A semicircle of radius a is in the first and second quadrants, with the center of curvature at t

step 1 Alright, so it should be recording now. Let's get right into the problem. So I'm just going to d

A semicircle of radius a is in the first and second...

Key Concepts

Electric Field from Continuous Charge Distributions

This involves calculating the electric field generated by a continuously distributed charge by integrating the contributions of infinitesimal charge elements. Each charge element produces an electric field according to Coulomb's law, and the cumulative effect is obtained by summing these contributions, typically in a vector fashion.

Superposition Principle

In electrostatics, the superposition principle states that the resultant electric field created by multiple charge distributions is the vector sum of the fields produced by each distribution separately. This principle is crucial when dealing with multiple regions of charges, as it allows independent calculation of each contribution before summing.

Symmetry in Charge Configurations

Use of symmetry can greatly simplify the calculation of electric fields. When charges are placed symmetrically, certain components of the field (like those in one direction) may cancel out, leaving only the contributions along the remaining axis. Recognizing these symmetries and cancellations reduces the complexity of the integration process.

Uniform Charge Distribution

A uniform charge distribution means that charge is spread evenly along the distribution's geometry, resulting in a constant charge density. This uniformity allows for a consistent treatment of the charge elements during integration and simplifies both setting up the integral and evaluating the total electric field at a given point.

Transcript

00:01 Alright, so it should be recording now.

00:06 Let's get right into the problem.

00:08 So i'm just going to draw my axes here.

00:12 I'm trying to make it as straight as possible.

00:19 Sorry for this drawing.

00:22 I don't have a mouse adapter for my mac.

00:25 So i just have to bear with me.

00:43 All right, so the problem is that we have a semicircle of a special charge distribution.

00:52 And essentially we want to know the magnetic field or not the the electric field at this at the origin.

01:01 So how do we go about this? well, from university physics, we should know that positive charges are going to emit an electric field.

01:14 So radially outwards, kind of like this.

01:22 Too close to the top there.

01:28 Radially outward.

01:33 While a negative charge will have an electric field of radially inward.

01:39 So if we look at the arc, if we say the charge is focused at the center of the arc, let's say the left of the y -axis, the positive charge.

01:50 We'll have an arrow here, and that can define our positive electric field contribution.

02:02 For the negative, we can do something similar.

02:04 And as i said, it was radially inward, right? so it should be here, e minus.

02:16 Right.

02:17 So as you can see, these are, these vectors are both in the x and y plane.

02:24 So if we write out the components for them, we'll see that only the x components arise because you have a positive y, x component for the negative charge, but you have a negative y x component for the positive charge.

02:43 So if you're not convinced of this right now, the math should convince you later, but that is what should happen.

02:52 We should see that there's only an x component of the electric field.

03:01 So let's look at that x component.

03:05 So if i just write ex here, generally an electric field will be written like kq over r.

03:18 Squared but in this situation we have a kind of special charge we don't have point charges we have a line of charge right so we need to use we need to use a special symbol lambda to define that we call the linear charge density so lambda is as you could guess you know charge per length so this is what we mean when we say linear linear charge density but it is a curved line right and so basically what we're going to have here is we have to try to figure out what l is right so let's say that we had a full circle so i'm just going to claim that if we're just focused on one part of the arc then well oh no it's not one part of that we want the whole arc, right? so this is a semicircle is going to be half of a circle, right? and so the length of a full circle is going to be the circumference.

04:38 But, oh yeah, we will focus on just this contribution, for example, just the positive contribution.

04:46 So this is a fourth of the circle, right? so if l is defined as the, well, if l of a full circle is the circumference, the length of the perimeter of the circle, 2 pi r.

05:02 And we take a fourth of that, like the positive q region.

05:06 So we just want to divide by 4.

05:09 And we should know that the radius is a, which i didn't write earlier, but that is in the problem as well.

05:15 Semi -circle of radius a.

05:17 Then r is going to change to a, and then we'll see that 2 and 4 cancel, and we'll have pi a over 2 as the length.

05:29 We plug that in for l and we should see 2q over pi a yeah 2k over pi a yeah i wrote this incorrect this not l divided by 4 this would be l equals c divided by 4 right circumference divided by 4 the length of 1 so this is the l is the length of one arc so half of the arc so we can start to look at the electric field so we can find out the magnitude and direction, right? so i'm just going to make a new page here.

06:21 Let's look at the x component of the field, the electric field.

06:31 So if we go back to the first page real quick, then we can see that we need to replace q over r in this electric field with a lambda, right, q over some distance.

06:41 So we can say electric field is going to be k lambda over r.

06:48 But as you could guess, there's electric fields coming from each angle, from all angles, at least from if this is, if we're saying this angle, the x -axis is zero, then this would be where the y -axis is, which would be pi over two, just following like the unit circle convention, reading theta from the x -axis, counterclockwise and then this would be theta equals pi so from pi to pi over two we're going to have positive charge electric field contribution and from pi over two to zero we'll have negative charge electric field contribution so with that in mind and that the angle is going to change so we need to account for that so we'll write our first part here k lambda over r but of course our radius is a, so i'm just going to write a here.

08:02 Then we actually want to have an integral, right? so we want to integrate cosine theta.

08:17 So integrate cost theta, d theta.

08:23 Now we have to figure out which.

08:25 So if this is going to be our positive contribution, right, i said it was going to be from, oops, that's not what i wanted.

08:33 Okay, this is going to be from pi over two to pie, right? so this is what our integration bound would be, pi over 2 to pi.

08:48 And we'll have something similar but negative for the negative charge contribution, right? so let's just write that out quickly.

08:59 Cos theta, d theta.

09:08 But now we're going to be going from 0 to pi over 2, right? because that's where the negative charges.

09:15 So these are pretty simple integrals to evaluate, right? so if we just pull k lambda over a out here and try to evaluate these integrals, we know that the integral of cosine is just the same thing as asking what function does, what function gives me cosine theta when i take the derivative of it, right? so that should be sign.

09:47 So we can write here, sine theta evaluated from pi, or from pi over 2 to pi.

10:02 And then we want to subtract the same thing, but now with different bounds.

10:12 0 to pi over 2.

10:19 Alright.

10:22 So this should be pretty simple here.

10:24 If we look at sine of pi, what is that going to be? well, on the unit circle, if theta is equal to pi, then there's no there's no y part, so it's going to be 0.

10:39 So and then sign of pi over two will be one, right? because it's the top of the circle here, as we can see...

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