direction like that. But there will be a part on the opposite side of the ring whereby symmetry. We'll call this E prime. By symmetry, the non X components would cancel out, and it was a sound argument, and we did get to the correct answer. But now that's where we work this problem. And this time we're going to be given nothing aside from what we're given already and, of course, the potential of a point charge, which for reference, I'll just put here this cake over our mhm. So now we break this ring into tiny little pieces, each of charge Dick, and we know that no matter where we are on a ring, we are at some distance square root of a squared plus X squared. So now our devi, for this little slice of the ring. It's simply K dick over that distance. And when we integrate, we get the full potential. And, of course, these terms are simply constant, so they come out of the integral. And when we integrate over Dick, we just get the total charge. Q. So we've got the correct potential that we expected. And now the electric field is simply the negative radiance of the potential. And there's only an ex dependence. So it's simply del vih del X in the X direction. And when we take this integral this partial integral with respect to X, we only need a next. We'll revisit our example of a charge string, which is simply a ring of wire, what some radius will call A and a total charge of Q that's distributed evenly around this wire now, before, when we calculated the electric field, we did so using by breaking up the ring in the little pieces of charge and say, we take this charge. We knew there was Samy in that direction, but when we took this charge, we had some me in this direction. And so as we go around the ring, we saw that the non X components would cancel out. We picked this piece here it have some y and Z components, but they'd be canceled by this piece on the opposite side of the ring. And our logic was sound. But let's now calculate this from the very beginning. And so all we'll all will be given is the potential of a point charge. And as we all call you point charge his cake over our. So now we can take any piece of this ring, it's gonna have some charged que and we know the distance from any point on the ring to our point of interest is the square root of a squared plus X squared. So our Devi is simply ACDC over that distance. Integrating this gives us the full potential. And of course, these terms air just Constance. So they come out. And when we integrate over the entire ring for Dick, we've got the full charge. Q. And so we've gotten the correct potential that we expected. Now the electric field is simply the negative radiant of V. And since there's Onley, an ex dependence in our potential, it'll be negative DLV del X in the X direction. So we take the derivative partial Jared with respect to X of our potential. And the cake was constant and taking the derivative the denominator. We get a factor of negative one half, so we'll change that to a plus sign. And this becomes a squared plus X squared to the three halves and then by the chain rule. We take the derivative this term and get two x on the extraction and what we simplify. We get que que x over a squared plus x squared to the three abs in the extraction. And so we recovered exactly what we expected to find for the electric field generated by a charge ring on it's central axis. Now we look again at the infinite charge sheet with some surface charge density of sigma. Now we can do what we just did where we first calculate the potential by breaking up breaking this up in a little point charges because this is infinite and we have infinite point charges which will give us infinite potential, and we might be able to work some tricks around that. But let's more worry about making sure that our method of calculating e from V works so here I've written what we know to be the potential of the sheet. And now we calculate the negative radiant of this potential in order to find the electric field. And since the radiant depends just on X, will Onley have an X component? And right away we see for X greater than zero, this is equal to all the negative signs will cancel each other out. All right, I like that to be a little less confusing. And the integral of X with respect to X is merely one in the rest of Constance. And of course, this is in the extraction, which we know is the correct field for X less than zero, then our electric field. Negative DDX of Sigma X or two Epsilon not in the X direction gives us negative Sigma over to Epsilon, not in the X direction. And so we've recovered the correct fields. They have the same magnitude but opposite directions on opposite sides of the sheet. Now we consider a charge line, and again, we'll do this problem from the very beginning, knowing only the potential of a point charge is cake over our We want to know the potential at some point X. That's a perpendicular distance away from the center of this. Well, we could start by calculating our potential. And our devi for each little slice of this is simply dick. Times K Over are the distance, and we'll note that are distance now is y squared plus X squared square rooted. So now we need to integrate this to find the total potential. And we, of course, integrate from negative all over to to positive all over to. We see this is an even function. So now we'll consider the potential from a finite charged wire. This has a length of L and a linear charge distribution of Lambda uniforms throughout the wire, and we want to find the point, the potential at a point and field at some point in X distance away perpendicular early from the center of the wire. So we're only given our general equation for the potential of a point charge, and so we'll need to cut this up into little pieces of thickness D y. So the total charge would be land at the Y. We'll note that each piece is a distance. Why squared plus X squared, where this distance is Why So our TV is Kadijk over X squared plus y squared, which is K lambda de y over X squared plus y squared To find the total potential, we just integrate and we'll do this from negative all over to toe all over to in the K Weiser Constant. And when we integrate one over radical terms such as this, we get Ln of why plus X squared plus y squared which will evaluate from negative all over to toe all over to and that gives us Okay, Lambda Natural Log all over two plus the square root of X squared plus l over two squared and in the denominator we get negative about over to plus square root of X squared plus all over two squared. And so we see we've arrived at the correct potential if you will call to doing this problem earlier. So now we want to calculate the electric field from this potential which I've saved here. But I've also written in this form here where to find these variables. This is just to make the algebra as I work through it easier. Not every course were required. It'll work through this by hand but it's good to be ableto do it, so we'll note. As always, the field is the negative radiant of the potential, and in this case, the potential only has an X component. So we take the partial derivative of the potential with respect to X in the X direction, and here we'll note they'll be del X equals zero in l a. Del X is equal to x over X squared plus b squared or, in other words, x over a. And I'm gonna call this term a prime just to make everything a little neater. So negative. DLV del X people the negative que lambda For the derivative logarithms we flipped the terms or invert the term. Then take the derivative of the original term and we'll use the product rule So we get a prime over a minus bi plus. But when I take the derivative of this, I'll get a minus sign. We have a plus B in the numerator still, and then I take the derivative of the original denominator, which is just a prime. So now we can continue first, we'll note we lose factors of a minus bi everywhere, and we can factor out this a prime. So this becomes negative. K lambda a prime times one over a minus bi. Hey, plus B. Sorry, minus one over a minus bi and again, we can use a common denominator difference of squares And on top of the A minus B minus a plus B. This just becomes negative to be. And so we get negative K or positive two K Lambda B A prime over a squared minus B squared and we're not done. Now let's put in these values to K Lambda via cell. Over to a prime is X over a, which is just X over the square root of X squared plus b squared. And then here we get X squared plus B squared, which is our a square term minus b squared. We see these be squares cancel and we lose a factor of X s A lose these factors of two and we end up with after all that work que lambda l over x times the square root of X squared plus b squared. And so let's move this aside and get ourselves just enough room to finish this problem. Find that vfx when we substitute our expression for Be Back in is K Lambda Times, al over X times the square root of X squared plus L over two squared in the X direction and this is the correct formula. So that took a lot of mathematical work. Where we see that thief theory is sound weaken. Take this any charge distribution calculated as finite charges and find the potential. And from that we just take the radiant of that potential and we find the electric field. We've seen that the one stumbling block we can have on this path is when we have on infinite charge, infinite in some direction, then becomes a little more difficult and something like gauze Islam might be preferable to find first the field and then the potential.

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