"Electric field due to a line of charge A total charge of Q is...

Question

"Electric field due to a line of charge A total charge of Q is distributed uniformly on a line segment of length 2L along the y-axis (see figure). The x-component of the electric field at a point (a, 0) is given by

Eₓ(a) = (kQa/2L) ∫-L L dy/(a² + y²)^(3/2),

where k is a physical constant and a > 0.

a. Confirm that Eₓ(a)=kQ / a √(a²+L²)

b. Letting ρ=Q / 2 L be the charge density on the line segment, show that if L → ∞, then Eₓ(a) = 2kρ / a.

Diagram showing a vertical line of charge from -L to L on the y-axis and a point at (a, 0) on the x-axis.

Accepted Answer

Welcome back, everyone. In this problem, a uniform sin rod of total mass M lies along the Y-axis from Y equals negative L to Y equals L. Let G be the gravitational constant. Find the X component of the gravitational acceleration at the point A0 with a greater than 0. By summing contributions from each mass element and passing to an integral, one obtains the formula G X A equals GMA divided by 2 L, multiplied by the integral from negative L to L of 1 divided by A2 + Y2 to the power of three halves with respect to Y. No, we are told that we, we can obtain this formula for GX of A. So essentially, that's what we would want to use to help us figure out the X component of the gravitational acceleration at the point A 0. But the problem is here that we will need to solve, OK. 4 GX of A in terms of a purely in terms of AL and or constant GM. So we'll need to integrate our expression here between negative L to L. So, let's try to integrate that by using trigonometric substitution, and then we can solve for GX of A in terms of those variables. So by trigonometric substitution, OK, by trigonometric. Substitution Let's let why. B equal to a multiplied by the tangent of theta, OK. Well, in this case, that means DUI. Is going to be equal to a 62 D theta. That's what happens when we differentiate with respect to theta and then multiply both sides by the theta. And if that's the case as well, since Y equals a theta, then A2 + Y2 will be equal to a plus a theta 2. And we can factor out A2 then to get A2 multiplied by 1 plus times square theta. 1 plus times square theta is 62 theta. So A2 + Y2 equals A2 multiplied by 62 theta. And with that, that means that a2 + Y2 to the poor of three houses will be equal to a 6 theta to the poor of three halves. And when we simplify, OK, then we end up with a cubed multiplied by 6 theta cub. No. If in this case, OK, we can use this to transform or integrate. Because if we let I be equal to our integral here from negative L. The Of the Y divided by A2 + Y2 to the 4 of 3 halves, OK. Then We could replace DUI divided by Square plus Y2 to the part of three halves with our terms that we've just found, OK? With DUI that we've found. And with a 2 + Y squared to the power of three halves. But the problem here is that we'll also need to change our limits. So when Y equals. Negative L. OK, when Y equals negative L, then based on what we know about Y being equal to a a theta, that means theta will be equal to negative alpha. On the other hand, when Y equals l. Then theta would be equal to positive alpha, OK, where we know that alpha is defined as the inverse tangent of L divided by a. So now with all this information, this implies, OK. That I is going to be equal to or integral between negative alpha. Eh? The alpha of a 62 theta with respect to the DUI divided by a cubed the cubed, which is A2 + Y2 to the power of three halves. OK. And now when we simplify that, we end up with 1 divided by Aqua. OK. Undivided by a squared. Multiplied by or integral between negative alpha to alpha of class theta with respect to theta, OK? Because notice here A goes into a cubed, a squared times, OK. And now 6 squared theta goes into 6 cube theta 6 theta times. And the reciprocal of 6 theta equals the cosine of theater. So that's how we ended up with this expression. So now we need to integrate cast the with respect to theater. When we do that Then I or integral will be equal to 1 divided by a quad multiplied by sine theta between negative alpha to alpha. And that would be equal to 1 divided by a quad multiplied by sine alpha minus the sine of negative alpha. Which leaves us with two sine alpha divided by a squared. So now we're almost there. All we have left to do is to express sine alpha in terms of A and L. Now, since Tan Alpha. Equals l divided by A. OK, then that means the sine of alpha would be equal to tan alpha divided by the square root of 1 plus 10 square alpha by definition of sine alpha, and that is going to be l divided by a divided by the square root of 1 plus. L squared divided by a squared. And that would simplify the L divided by the square root of A2 plus L2. So this means then that our integral is going to be 2 divided by a 2 multiplied by L divided by the square root of a2 plus 12, and that will be equal to 2 l divided by Aqua multiplied by the root of aqua plus L2. Know that we have our integral, we can substitute this back into GXF A. And simplify because no, this means that G X of A remember we were told earlier that we we got the formula by summing contributions from each mass element and passing it to an integral. And now that we have this integral, then g x of A is going to be equal to GMA divided by 2l multiplied by the result of our integral. That's 2 l divided by a squared multiplied by the square root of a 2 plus L2. And now notice two L cancers. 2 L A goes into a squared A times and that leaves us then with GM divided by a multiplied by the square root of Aqua plus L2. Thus, this is going to be the X component of the gravitational acceleration at the point A 0 with a greater than 0. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Electric Field Due to a Continuous Charge Distribution

Definite Integral and Substitution in Calculus

Limit Process and Charge Density Concept

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