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A charged rod of length $ L $ produces an electric field at point $ P(a, b) $ given by$$ E(P) = \int_{-a}^{L - a} \frac{\lambda b}{4 \pi \varepsilon_0 (x^2 + b^2)^{\frac{3}{2}}}\ dx $$where $ \lambda $ is the charge density per unit length on the rod and $ \varepsilon_0 $ is the free space permittivity (see the figure). Evaluate the integral to determine an expression for the electric field $ E(P) $.

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 3

Trigonometric Substitution

Integration Techniques

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

07:50

A charged rod of length $L…

03:36

04:29

05:49

The charge per unit length…

06:06

A rod of length $L$ lies a…

04:22

A thin rod extends along t…

02:12

05:05

An infinitely long rod car…

11:41

A thin rod of length $\ell…

08:33

A uniformly charged rod of…

05:35

A thin rod of length $L$ l…

10:05

a charge through out of length. L produces an electric field at the point p, given by the following. Integral. And here's a picture of what's going on we have Here's our Rod and Blue of length l There's a point Pete here. We just need to evaluate the integral and just find another expression for the electric field. So really here. Let's just try to go ahead and use the tricks up and just evaluate this integral. See what we get inside the radical and the denominator. We see X squared plus B square so we can go ahead and choose X equals B tan data. That should be our tricks up. Taking a differential on the side d X equals B C can't square. Then let's go ahead and rewrite this. First. Let's actually notice that we can simplify the denominator X squared plus B square Scott and simplify this now. So we had B square tan square plus B square factor about that B square tan squared plus one in the parentheses. And we know that sequins were and then we'LL go ahead and raise this the three house power. So let me write this over here somewhere. So here will just go ahead and multiply too. Two, three halfs. And we just couldn't be cubed seeking cute. So let me go to the next page and simplify this integral. So originally, let me write the original Integral. That's the land of B for pie. Absalon Subzero X squared plus B square to the three halves the ex. Now, after we do a tricks of these limits of integration may change. We don't know what they are yet. Let's just call them fate a one and a two. And then we could actually go ahead and start factoring out. So let's simplify. We have let me actually make some more room here, pull up the constants. We have pi times bees going Pull the bot for pie. Absalon, Subzero. So, basically, just pulling this out just a constant. No exes there. Then we have the integral. They don't want to stay, too. We pulled out the Lambda, so we're left over with the eggs, which is B sequence, where data data and on the bottom, we're just left over with. We simplified this in the previous page. It was just be cute seeking cubes there. Now, here we have b squared on top. Be cute on the bottom. So we're left over with one over b. So that will give us a B here in the bottom. And then we're just left with the integral fatal one data too. And then we just have one over C camp. We know one over. Seek it. This is just coast on data in the general. That is your sign. Spread those constants absolutes of zero. And then we have the integral co sign, which is signed. Then we should technically put our Fada one and data to. But since we don't know what those data values are, we can draw the triangle, use the triangle to go from sign back into the variable X. And then we could use our original limits, which for? Minus a and L minus a. So here, Lester on the triangle. We had originally X equals B tan data. That was our tricks up. Sulfur. So here's our data. So tangent opposite over a Jason X over b Call the hype on his age. Then by Pythagorean serum we have a square is X square plus B square and then solving for age. It's square room escort. These four. So now sign is just X divided by H so we can go ahead and replace scientific with this expression here. And then we can go ahead and go back into our original limits. I'm going to say and minus a So let's go ahead and write that on the next page. You still have our constant out here. And then we evaluated, signed back in terms of excusing the triangle and then were a lot to go back into our original limits since we're back in the Variable X and then out of the last thing to do is just to plug in thes leper. Lower and upper limits is X. There's our land. Ah, for pie trio Excellent, Obie. And then plugging in l minus a. We have l minus a up top square room, a minus a square plus B square minus. But then I'm subtracting on minus eight or so. Let me go out and actually turn that into a plus. Because of that double negative, Then I have a A squared plus B squared in the bottom on the radical. And so this is a different expression for the electric field. So really, we just use the trucks have been evaluated the integral, so I would stop right here

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