Positive charge Q is distributed uniformly along the positive y -axis between y=0 and y=a A nega

Positive charge Q is distributed uniformly along the...

Key Concepts

Far-field Approximation

The far-field approximation involves simplifying the expressions for the electric field and force when the observation point is at a distance much larger than the size of the charge distribution. Under these conditions, the charge distribution appears nearly point-like, leading to simplified expressions that capture the dominant behavior of the field with higher-order corrections reflecting the spatial extent of the distribution.

Force on a Test Charge due to an Electric Field

Once the electric field from the charge distribution is determined, the force acting on any test charge placed in the field can be computed directly using the relation F = qE. This principle is used to calculate both the magnitude and direction of the force acting on the point charge due to the distribution.

Coulomb's Law

Coulomb's Law provides the fundamental relation for the electric field created by a point charge. In problems involving charge distributions, this law is used in its differential form to calculate the contribution of an infinitesimal element of charge to the overall electric field, forming the basis for setting up the integration across the entire charge distribution.

Continuous Charge Distribution

A continuous charge distribution replaces discrete charges with a density function spread over a region, requiring integration to obtain quantities like electric field or force. In this context, a uniform linear charge is considered, and the integration of the field contributions from each differential charge element yields the total field at the observation point.

Decomposition of Field into Components

Since the electric field is a vector, it must be decomposed into its constituent components (typically along the x and y axes) for practical calculations. In this problem, the contributions from each differential charge element are broken down into x- and y-components using geometry and trigonometry, which then allows for integration over the length of the charge distribution.

Transcript

00:01 Hello everyone.

00:01 In this problem where s defined the electric field, the electric force, and the electric force in a given limit that a vertically oriented linear charge distribution exerts on a point charge along an axis perpendicular to its orientation.

00:20 So here's the picture that we have.

00:23 So we have a point charge minus cube along the x -axis over here.

00:31 And then we have a positive charge distribution with the total charge, positive q, along the y -axis, and it has a length of a, and the minus q low charge is located at a distance x away from the origin.

00:48 So what we want to find is we want to find the x component first and the y component of the of this charge distribution on the low charge minus q.

01:03 So that's part a.

01:04 So how do we do this? so first of all, notice that for any given line segments, for any given segment on this line, right, the net electric field is going to be pointing along the line connecting the element and the charge minus q.

01:27 And the reason for that is that the charge on the, the line or the rod is positive, whereas the charge of the small charge is going to be negative.

01:38 So the negative charge is going to be attracted towards the positive charge.

01:43 So then this electric field component or this electric field contribution is going to make an angle theta with the horizontal.

01:54 Let's call it.

01:56 So if we take this tiny element to be a distance.

02:02 Away from the origin along the y -axis, then the distance, so the distance are between q and so the charge distribution, so the little element d -cube and the point along the x -axis small q is going to be the pythagorean sum of y and x.

02:26 So it's going to be the square root of x squared plus y squared, which just means, of course, that r squared, which is what we're primarily interested in, is going to be x squared plus y squared.

02:39 So there's that.

02:41 Another thing that this tells us is that the magnitudes of the x and y components are going to be constrained by the angles.

02:55 So we're going to have that the, if we draw the similar triangle, so if we draw a triangle that has de as they had pot in use, then this is going to be made up of a small component along the x -axis, d -e -x, and a component along the y -xs, d -e -y, which make the same angle between, so d -e -e makes the same angle with d -e -x as the line connecting the charge element d -q to q does, with the x -axis.

03:36 So given this, we're going to see that d .e.

03:42 X is actually going to be cosine theta.

03:46 So it's going to be d .e.

03:48 So the magnitude of d .e.

03:50 Times cosine of theta.

03:53 And d -e -y is going to be the magnitude of d .e.

03:59 Times sine theta.

04:03 And then the thing about the x component is that it's going to be pointing in the negative x direction, as you can see.

04:08 And the y component is going to be pointing in the positive one direction.

04:12 So we can also express the angles cosine theta and sine theta in terms of these distances, right? so we can say that cost theta is equal to, let's see, so cosine theta will be adjacent over hypotenuse.

04:35 So it's x over r, which is equal to x over the square root of x squared plus y squared.

04:46 And sine theta is equal to y over r, so opposite over high partners, which means that this is y over the square root of x squared plus a y squared.

05:04 So now all we have to do is put everything together.

05:08 And so one more thing that we know is that the charge is, linearly distributed over this rod.

05:16 So the linear charge density is q over a.

05:20 This means that the small element dq is going to be equal to lambda times d, y.

05:29 So this element has a length of dy.

05:34 And so this is going to be equal to lambda times d .y.

05:36 All right.

05:38 So now we are in a position where we should be able to evaluate everything.

05:43 So let's see.

05:45 So de as a vector, or rather as a magnitude, i should say, we're going to take care of the vectors later, is going to be equal to k times the little charge dq divided by the distance between the element and the charge squared.

06:07 So this is going to be r squared.

06:09 So now notice that we're explicitly finding the direction at the position of minus q instead of, you know, finding it along some general coordinate along the x -exes.

06:20 So we're going to say that this is k times dq over r squared.

06:26 And we're going to say that dq is equal to lambda times d -y.

06:36 And r -squared is, of course, x squared plus y -squared.

06:41 So this is what we have right now.

06:44 And then what we're going to do is we're going to put in, we're going to resolve this, towards the different components.

06:55 So we're going to say that the x is going to be, you know, magnitude of de, which is literally what we just wrote.

07:01 So k times q over a times dy over x squared plus y squared and times cosine theta.

07:14 But cosine theta, remember, is just x over x squared plus y squared.

07:20 So it's going to be x over x squared plus y squared.

07:28 And, sorry, square rooted there, right? so altogether, this becomes kq over a times x times dy over x squared plus y squared to the three halves.

07:55 Okay, so we're going to have that.

07:58 That will be one contribution.

08:00 This will be the contribution in the x direction.

08:05 Then we can actually, yeah, we can go ahead and evaluate this if you want.

08:10 So let's say that the only thing we really want to do is just integrate this.

08:16 So we're going to have this is equal to k times bicku over a times x.

08:24 Right, notice that x is a constant, right? so it comes out of the integral.

08:27 So we're going to have to do the integral from 0 to a of dy over x squared plus y squared to the three halves now you should just put this into your favorite solver because actually carrying this out will probably be very lengthy so let me just find my favorite solver and so i'm going to integrate um one over squared plus y squared power off three over so i get that the result of this integral is going to be equal to k.

09:32 So k times q over a times x.

09:40 And then the integral is going to be, so y is going to be y is going to be y over x squared, of the square root of x squared plus y squared.

10:01 So you're wondering where this comes from.

10:02 This probably comes from some trigonometric substitution.

10:07 Anyway, we're evaluating this at m a and zero.

10:11 So this will be equal to k, times q over a times x.

10:21 And evaluating it so we have one over x squared.

10:25 And then we have a over the square roots of x squared plus a squared.

10:32 So remember we're replacing y.

10:34 That was the thing that we integrated, minus 1 over, sorry, minus 0.

10:45 So that's it, right? so we have, because we have minus 0 over the square root of x squared plus 0 squared, which of course is 0.

10:57 So we're left with x being equal to whatever this is, right? so we noticed that one of the a's, the is cancel here, one of these, x is canceled.

11:08 So we have k times q over over x times the square root of x squared, x squared plus a squared.

11:28 So that's what we have here.

11:32 And in case, you know, you're wondering which way this points, we can say that this is pointing in the negative x head direction.

11:45 So we're going to put a minus there.

11:47 So that's just to indicate the direction.

11:50 So that's x.

11:51 And then ey kind of proceeds along similar lines.

11:54 That's a slightly easier integral to do, though.

11:57 So you're going to have that ey is equal to k times q over a.

12:05 And let's see.

12:06 So we had that the magnitude of the e was over x squared plus y squared.

12:14 So k times q over a times dy.

12:21 Over x squared plus y squared and now you're multiplying this by sine theta right so sine theta it was y over the square root of x squared plus y squared so this becomes the integral of that it becomes zero and a so this is going to be k times q over a and i notice that we can't take the y out right because y is something that we're actually iterating over so we're going to go from 0 to a here of d .y, y over x squared plus y squared to the three halves.

13:08 Now this is an integral that you can actually do quite easily.

13:11 So you're going to say that u is equal to x squared plus y squared, which means that du is going to be equal to 2 times y times d u, sorry, dy, dy, yeah.

13:31 Then this tells you that du is, sorry, dy is equal to du over 2y.

13:43 Okay, that's one thing.

13:44 The other thing is you're going to evaluate the limits...

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