Two 1.20 m nonconducting rods meet at a right angle. One rod carries +2.50 μC of charge distrib

Two 1.20 m nonconducting rods meet at a right angle. One...

Key Concepts

Relation Between Electric Field and Force

This concept explains that the force experienced by a charge placed in an electric field is given by the product of the charge and the electric field strength, expressed as F = qE. This relationship is especially important when predicting the behavior of charged particles in externally applied fields.

Coulomb’s Law

Coulomb’s Law provides the magnitude and direction of the electric force between two point charges. In the context of continuous charge distributions, Coulomb’s Law is applied in an integrated form to sum the contributions from each infinitesimal charge element, thereby providing the resulting electric field at a point.

Uniform Charge Distribution

A uniform charge distribution means that the charge is evenly spread along the length of the object, which simplifies the calculation of the electric field since the linear charge density is constant. This concept is essential for setting up and solving integrals that describe the electric field created by the distributed charge.

Electric Field from a Continuous Charge Distribution

This concept involves calculating the electric field generated by a charge that is distributed along a continuous object, such as a rod, instead of a point charge. The calculation typically involves integrating the contributions from each infinitesimal element of the charge distribution to find the net electric field at a given point in space.

Superposition Principle

The superposition principle states that the net electric field produced by multiple charge distributions is the vector sum of the electric fields produced by each individual charge distribution. This principle is fundamental in analyzing systems where several charged objects contribute to the overall field at a point.

Transcript

00:01 Hello everyone.

00:02 In this problem, we're asked to find the magnitude and direction of the electric field produced by two rods that meet at right angles.

00:10 And then we're also asked to find that if an electron is released at a point that is halfway between, there's basically on the bicecture of both rods, then what is the direction and the magnitude of the networks that this electron experiences? so here's the situation.

00:29 So we have a negatively charged rod over here, and we have a positively charged rod over here.

00:41 Both of the rods have the same length.

00:45 This is l, this is l here as well.

00:49 L is 1 .20 meters.

00:53 Then we also have that the charge here and the charge here on the top part are the same, except one is positive, and the other one's negative.

01:04 So this is plus q and this is minus q, where q is equal to 2 .50 micro -culems.

01:15 So that's 2 .50 times 10 to the minus 6 couloms.

01:22 Okay.

01:23 So that's the charge of these rods.

01:26 And so the point that we're interested in is located halfway, is located on the bicector of both of the rods.

01:33 So if i kind of draw the bisector as this dashed line, then there's going to be a point in the middle, and that is what we're interested in finding.

01:46 That is the position where we're interested in finding the magnetic, or so the electric fields, and then the force if we place an electron over there.

01:56 So what do we know? well, first we have to find e.

02:01 So what is the electric field at this point? that is the question.

02:11 Okay, so first we should be able to find what is the electric field generated by a current distribution that is just a line.

02:19 So that's the first step.

02:20 We have to find what the electric field generated by just a simple line of charge is.

02:26 So let's say that we have a line of charge like this that has a size of of a.

02:37 So imagine that the y -axis is up here.

02:45 It's going up there.

02:46 That's the y -axis.

02:50 And then we have the x -axis that's going along this way.

02:56 So the top of the, so the length of this entire charge distribution is a.

03:01 It has a total charge.

03:03 Let's call that capital cube in red.

03:07 Okay.

03:08 And so we're interested in finding the electric field along the line that is bisecting this charge distribution, so such that this distance is a over 2, and distance to any point over here is arbitrary.

03:25 So we're going to say that that is the point x.

03:30 Okay.

03:31 So the way we do this is we're going to say that to this point, let's call that p, there is a, there's going to be a line element that we're going to consider dq, that is going to be some distance away from the origin.

03:55 So let's call that y.

04:02 And that is going to make a distance, or that it's going to have a distance from the point of interest that we're going to call r.

04:12 Okay.

04:14 So now if you kind of look at what's happening here, we're going to have that dq, the dq element gives us an electric field, de, that we can calculate using kdq.

04:32 So we're going to use just the kulam's law to calculate what this electric field is.

04:38 So that's the e.

04:43 And so this is the magnitude of the electric field at that point.

04:47 And this is pointing in the same direction as e will be pointing in.

04:52 So we don't necessarily know what that direction is a primary, but let's just keep an open mind for that one.

04:58 So another important thing that we need to note here is the angle down here, which is theta.

05:06 Right.

05:06 So this dq, the line of dq to the point p makes an angle of theta with the vertical.

05:14 So given that information, we can work out what the various sides are.

05:21 So if this total distance to the bicector is a over 2 and the distance to the element is y, then this side of the triangle that we're interested in, remember this is a right -handed triangle, then that is going to be a over 2 minus y.

05:45 So that tells us that r is going to be equal to the square root of x squared plus a over 2 minus y squared.

06:01 That's one of the things that this tells us.

06:03 The other thing that this triangle tells us is that we have that the sign theta is going to be equal to opposite over hypotenuse, right? so that's a over 2 minus y over r.

06:28 And the cosine is going to be equal to the adjacent over the hypotenuse, which will be, oh, excuse me, sorry, this will be x.

06:39 This is x over here, right? the opposite is x.

06:44 Opposite size is x.

06:46 Just put that there.

06:47 So it's x over.

06:48 And then this cosine theta is going to be a over to minus y.

06:53 So that's the adjacent divided by.

06:58 Now, a couple of things.

07:00 This de component is of course a vector.

07:05 So it's going to have an x and a y component.

07:08 So the x component is going to be, so if that angle is theta, then we're going to have that this angle down here is also theta.

07:24 So dex is going to be actually sine theta.

07:28 So this is going to go with, this is going to go with the size.

07:32 Of the e times sine of theta.

07:41 So the size of de is just this quantity up here.

07:46 So this is the k -d -k u -r -squared -squared.

07:52 So this is going to be k -d -k -u -over -r -squared times sine theta, which we just agreed was x over r.

08:03 And since this is a linear uniform charge distribution, we also know the d -q is going to be the linear charge density times dy, which is the size of the little charged element.

08:16 And since we know that a total charge is q and the total size is a, or the total length is a, lambda is just q over a.

08:24 And so dq is q over a times dy.

08:27 So all in all, d -e -x becomes k, b -q over a times d -y, times d -y -x over the square root of y -square.

08:44 Sorry x squared plus a over two minus y squared to the three halves so that is that is the x component of the electric field generated by this linear charge distribution along the bi sector by this little charge element dq so what's the next step well the next step is we are going to integrate this this equation but notice that this way of setting up this distance of the triangle, so setting up this side only works on this bottom half.

09:35 For the top half, we're going to need an other quantity, and we will discuss that shortly.

09:41 But for now, let's just do this.

09:43 Let's just do this integral.

09:45 So we're going to say that d .e, sorry, we're going to say that e, x, coming from the bottom half, is going to be the integral, k times q over a, times the integral.

10:02 Of d .y over x squared plus a over 2 minus y squared to the three halves times x.

10:21 Okay.

10:23 Now we can kind of simplify this equation a little bit, so we can say that w is equal, and of course this goes from 0 to a over 2, right? y goes between 0 and a or 2.

10:35 And w, we're going to say is equal to just a over 2 minus y.

10:49 Okay? so that's a little, that's a small simplification.

10:54 The w is equal to minus the y.

10:57 The limits are w at zero.

11:01 So when y is zero is a over two, w at a over two is equal to zero.

11:08 So they kind of flip.

11:10 So we're going to have k times, q over a times the integral from a over 2 to 0...

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