WEBVTT physics/ap-physics-c-electricity-magnetism/fullerton
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Hi, everyone, and welcome back to www.educator.com.
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I'm Dan Fullerton and in this lesson we are going to talk about electric fields.
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Our objectives are going to be to calculate the electric field due to 1 or more point charges.
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To analyze electric field diagrams.
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To calculate the electric field by various continuous charge distributions by integration and this is going to get a little bit involved for our second lesson.
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This is one of the trickier parts of the course.
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Take some time here, do not always expect to get that on the first try.
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And then apply Gauss’s law to find the electric field for planar, spherical, and cylindrical symmetric charge distributions.
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We are going to set some of the base line up in this lesson and then on the following lesson on Gauss’s law specifically.
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We are really going to dive to that with a bunch of examples and finish up that once.
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We will not finish this fourth objective here in this lesson itself.
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As we get started, I would like to mention that a lot of this work in AP Physics C is fairly complicated especially the first time you see it.
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The notation is a little tricky, there is a lot to think about, a lot to do.
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You are not going to get all of these the first time through.
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Doing this has to be an active pursuit.
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As we come to the example problems, I recommend hitting the pause button and seeing if you can do the problems.
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If you get stuck, play it a bit more, and if you can get through it, do so and then check your answer with what you see in the video.
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There is a lot more active learning that needs to be done to really ingrain some of these concepts.
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Please take the time to do that as you work through our example problems.
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Moving on, what are electric fields? if the electrostatic force is a non contact or field force, it operates across some distance.
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The property of space that allows a charged object to feel a force is called the electric field.
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It is a made up concept to help us understand how that force works.
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You can detect the presence of an electric field by placing a positive test charge,
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add various points in space and then measuring the resulting force on that test charge.
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The electric field strength factor which we symbolize with a E and you will see it
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in textbooks sometimes with bold or with an arrow over top of it, it is known as a vector.
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It is the amount of electrostatic force observed by a charge per unit of charge.
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The electric field vector is the force it feels ÷ the amount of charge.
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The direction of the electric field vector is a direction of a positive test charge would feel a force.
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As we talk about electric fields, it is important to remember that these have units and that they are going to have some meaning.
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The standard units of the electric field are units of force N/ charge N/ C.
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We are going to find out later that is also equivalent to volt per meter but standard units you are going to see most of the time are N/ C.
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Let us start off with a simple example and then we will work our way into more and more complex examples.
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2 oppositely charged parallel metal plates 1 cm apart exert a force of 3.6 × 10 ⁻15 N on electron placed between plates.
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Calculate the magnitude of the electric field strength, we want to know how strong the electric field is between the plates.
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We can do this in a pretty straightforward manner saying that the electric field is = electrostatic force ÷ the amount of charge.
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That is going to be 3.6 × 10 ⁻15 N ÷ our charge, it is the charge that we have on an electron
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because it is an electron between those plates, the magnitude of the charge on electron is an elementary charge or 1.6 × 10⁻¹⁹ C.
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When I go through and do this with my calculator, I come up with about 2.25 × 10⁴ N/ C.
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Hopefully, a pretty straightforward example to get us started.
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Let us talk about how we could visualize this electric field.
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Since, you can actually see the electric field you can visualize it by drawing
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what we call field lines to show the direction of the electric force on the positive test charts that we would put somewhere in space.
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Electric field lines always point away from positive charges and toward negative charges,
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away from the positives and toward the negatives.
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Electric field lines never cross each other and they always intersect conductors at right angles to the surface.
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Stronger fields have closer and denser lines, weaker field the lines are further apart.
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The field strength and line density decreases as you move away from the charges.
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To do just a simple one to begin with.
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If we were to take a look at a positive point charge, we have electric field lines moving away from it, moving radially away.
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Electric field is the same at all points that are in equal distance from the center of that point charge so you have the same density of lines.
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You have the symmetry.
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You can almost think of a positive charge when we draw electric field lines as kind of a magic blow dryer.
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It is blowing air out from that point in all directions and what you are drawing is the vectors of the air coming out of it.
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On the other hand, a negative charge works the same way but all the lines are coming into it.
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A positive charge would be sucked in this direction.
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You can almost think of these as the magic vacuum cleaner in space, it is sucking all the air in and you can draw just the wind vectors there.
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Away from positive charges in the negative charges.
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Also note that, as you get further away the distance between these increases quickly showing that the electric field strength drops off quickly.
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We have another example of an inverse square law here, just like we had for the electrostatic force.
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If we put these charges together to make what is called the dipole to charges, positives go toward negatives,
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the electric field lines begins at the positive and ends at the negative.
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They are symmetric and if we were to have a particle hearsay, positive test charge somewhere sitting right there for example,
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it is going to feel a force in the direction of the field lines which we do not show all of them there but it would feel a force in roughly that direction.
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Or if we put a positive charge over here, it would feel a force in that direction.
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What if you have a negative charge?
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Let us take and do an example with a negative charge.
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If we put a negative charge, let us say we put it right over here that purple spot,
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it is going to go in the opposite direction of the field line that would feel a force that way which only makes sense,
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it would be attracted by that positive test charge and move to the right.
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If we happen to have 2 positive charges or 2 negative charges, you get this other pattern as well,
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where there is a point in the middle of those, where there is no force or everything cancels out.
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Some examples of how you can draw electric field lines.
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We did a couple point charge diagrams, let us talk about the electric field due to point charges in a little bit more detail.
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Electric fields are caused by electrical charges, of course.
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The electric field due to a point charge can be derived from the definition of the electric field and we just learned about in our previous lesson, Coulomb’s law.
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If the electric field is the force ÷ the charge and the magnitude of the force is given by Coulomb’s law here.
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Or if you want it to, we can go to our other notation and say that the force is 1/4 π E₀, remember that is equivalent to the K × Q1 Q2/ R².
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We can start to put these together and say for example the electric field here is the force K Q1 Q2/ R² ÷ Q.
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We end up with one of the Q cancels out makes a ratio of 1, we end up with KQ/ R².
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Or if you are looking at it in this notation here, the electric field would be 1/ 4π E₀ × Q/ R².
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We have our formulas for the electric field due to a point charge.
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To find the electric field due to a multiple point charges, what we do is we take the vector sum of
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all of the electric fields due to the individual point charges.
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We use the law of super position to just add all of those up.
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You will notice a bunch of comparisons electricity to gravity.
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The formula electric forces K Q1 Q2/ R² compared to gravity G M1 M2/ R².
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The only major difference mass vs. charge and the constant, the fudge factor.
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The electric field strength is the force ÷ the charge.
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With gravity, the gravitational field strength was the gravitational force ÷ the mass.
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We can also write this, our formula is KQ/ R² for electricity, for gravity GM/ R², see they are parallels here.
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The constants are different but that is just because we are trying to make the math workouts so our units come out to something that makes sense.
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The charge units are Coulomb’s, the mass units are kilograms.
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The math has tons of similarities.
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One that we have not pointed out yet though that is a very important, gravity only attracts.
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The electrostatics can attract and repel.
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This is very important.
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Let us do a problem where we have got the electric field from a couple of point chargers.
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Find the electric field at the origin due to the three charges shown in the diagram.
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We got a charge here at 0, 8 m which is 2 C, a huge amount of charge.
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Over here at 2, 2 we have got a + 1 C charge and here at 8, 0, 8 m on the X axis we have -2 C charge.
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How can we do that?
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What I'm going to do is I'm going to find the electric field of the origin due to each of these 3 charges individually and then take their vector sum.
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First thing is let us find the electric field at the origin due to do this green charge up here.
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Since, this is a positive charge we should be able to see that down here at the origin,
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the field lines go away from positive charges our electric field should be in that direction.
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The electric field due to that + 2 C point charge at the origin is going to be KQ/ R² where K is 9 × 10⁹.
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Our charge is 2 C ÷ our distance from the origin 8 m that we have to square or I get about 2.81 × 10⁸ N/ C down which is in vector form 0, -2.81 × 10⁸ N/ C.
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There is the electric field at the origin due to this green charge up here.
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Let us do it now for the red charge.
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We can see if this is a -2 C charge, the electric field is going to go toward the -2 C charge so they should feel that the origin is going to go toward the right.
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We already know the direction.
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The magnitude then is given by KQ/ R² which is 9 × 10⁹ Nm²/ C² × 2 C charge.
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I’m not going to worry about the magnitude since I already know the direction, ÷ the square of the distance 8 m between those.
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Or again, we end up with 2.81 × 10⁸ N/ C to the right or in vector notation that is going to be 2.81 × 10⁸ N/ C in the x and nothing in the y, 0.
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We have got to figure out our portion due to this blue charge is 1 C charge at 2, 2.
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By inspection, we should be able to see that the field lines are going to get away from the positive charge so
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the origin that is going to be going in that direction.
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To find its magnitude however, our same formula E = KQ / R² where K is 9 × 10⁹.
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Our charge now is 1 C and the square of this distance.
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To do that, we got to do a little bit of math.
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I'm going to make a right triangle realizing that this is 2, this is 2, so I can use that the Pythagorean Theorem to find out that distance that is 2 and 2.
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This is going to be the √2² + 2² so R² is √2² + 2² is just √2² + 2²² or 8.
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That is 9 × 10⁹/ 8 which will be 1.13 × 10⁹ N/ C down into the left in a 45° angle.
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What we have been doing this in vector notation so what we do with that?
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We do not have a straight x and y component right from our answer yet.
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We got to do a little bit of math here and realize that that is going to be equal to.
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The x component is going to the left so that is going to be -1.13 × 10⁹ N/ C in the x component cos 45°.
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Our y component is going to be going down as well so that will be negative.
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I end up with -1.13 × 10⁹ N/ C sin 45° or when I do the math there that is = -7.95 × 10⁸ N/ C in the x and -7.95 × 10⁸ N/ C in the y.
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Just taking a minute to step back.
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We now have the magnitude and the broken down vector notation of the electric fields due to each of these 3 charges.
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To find the total electric field, we need to add these keeping in mind of their vectors.
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We cannot just add their straight magnitudes, we have to add vector components.
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To find the total electric field, we are going to add up all of the x components and all of the y components.
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Our x components we have a 0 from the green one.
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We have 2.81 × 10⁸ from our red charge and we have a -7.95 × 10⁸ N/ C from our blue charge.
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We have these three to add up for our x and for the y component we are going to do the same basic thing.
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We have -2.81 × 10⁸ N/ C due to our green charge.
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We have no contribution in the y direction due to our red charge and we have -7.95 × 10⁸ contribution from our blue charge.
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We add up the x, we add up the y to find out that our total electric field is going to be,
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when we add all those up we are going to get about -5.14 × 10⁸ N/ C in the x direction and about -1.08 × 10⁹ N/ C in the y.
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Our total electric field is going down into the left, you can figure out the angle if we wanted to.
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But this gives us the vector sum of these 3 electric field contributions to give us our total.
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Let us make sure we highlight that after all that work.
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There is our total electric field at 0, 0 due to those 3 point charges.
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It is kind of complicated but if you take it step by step, it is really not bad.
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Let us take a look at an example where we are looking to find out where the electric field is 0.
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Here on this line, we have 2 point charges, we have a + 1 C charge at x = -6 m and the + 2 C charge at x = 5 m.
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The electric fields from both of these go away from positive charges so we got an electric field to the right here and to the left on that side of them.
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Due to the red charge, we have an electric field to the right, electric field to the left.
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There should be some point on this number line where the red and the blue exactly cancel out and there is no electric field.
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By inspection, as I look at this, we have a smaller charge here so I would expect that
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we would have the point at 0, maybe a little bit to the left of the middle of those 2 charges.
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For 5 and 6, about -.5 would probably be halfway between them so I'm expecting we would probably get an answer that is somewhere over in this region.
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Before we even go into the math, let us qualitatively take a look, make something of a guess and
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then we will go see if everything matches up when we go through our math here.
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What we are going to do to find this out is, let us say that this is our point and they do not know exactly where it is
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but we will define the distance between our 1 C charge and our point where the electric field is 0, let us call that distance R.
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If we call that R, then the distance from here to our red charge must be 11 -R because the total distance between them is 11 so R + 11 – R is 11 m.
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We can go in and we can start doing our math to find out exactly where that point resides where the electric field is 0.
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Looking at the blue charge first, the electric field due to that 1 C charge is K Q1/ R².
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If we look at our red charge over here, the electric field due to that is going to be K Q2/ our distance is going to be 11 - R².
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The total electric field we know has to be 0 at this point.
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The total electric field which is going to be our electric field due to the blue charge
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which is K × 1 C ÷ R² - this one was going to the left our red electric field contribution
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K × 2 ÷ 11 - R² and we know all of that has to = 0.
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We have got an algebra problem.
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We have got K on the left, K on the right, 0 over here, so what we can do is we can factor out the K's
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which implies 1/ R² from the left, after we cancel out our K’s must be equal to,
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I’m going to move this portion to the right hand side.
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We have 2 ÷ this 11- R it is kind of awkward to use that way.
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I’m going to multiply that out 11 - R² is 11 -R × 11 - R or 121 11² -11 R -11 R is -22 R + R².
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I’m trying to solve this for R so what I might do here is maybe I will cross multiply this and just make it look a little simpler,
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to say then that we will have 2 R² = we still have our R² -22 R + 121
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which implies then if I subtract R² from both sides, I get R² = and then we have R² =.
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Let us do it a little bit differently.
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I think we can save a step here.
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We are going to put this into a quadratic equation so I could bring all of this over to the left, subtract all of this and say that R² + 22 R -121 = 0.
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And that is a little bit slicker.
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I'm taking 2 R² - R² to get R² -22 R or adding 22 R to both sides and then subtracting the 121 to rearrange.
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We got a system that we can put in a quadratic formula because I do not see a very easy way to factor that in any other direction.
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If we use the quadratic formula, we can then say that our R must be equal to, quadratic formula -B ± √B² - 4 AC/ 2 A.
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In this case, -B is going to be our -22 ± √22² -4 A × C 1 × 121 is just going to be that -121/ 2 A and A is 1 so that is ÷ 2.
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A little bit of calculator work and you should come up with 2 possible answers, either R = 4.56 or R = -26.6.
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In here, we got to use a little bit of common sense and say which one of these make sense?
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If R is 4.56 that means this distance over here, pretty close to our guess a bit, that make sense.
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If R was -26.6 we are way to the left of the screen and that is not going to make any sense.
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We have got a left vector from this electric field, left there and they are going to add together.
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You are not going to get 0.
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We can cross that one out as not making sense for our problem.
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Therefore, we know that R must equal 4.56.
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Where on this diagram is that 0 point?
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If we start at -6 and we go 4.56 to the right that tells us that our x position
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must be -6 + 4.56 or about -1.44 m on our number line.
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The exact answer should be right around there.
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Finding the electric field due to a couple of point charges we can use the law of super position.
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Let us take a look at a little bit more straight forward example for a moment, stepping back.
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A distance of 1 m separates the centers of 2 small charge sphere, they exert a gravitational force FG and electrostatic force FE on each other.
00:24:45.200 --> 00:24:51.300
If the distance between the center of the spheres is increased to 3 m, we are tripling that distance,
00:24:51.300 --> 00:24:55.500
what happens to the gravitational force and electrostatic force?
00:24:55.500 --> 00:25:02.900
If we triple the distance, they get further apart, of course the force has to get smaller so we can get rid of those choices.
00:25:02.900 --> 00:25:11.700
Again, this is an inverse square law relationship for both the gravity and the electrostatic force.
00:25:11.700 --> 00:25:17.500
Inverse square law, triple the distance 1/9 of the force.
00:25:17.500 --> 00:25:23.500
The correct answer here must be A.
00:25:23.500 --> 00:25:27.000
The direction of the electric field due to a point charge.
00:25:27.000 --> 00:25:31.800
In the diagram here below, P is a point that is near a sphere with charge -2 C.
00:25:31.800 --> 00:25:35.500
What is the direction of the electric field at point P?
00:25:35.500 --> 00:25:45.100
If you remember our rules for electric field lines, electric field lines go in to negative charges.
00:25:45.100 --> 00:25:50.400
Over here at P, we must have a force to the left.
00:25:50.400 --> 00:25:57.300
The electric field there must be pointing to the left, assuming that would be a positive test charge.
00:25:57.300 --> 00:26:04.700
To the left would be the correct answer.
00:26:04.700 --> 00:26:10.400
Moving on, let us see if we can get into a little bit deeper derivations.
00:26:10.400 --> 00:26:14.100
To do that, we are going to have to talk for a moment about charge density.
00:26:14.100 --> 00:26:18.600
Let us assume we have something like a line that is uniformly charged.
00:26:18.600 --> 00:26:23.300
The total charge ÷ the length of that line would give you the linear charge density which
00:26:23.300 --> 00:26:30.000
we are going to give the abbreviation the symbol λ Λ, it is amount of charge per unit length.
00:26:30.000 --> 00:26:35.300
In other cases, we may be looking at a distribution of charge that is on some sort of surface.
00:26:35.300 --> 00:26:42.200
A linear charge density does not make sense but instead we are going to look at total charge per unit area
00:26:42.200 --> 00:26:48.000
that we are going to give the symbol sigma Σ, a surface charge density.
00:26:48.000 --> 00:26:55.700
We may also be looking at objects that are three dimensional that have some volume.
00:26:55.700 --> 00:27:04.900
When we have charged uniformly distributed through a volume, it can have some volume charge density taking the total charge ÷ the unit volume.
00:27:04.900 --> 00:27:07.600
And that we are going to give the symbol ρ Ρ.
00:27:07.600 --> 00:27:11.500
We are going to use these definitions fairly regularly throughout the rest of the course.
00:27:11.500 --> 00:27:14.700
A linear charge density, charge ÷ length.
00:27:14.700 --> 00:27:18.100
An area charge density, charge ÷ area.
00:27:18.100 --> 00:27:23.700
The volume charge density, charge ÷ volume.
00:27:23.700 --> 00:27:31.700
Let us put this in to play as we try and find the electric fields due to a semicircle of charge.
00:27:31.700 --> 00:27:40.100
We have a thin insulating semicircle of charge total charge Q at some radius R centered around at point C.
00:27:40.100 --> 00:27:45.200
We want to find the electric field at point C due to the semicircle of charge.
00:27:45.200 --> 00:27:49.400
Before I even begin the math, the first thing I’m going to do is observe for a minute.
00:27:49.400 --> 00:27:57.300
If this is symmetrical around C and it is uniformly charged, there should not be any horizontal component of electric field.
00:27:57.300 --> 00:27:59.100
That should all cancel out.
00:27:59.100 --> 00:28:03.300
We should only have a vertical component that we need to deal with.
00:28:03.300 --> 00:28:08.600
By symmetry, we can simplify the problem and work in just the horizontal direction.
00:28:08.600 --> 00:28:15.000
As we do this, the first thing I’m going to do is I'm going to define some linear charge density where our linear charge density,
00:28:15.000 --> 00:28:23.500
the charge per unit length on this, is the amount of charge on that length ÷ its length,
00:28:23.500 --> 00:28:30.400
which in this case is going to be its total charge Q ÷ the length of that semicircle.
00:28:30.400 --> 00:28:39.500
If the distance around an entire circle is its circumference 2 π R, half of that for a semicircle must be π R.
00:28:39.500 --> 00:28:45.300
Our linear charge, hence the Λ must be charged ÷ π R.
00:28:45.300 --> 00:28:53.500
As we do this, we are also going to break this up and our strategy is going to be to take and look at little tiny pieces of the semicircle.
00:28:53.500 --> 00:29:00.100
And as we go through it, we are going to find the electric field due to each of those pieces and then add them all up.
00:29:00.100 --> 00:29:11.000
I'm going to define this little piece and draw a couple of lines here just to make this a little bit clear.
00:29:11.000 --> 00:29:15.000
It is not quite as clear as we wanted there.
00:29:15.000 --> 00:29:17.400
It looks something like that.
00:29:17.400 --> 00:29:29.600
What we are going to do is we are going to call this R Θ so that this angle right there is a little differential of Θ, a little tiny piece of Θ.
00:29:29.600 --> 00:29:38.300
The amount of charge that we are enclosing up here in this little piece, that length is going to be RD Θ.
00:29:38.300 --> 00:29:43.600
The amount of charge that is enclosed there which we are going to call DQ, a little tiny bit of the charge
00:29:43.600 --> 00:29:53.700
is going to be that length RD Θ × the linear charge density in that area which we just defined as Λ.
00:29:53.700 --> 00:30:01.900
The differential of charge, the tiniest little bit of charge and that little bit is RD Θ × Λ.
00:30:01.900 --> 00:30:08.500
To find then the electric field due to that just that little bit of charge, we will find the differential.
00:30:08.500 --> 00:30:17.600
The little tiny piece of electric field which we are only looking in the y direction is going to be, we could use KQ/ R².
00:30:17.600 --> 00:30:32.200
I’m going to use 1/ 4π E₀ version, 1/ 4π Ε 0 is the same as K × our charge DQ ÷ the distance between them R²
00:30:32.200 --> 00:30:37.800
and we are looking at just the y components, so sin Θ.
00:30:37.800 --> 00:30:41.200
Once we have that we can start to get set up and do all of or math.
00:30:41.200 --> 00:30:46.300
The hardest part of these problems is just getting into the habit learning how to set these up.
00:30:46.300 --> 00:30:54.200
There is no better way than to just do a problem after problem after problem and check as you go on.
00:30:54.200 --> 00:30:57.900
Let us take this a little bit further.
00:30:57.900 --> 00:31:06.100
We have our RD Θ here, our differential of the angle if that is Θ and we just said that the differential of the electric field
00:31:06.100 --> 00:31:21.000
in the y direction due to this little bit of charge was 1/ 4π Ε 0 DQ/ R² sin Θ which implies then since we know that DQ = Λ RD Θ.
00:31:21.000 --> 00:31:42.900
We just said that on our previous screen that the differential EY = 1/ 4π E₀ and replace DQ with Λ RD Θ.
00:31:42.900 --> 00:31:58.200
Λ RD Θ we will make that stand out ÷ R² × sin Θ.
00:31:58.200 --> 00:32:05.100
To add up all these little bits of the electric field what we are going to do is an operation called integration.
00:32:05.100 --> 00:32:08.900
It is adding all of these tiny little pieces up to get a whole.
00:32:08.900 --> 00:32:16.800
We will integrate on both sides, do the same thing to both sides you get the same thing.
00:32:16.800 --> 00:32:26.600
We are going to integrate, to get the entire electric field we need to integrate from Θ = 0 all the way across until we get to Θ = π.
00:32:26.600 --> 00:32:32.400
Because once around an entire circle is 2 π, halfway around the semicircle would be π.
00:32:32.400 --> 00:32:39.100
We are going to integrate from Θ = 0 to π.
00:32:39.100 --> 00:32:44.100
What we can do, what is nice over here is you can pull constants out of the integration.
00:32:44.100 --> 00:32:51.800
1/ 4π E₀ that is all a constant anywhere in our problem, that is not going to change.
00:32:51.800 --> 00:32:59.500
Let us rewrite this again and say that this then is, here I see a simplification.
00:32:59.500 --> 00:33:02.900
R and R² we can simplify there.
00:33:02.900 --> 00:33:11.400
We can write this as integral of DEY = the integral from Θ= 0 to π.
00:33:11.400 --> 00:33:15.100
We have 1/ 4π E₀.
00:33:15.100 --> 00:33:25.500
We got R in the denominator, we have got Λ sin Θ D Θ.
00:33:25.500 --> 00:33:30.800
We write it that way before we really get into the math, it is like simplifying as much as possible.
00:33:30.800 --> 00:33:35.000
What we can do is we can pull all of the constants out of the integral.
00:33:35.000 --> 00:33:43.000
The left hand side is the integral of all these little pieces of tiny pieces of the y give us the total EY.
00:33:43.000 --> 00:33:49.000
The electric field in the y direction or the y component of that is going to be equal to, pulling other constants,
00:33:49.000 --> 00:33:54.600
Λ is a constant 4π E₀ and R are all constants for the purposes of this problem.
00:33:54.600 --> 00:34:20.800
That will be Λ / 4π E₀ R integral from Θ = 0 to π of sin Θ D Θ = Λ/ 4π E₀ R, the integral of the sin is the opposite of the cos.
00:34:20.800 --> 00:34:31.400
We have the opposite of that cos of Θ evaluated from 0 to π which implies then the y component of the electric field
00:34:31.400 --> 00:34:49.400
that is going to be Λ/ 4π E₀ R and then we have – cos π - cos of 0.
00:34:49.400 --> 00:34:57.900
Cos of π is -1 so --1 is going to give us a 1 here, 1 – cos 0 is 1.
00:34:57.900 --> 00:35:00.300
Two negatives we will get 1 there.
00:35:00.300 --> 00:35:12.300
What I end up with is the y component of my electric field is going to be Λ/ 4π E₀ R.
00:35:12.300 --> 00:35:19.200
All of this is just 2.
00:35:19.200 --> 00:35:27.900
Therefore, the y component of our electric field becomes 2 Λ / 4π E₀ R.
00:35:27.900 --> 00:35:38.500
2/4 I can reduce to 1/2 or Λ/ 2 π E₀ R.
00:35:38.500 --> 00:35:44.100
There we have our y component of our electric field but we can even take that a bit further.
00:35:44.100 --> 00:35:51.200
That is a good answer but we know the total charges so we can substitute that and get rid of that Λ knowing that
00:35:51.200 --> 00:36:17.200
we already defined previously Λ = Q/ π R to say then that our y component of the electric field is Q / π R × whatever we had before 2 π E₀ R,
00:36:17.200 --> 00:36:32.900
which implies then that the electric field, the y component is going to be, we got a π² and R², so that is going to be our total charge Q ÷ 2 E₀.
00:36:32.900 --> 00:36:42.900
We have got π² and then R².
00:36:42.900 --> 00:36:47.700
Quite a bit of math involved but our strategy is pretty straightforward.
00:36:47.700 --> 00:36:51.900
Find and break up our little bit of charge.
00:36:51.900 --> 00:36:59.300
Find the electric field due to that little bit of charge and then we just add that up for all of those little bits of charge to make the whole.
00:36:59.300 --> 00:37:04.900
That is what we are really did with all of this math.
00:37:04.900 --> 00:37:08.900
Probably worth doing another couple of these.
00:37:08.900 --> 00:37:19.200
Thin, straight, uniform line of charge, find the electric field at distance D from a long straight and slating rod of length L at a point P
00:37:19.200 --> 00:37:25.000
which is perpendicular to the wire and equidistance from the end of the wire, assuming the wire is uniformly charged.
00:37:25.000 --> 00:37:33.500
We have got this point P lined up right with the center of this length L of an insulating rod that is uniformly charged.
00:37:33.500 --> 00:37:36.700
Our strategy again is going to be very similar.
00:37:36.700 --> 00:37:40.900
Let us even write that down to make sure we have got it nice and clear.
00:37:40.900 --> 00:37:55.500
First thing we are going to do is divide the total charge Q into smaller charges Δ Q.
00:37:55.500 --> 00:38:08.000
We are going to find the electric field due to each Δ Q.
00:38:08.000 --> 00:38:27.300
Once we have done that, we are going to add up the electric field due to each Δ Q to get our total electric field.
00:38:27.300 --> 00:38:33.100
And finally, as I look at this problem it looks like we can make another symmetry argument and should be pretty obvious that
00:38:33.100 --> 00:38:39.800
if this is centered on our wire then we are not going to have to worry about any vertical components of the electric field.
00:38:39.800 --> 00:38:41.400
It is all going to be horizontal.
00:38:41.400 --> 00:38:51.700
We are going to use symmetry to show that the y component of the electric field is 0 or therefore we are only going to worry about EX,
00:38:51.700 --> 00:38:56.700
the X component due to symmetry.
00:38:56.700 --> 00:39:02.300
It is always nice to simplify these wherever you can because they are complicated on their own.
00:39:02.300 --> 00:39:12.600
We have some uniform charge on this insulating rod, if they are uniformly charged we can talk again about a linear charge density.
00:39:12.600 --> 00:39:17.400
Λ was going to be our total charge Q divided by our total length of the wire L.
00:39:17.400 --> 00:39:22.400
Before we did the math, we are going to set this up as carefully as possible.
00:39:22.400 --> 00:39:31.800
If this total length is L, I'm going to call this Y = - L / 2 and we will call this end L/ 2.
00:39:31.800 --> 00:39:39.500
I have got our distance D toward point P and let us take just a little bit of charge, a little bit of length of that wire and
00:39:39.500 --> 00:39:45.600
we will find out what that is going to look like as far as the electric field contribution.
00:39:45.600 --> 00:39:52.200
I'm going to draw a line.
00:39:52.200 --> 00:39:59.400
There is our electric field due to that little bit I, we are only going to be worrying about the x component of that.
00:39:59.400 --> 00:40:04.400
If we do this, we will define this angle as Θ to that little piece I.
00:40:04.400 --> 00:40:16.400
We will have some y component, some y value, y distance from our center line which will call yi to that little piece of line i with charge Qi or DQ.
00:40:16.400 --> 00:40:19.300
As we do this, the length of this is probably going to be important.
00:40:19.300 --> 00:40:23.200
The distance from our charge due to a point where we want to know the electric field.
00:40:23.200 --> 00:40:37.800
We will call that Ri and by the Pythagorean theorem I can see that that is going to be equal to the √D² + yi².
00:40:37.800 --> 00:40:40.200
All right, that looks pretty good.
00:40:40.200 --> 00:40:49.200
Finally, let us write our equation for the x component of the electric field due to this little bit I before I get heavy into the math.
00:40:49.200 --> 00:41:06.400
The electric field due to this little portion I in the x direction is going to be 1/ 4π E₀, our K value again × the charge DQ ÷ RI²
00:41:06.400 --> 00:41:14.400
the distance from our charge to our point × the cos of Θ I because we are after the x component.
00:41:14.400 --> 00:41:18.400
I think we have got everything we need for our setup.
00:41:18.400 --> 00:41:24.500
It is time to start getting into the math and plugging through this thing.
00:41:24.500 --> 00:41:27.100
Let us dive in.
00:41:27.100 --> 00:41:48.900
The electric field in the x direction = 1/ 4π E₀ × Δ Q in that little piece ÷ RI² × cos of Θ I which implies then
00:41:48.900 --> 00:42:02.400
we know that RI = √y I² + D² and the cos of Θ I, as we look here cos SOHCAHTOA.
00:42:02.400 --> 00:42:07.100
Adjacent / hypotenuse is going to be D/ RI.
00:42:07.100 --> 00:42:26.500
cos Θ = D/ RI and we can write this as EI in the x direction is 1/ 4π E₀ Δ Q/ y I² + D² is RI².
00:42:26.500 --> 00:42:40.500
We can get rid of the square root sign × cos Θ I which is D/ RI which is y I² + D² √ ½.
00:42:40.500 --> 00:42:45.300
Let us get rid of that Δ Q now.
00:42:45.300 --> 00:42:54.800
Our Δ Q we know is Q/ L × our little bit of differential of y up and down that length.
00:42:54.800 --> 00:43:04.900
Then the x component due to that little piece I is 1/ 4π E₀.
00:43:04.900 --> 00:43:14.700
We have our Q, we have our D, we have our Dy, be careful because D and Dy are different.
00:43:14.700 --> 00:43:16.800
D does not mean D multiplied Dy.
00:43:16.800 --> 00:43:21.600
Dy here is differential of y, little tiny piece of Dy.
00:43:21.600 --> 00:43:26.600
Q D Dy/ L, capital L is still down there.
00:43:26.600 --> 00:43:36.100
We still have our YI² + D² but because we have y² + d² × y² + d² ^½.
00:43:36.100 --> 00:43:47.800
I can write that as 2³/2, which implies then that the EI in the x direction.
00:43:47.800 --> 00:43:51.800
Let us do the whole thing, let us integrate both sides.
00:43:51.800 --> 00:44:00.900
Let us say that just by looking at that little piece, the x component is going to be the integral from y = - L / 2 to L / 2.
00:44:00.900 --> 00:44:17.300
We are going to add up from here all the way to the top, all of those little pieces of 1/ 4π E₀ × we have got Q × D/ L.
00:44:17.300 --> 00:44:31.300
We also have multiplied by our DY/ y I² + D²³/2,
00:44:31.300 --> 00:44:40.600
which implies then that the x component of our electric field is, we can pull other constants.
00:44:40.600 --> 00:44:42.200
Q and D are not going to change.
00:44:42.200 --> 00:44:45.900
Our L is not going to change and 4π E₀ is not going to change.
00:44:45.900 --> 00:44:50.900
Those are all constants for the purposes of our problems so they can come out of the integral sign.
00:44:50.900 --> 00:45:17.700
We have QD / 4π E₀ L × the integral from - L/ 2 to L/ 2 of DY/ yI² + D²³/2.
00:45:17.700 --> 00:45:22.400
This is a tricky integral, not one that I expect you to be able to do with that.
00:45:22.400 --> 00:45:26.900
When we get to an integral like this where I like to go is the back of a calculus book, the front.
00:45:26.900 --> 00:45:32.000
It is a great place to go and look up some of those integration formulas.
00:45:32.000 --> 00:45:36.500
If you can find a form that this fits where it solved for you, let somebody else do the work.
00:45:36.500 --> 00:45:41.000
Use the formula that gives you the integration and that is exactly what I'm going to do here.
00:45:41.000 --> 00:45:43.900
I have looked this up already.
00:45:43.900 --> 00:46:10.900
This implies then using the formula that I looked up, the integral of some Dx/ A² + x²³/2 = x ÷ a² × √a² + x² + constant of integration.
00:46:10.900 --> 00:46:13.000
This fits that same format.
00:46:13.000 --> 00:46:19.900
Instead of Dx, we have got Dy, our x² we got yi², a² we got D².
00:46:19.900 --> 00:46:25.500
We can use this formula to do our integration.
00:46:25.500 --> 00:46:38.500
Taking the next step using this formula, then our electric field x component is Q D/ 4π E₀ L.
00:46:38.500 --> 00:46:47.200
I'm very careful to make sure that my solution is using the same piece as the same correlation with this formula.
00:46:47.200 --> 00:46:50.100
I use green to help highlight what we have done here.
00:46:50.100 --> 00:47:10.500
Instead of x we have y/, instead of a² we have our constant D², then we have got √a² + x² which for us will be √y² + D² ^½ .
00:47:10.500 --> 00:47:16.600
We are evaluating this whole integration as for - L/ 2 to L/ 2.
00:47:16.600 --> 00:47:24.300
Evaluated from - L/ 2 to L/ 2 which is equal to,
00:47:24.300 --> 00:47:34.900
Our left hand side over here still stays the same Q × D/ 4π E₀ L.
00:47:34.900 --> 00:47:45.000
We have got D² here and the D up here, we can do some simplification and leave us with just a D in the denominator.
00:47:45.000 --> 00:47:51.900
We will put a D down here and that one can go way.
00:47:51.900 --> 00:47:57.700
We also now have to deal with all of this mess.
00:47:57.700 --> 00:48:03.000
Let us just go right in, I will do this over here to give us some room.
00:48:03.000 --> 00:48:28.700
L/ 2/ L / 2² + D² ^½ - -L/ 2 ÷ -L / 2² + D² ^½.
00:48:28.700 --> 00:48:45.400
A little bit more simplification, that is = Q/ 4π E₀ DL this piece, × L/ 2 - -L/ 2 that is going to be the same denominator when you square that.
00:48:45.400 --> 00:48:59.200
It is just going to give us one total L so that is × L ÷ L/ 2² + D² ^½.
00:48:59.200 --> 00:49:04.200
Putting all of this together, we got an L down here and L up here.
00:49:04.200 --> 00:49:28.800
We can simplify that, I end up with the x component of the electric field at point P due to this line of charge as Q/ 4π E₀ D × L/ 2² + D²¹/2.
00:49:28.800 --> 00:49:33.400
We will make it nice and pretty and put in the square root sign there.
00:49:33.400 --> 00:49:38.100
There is our answer.
00:49:38.100 --> 00:49:40.800
Almost feel I need a nap after that.
00:49:40.800 --> 00:49:43.800
There is a lot involved, step by step.
00:49:43.800 --> 00:49:50.100
Each little step is not that difficult but there is a lot of bookkeeping, a lot of keeping track of the details.
00:49:50.100 --> 00:49:55.200
When you see these problems, take your time, do not let them intimidate you but it is all about practice.
00:49:55.200 --> 00:49:59.900
Just doing it again and again and again, nobody gets it on the first time.
00:49:59.900 --> 00:50:08.400
It is scary the first couple of times you look at it and it is supposed to be, stick with it.
00:50:08.400 --> 00:50:12.100
Let us take the solution and do a little bit more with a lower here though.
00:50:12.100 --> 00:50:18.000
Let us see what would happen if this line was infinitely long?
00:50:18.000 --> 00:50:28.600
In that case, the electric field in the x direction we will find that out by taking the limit as our length of that wire
00:50:28.600 --> 00:50:48.200
approaches infinity of our answer from before, Q/ 4π E₀ D × L/ 2² + √D².
00:50:48.200 --> 00:50:50.800
That can give us the answer.
00:50:50.800 --> 00:50:56.900
You got to be careful as we did the limit because as L gets very big you would think that this is just going to dominate.
00:50:56.900 --> 00:51:01.200
You also have to remember that Q increases and goes toward infinity as L does.
00:51:01.200 --> 00:51:03.300
We cannot do it quite that simply.
00:51:03.300 --> 00:51:13.100
Instead, we can look at this and say you know as L approaches infinity, L/ 2² is going to get so much bigger than D² that this D² really is not going to matter.
00:51:13.100 --> 00:51:19.200
The √L/ 2² is going to simplify to L/ 2 in the limit.
00:51:19.200 --> 00:51:38.300
We would get Q/ 4π E₀ D L/ 2 but we also know if you recall Λ is Q/ L or Q = Λ L.
00:51:38.300 --> 00:51:42.800
As L gets very big having a set charge is going from infinity does not make sense.
00:51:42.800 --> 00:51:45.700
We got to put it in terms of a linear charge density.
00:51:45.700 --> 00:52:00.500
We would write this as Λ L/, we have here L / 2 × 4 is just going to give us L/ 2 L.
00:52:00.500 --> 00:52:09.300
We will have 2 π E₀ DL and our L make a ratio of 1.
00:52:09.300 --> 00:52:22.500
We can state then that the electric field in the x direction due to that infinitely long line of charge would be Λ/ 2 π E₀ D.
00:52:22.500 --> 00:52:25.000
That looks a whole lot nicer.
00:52:25.000 --> 00:52:28.800
We are going to come back to a problem like this and solve it in a different way,
00:52:28.800 --> 00:52:34.100
a much smoother way in our next lesson when we talk about Gauss’s law.
00:52:34.100 --> 00:52:37.600
How about the electric field if the distance D is infinite?
00:52:37.600 --> 00:52:41.500
Instead of an infinite line, we are an infinite distance away.
00:52:41.500 --> 00:52:48.100
We can follow the same basic strategy, the horizontal component of an electric field
00:52:48.100 --> 00:53:09.300
is going to be the limit, as D approaches infinity of Q/ 4π E₀ D × √L/ 2² + D².
00:53:09.300 --> 00:53:14.900
As D approaches infinity now, this L/ 2 piece is not going to matter.
00:53:14.900 --> 00:53:24.400
The √D² is going to be D so we are going to get Q/ 4π E₀.
00:53:24.400 --> 00:53:29.200
We had 1 D, we got a D from this piece, D².
00:53:29.200 --> 00:53:33.600
That look pretty familiar, that looks like it is a point charge.
00:53:33.600 --> 00:53:35.800
Instead of D, think of R, how far you are.
00:53:35.800 --> 00:53:40.800
As you get infinitely far away from this wire, it is going to start looking like a point charge.
00:53:40.800 --> 00:53:43.700
Imagine you are zillions of miles away, you cannot tell it is a line.
00:53:43.700 --> 00:53:47.200
Electrically you cannot tell either, it starts to act like a point charge.
00:53:47.200 --> 00:53:49.600
That answer should make sense as well.
00:53:49.600 --> 00:53:58.700
You are very long distance away that piece of wire starts to act like a point charge.
00:53:58.700 --> 00:54:00.800
Let us do another derivation here.
00:54:00.800 --> 00:54:03.500
I think we need all the practice we can get.
00:54:03.500 --> 00:54:09.600
We are going to try and find the electric field at a point on the axis here P that is perpendicular to a ring,
00:54:09.600 --> 00:54:15.100
a thin insulating ring that is uniformly charge at some radius R.
00:54:15.100 --> 00:54:21.200
We are going to use the same strategies again, all of the real physics work is in the setup.
00:54:21.200 --> 00:54:26.800
By symmetry as I look at this, I can see that the only electric field is going to be in z direction.
00:54:26.800 --> 00:54:32.100
Everything else along the x and the y cancel out so that will help a little bit.
00:54:32.100 --> 00:54:40.300
We can also define for this problem a linear charge density Λ which is Q/ L.
00:54:40.300 --> 00:54:49.600
It is going to be Q/ the length of our ring is a circumference 2 π R.
00:54:49.600 --> 00:55:06.000
Or Δ Q as we look at some portion of the ring is going to = our linear charge density × our radius × the arc length that we go through here.
00:55:06.000 --> 00:55:13.700
I'm going to call this as we go up to here, let us call this direction φ Φ.
00:55:13.700 --> 00:55:19.800
Times some little bit of Φ, differential of Φ, D Φ.
00:55:19.800 --> 00:55:24.800
Once we have done that, we can start looking at some of the geometry of the problem here.
00:55:24.800 --> 00:55:29.900
We need to know the electric field at point P due to this little bit of charge.
00:55:29.900 --> 00:55:42.600
I will draw my vector distance here being nice happy green color.
00:55:42.600 --> 00:55:52.700
This will be our electric field due to that little bit of charge.
00:55:52.700 --> 00:56:08.600
This distance we will call Ri which in this case we can use the Pythagorean Theorem, it is going to be we have got 1 piece of the triangles Z, the other piece R.
00:56:08.600 --> 00:56:19.200
This will be the √z² here is the component + R² the constant, and the radius of our hoop here, our ring of charge.
00:56:19.200 --> 00:56:24.700
We will call our angle here Θ i.
00:56:24.700 --> 00:56:27.100
What else do we need to define?
00:56:27.100 --> 00:56:33.700
Usually we can get that little piece of electric field defined before we really dive in.
00:56:33.700 --> 00:56:43.600
Let us say that the electric field due to that little piece we are looking at just as z component 1/ 4π E₀.
00:56:43.600 --> 00:56:49.300
We have got our charge Δ Q ÷ our distance Ri².
00:56:49.300 --> 00:56:59.700
It looks like we are dealing with the cos of Θ I, the horizontal component as we have set this up currently.
00:56:59.700 --> 00:57:01.400
We can do a little bit more work here.
00:57:01.400 --> 00:57:05.700
We know that RI is z² + R² we already defined that.
00:57:05.700 --> 00:57:14.200
RI = √z² + R² and cos of Θ i.
00:57:14.200 --> 00:57:27.700
We also look at the cos that is the adjacent / the hypotenuse that is going to be, our adjacent side is our z component and our hypotenuse is Ri again.
00:57:27.700 --> 00:57:43.600
Rewriting this, Ei in the z direction is 1/ 4π E₀ × our charge Δ Q and pulls up there ÷ Ri²
00:57:43.600 --> 00:58:01.600
is going to be z² + R² × cos Θ i which is going to be z / √z² + R².
00:58:01.600 --> 00:58:11.400
If you want to simplify that before moving on which is probably a good idea, we have 1/ 4π E₀.
00:58:11.400 --> 00:58:29.700
We will throw in our z, we have got z/ z² + R²³/2 × our Δ Q.
00:58:29.700 --> 00:58:34.000
Before you get too nervous, this one is actually a little simpler than the last one.
00:58:34.000 --> 00:58:41.700
Let us see if we can dive into the math now and solve this one.
00:58:41.700 --> 00:59:01.500
Electric field due to a little bit i in the z direction is going to be 1/ 4π E₀ z/ z² + R²³/2 Δ Q.
00:59:01.500 --> 00:59:06.600
But we know what Δ Q is.
00:59:06.600 --> 00:59:16.900
Δ Q is just going to be our linear charge × Λ × the radius × D Φ.
00:59:16.900 --> 00:59:35.700
Then we get that E in the z direction is going to be the integral from Φ = 0 all the way around the circle this time so from Φ= 0 to 2 π of 1/ 4π E₀.
00:59:35.700 --> 00:59:49.800
We have our z/ z² + R²³/2 Λ R D Φ.
00:59:49.800 --> 00:59:57.200
Next up, electric field in the z direction = let us pull all of our constants out of here.
00:59:57.200 --> 01:00:09.000
4π E₀ a constant in our problems z constant, z² must be a constant, R a constant, Λ is a constant, R again is a constant.
01:00:09.000 --> 01:00:32.100
We can pull most of these out of the integral sign so we have Λ R/ 4π E₀ z / z² + R²³/2 integral from 0 to 2π of D Φ.
01:00:32.100 --> 01:00:34.400
That looks so much friendlier.
01:00:34.400 --> 01:00:39.500
The reason we can do that is none of these depend on our variable of integration here Φ.
01:00:39.500 --> 01:00:44.000
They are all constants with respect to any changes in Φ.
01:00:44.000 --> 01:00:46.900
The integral of DΦ is just going to be Φ.
01:00:46.900 --> 01:00:56.200
Therefore, we can say that Ez = the integral of DΦ is Φ from 0 to 2 π that is just going to be 2 π.
01:00:56.200 --> 01:01:22.400
We have 2 π from this × Λ R ÷ 4π E₀ × z ÷ z² + R²³/2.
01:01:22.400 --> 01:01:24.500
This is much better.
01:01:24.500 --> 01:01:35.600
Our Λ we know what that is equal to, so this implies then knowing that Λ we already defined as Q/2 π R.
01:01:35.600 --> 01:01:53.900
We can say that the z component of our electric field is = we have 2 π Λ was Q/ 2 π R.
01:01:53.900 --> 01:02:09.000
We still have R up here and the 4π E₀ down here, a z/ z² + R²³/2.
01:02:09.000 --> 01:02:11.700
It looks like we can make a couple of simplifications here.
01:02:11.700 --> 01:02:15.100
2 π/ 2 π does make a ratio of 1.
01:02:15.100 --> 01:02:22.600
We have got R and R, those make a ratio of 1 and that looks pretty good.
01:02:22.600 --> 01:02:33.100
We can rewrite this then the electric field in the z direction is going to be 1/ 4π E₀.
01:02:33.100 --> 01:02:51.700
We got a Q and a z ÷ we are left with z² + R²³/2.
01:02:51.700 --> 01:02:54.900
Hopefully, you are starting to see a pattern in these.
01:02:54.900 --> 01:02:59.900
The same sort of strategy applied again and again and again.
01:02:59.900 --> 01:03:08.700
If we take a single ring, we just solved for the electric field there but what happens if we start solving for electric field due to a bunch of rings?
01:03:08.700 --> 01:03:12.100
That is going to be our next problem.
01:03:12.100 --> 01:03:17.200
A uniformly charged disk, define the electric field due to a uniformly charged
01:03:17.200 --> 01:03:23.800
and slating disk of some radius R to point B perpendicular to the disk as shown on the diagram.
01:03:23.800 --> 01:03:27.800
To do this when we are going to rely heavily on what we just did.
01:03:27.800 --> 01:03:43.400
Realizing we already talked about how you find a ring of charge, our strategy is going to be to take our ring that we just solve for.
01:03:43.400 --> 01:03:48.200
We know the electric field due to that ring and just say let us start with a little ring at the center,
01:03:48.200 --> 01:03:54.000
find its electric field and make a bigger ring and add it, and a bigger ring.
01:03:54.000 --> 01:04:03.000
We are going to integrate from the radius is 0 all the way to R, of all of these little rings to get the total electric field at point B.
01:04:03.000 --> 01:04:06.500
Relying heavily on what we have done previously.
01:04:06.500 --> 01:04:12.800
Again, we can make a symmetry argument that we only have to deal with the z component because we are centered on this disk.
01:04:12.800 --> 01:04:21.000
This time we are going to define a surface charged area of Σ which is total charge Q ÷ A.
01:04:21.000 --> 01:04:29.500
In this case, our area of the disk is π R².
01:04:29.500 --> 01:04:38.000
If this is our ring here in purple, the charge of that ring is Δ Q on our ring there.
01:04:38.000 --> 01:04:43.600
We will call the radius of our ring some Δ,
01:04:43.600 --> 01:04:47.800
Let me make that a lowercase just to help differentiate what we are doing.
01:04:47.800 --> 01:04:55.200
A Δ R, as we go out, we will call this little r, this variable that we are going to change.
01:04:55.200 --> 01:04:59.800
We can look at the charge contained Δ Q.
01:04:59.800 --> 01:05:05.600
Δ Q is going to be that surface charge density × the area of this ring.
01:05:05.600 --> 01:05:13.000
To find the area of the ring, we have got to consider here is our ring, what happened if we cut it 1 point and then spread it out to make a line.
01:05:13.000 --> 01:05:15.100
We get a little skinny rectangle.
01:05:15.100 --> 01:05:33.700
The length of that rectangle is going to be Σ × the area, the length of that is going to be 2 π × Ri and the thickness of that rectangle is going to be Δ R.
01:05:33.700 --> 01:05:43.700
With that in place, we can start to find our electric field due to some little bit of our disk, our little bit being that single ring.
01:05:43.700 --> 01:06:13.400
The electric field in the z direction due to that little ring is 1/ 4π E₀ again × we have our z Δ Q due to that little piece of i/ z² + Ri²³/2.
01:06:13.400 --> 01:06:17.900
Coming from the solution to our previous problem, we just did that work.
01:06:17.900 --> 01:06:37.000
This implies then that Δ Q, using Δ Q = Σ × 2 π Ri Δ R that the electric field due to that ring i in the z direction
01:06:37.000 --> 01:06:47.300
is 1/ 4π E₀ × z × Δ Q i which we just define over hear in green.
01:06:47.300 --> 01:07:04.500
Σ 2π Ri Δ R ÷ z² + Ri²³/2.
01:07:04.500 --> 01:07:07.000
There is our starting point then we are going to integrate this,
01:07:07.000 --> 01:07:10.700
we are going to add these all up as we go from a little tiny ring here in the center
01:07:10.700 --> 01:07:17.900
and expand it outwards and upwards until we get to the entire width of disk.
01:07:17.900 --> 01:07:21.300
I think we are ready for the math.
01:07:21.300 --> 01:07:48.900
Electric field in the z direction is going to be the integral adding all these up of 1/ 4π E₀ × z Σ 2π Ri DR / z² + Ri²³/2.
01:07:48.900 --> 01:07:52.600
Let us pull out our constant and see if we can simplify this a little bit.
01:07:52.600 --> 01:07:59.000
Which implies that Ez = Σ and z can both come out, those are constants.
01:07:59.000 --> 01:08:12.000
Our 2 E₀ can come out and this 2 and 4, we can do a little bit division there and say that we end up with Σ z/ 2 E₀.
01:08:12.000 --> 01:08:20.400
A π in a π cancel out, integral from r = 0 all the way to the R, the entire radius of our disk.
01:08:20.400 --> 01:08:32.400
We are left with r DR / z² + R²³/2.
01:08:32.400 --> 01:08:36.300
To integrate this I'm going to do a little bit of a substitution trick.
01:08:36.300 --> 01:08:41.600
This is not quite so obvious to me how to do so, I like to simplify things as much as possible.
01:08:41.600 --> 01:08:49.600
I'm going to say, I’m going to define a new variable U and say that u = z² + R².
01:08:49.600 --> 01:08:52.600
This would be u³/2.
01:08:52.600 --> 01:09:01.300
If I do that the differential of u, our only variable here is the R so that is going to be 2 RDR.
01:09:01.300 --> 01:09:06.100
To make all of this fit that form, I have got u down here.
01:09:06.100 --> 01:09:09.000
I almost have Du up here.
01:09:09.000 --> 01:09:13.000
If I had a 2 here this would be Du/ u.
01:09:13.000 --> 01:09:15.300
Let us multiply it to here and change the values.
01:09:15.300 --> 01:09:19.000
I have to divide by a 2 as well out there.
01:09:19.000 --> 01:09:31.500
What I have is a constant × the integral of Du/ u³/2 and that I know how to integrate.
01:09:31.500 --> 01:09:37.600
Not Du/ u, Du/ u³/2 and that I know how to integrate.
01:09:37.600 --> 01:09:50.800
We can rewrite this electric field in the z direction, we have Σ z/ 4 E₀ integral.
01:09:50.800 --> 01:09:54.500
I have changed my variable of integration from R to u.
01:09:54.500 --> 01:10:01.200
Now we went from r = 0 corresponds if r is 0, u is z².
01:10:01.200 --> 01:10:14.400
We are integrating from u = z² to r = R so this must be u = x² + R².
01:10:14.400 --> 01:10:27.100
Our simplified version of what we are integrating now becomes u⁻³/2 Du, much easier for me to integrate.
01:10:27.100 --> 01:10:49.100
That becomes Σ z/ 4 E₀ × -2/ √u¹/2 evaluated from z², 2z² + R².
01:10:49.100 --> 01:10:51.900
There is the integration I have to substitute in here.
01:10:51.900 --> 01:11:23.000
We end up with Ez = Σ z/ 4 E₀ - 2/ z² + R²¹/2 - -2/ u ^½ which in this case is z² ^½ or z.
01:11:23.000 --> 01:11:49.200
A little bit of algebra here that is equal to Σ z and we can pull out a 2 here from both sides ÷ 4 E₀ × that is going to be 1/ z-1/ √ z² + R².
01:11:49.200 --> 01:12:08.200
If I have a 2 here and the 4 here, we can simplify there the electric field in the z direction will be 2 there, 2 there, Σ/ 2 E₀.
01:12:08.200 --> 01:12:25.900
I can take out the z and say that we have 1 - z/ z² + √R².
01:12:25.900 --> 01:12:29.900
This is just a little bit of algebra going from here to here and simplifying that out.
01:12:29.900 --> 01:12:33.900
There is our answer.
01:12:33.900 --> 01:12:39.600
I think I’m happy with that, not really feel like simplifying any further.
01:12:39.600 --> 01:12:42.300
What if that disk or an infinite disk?
01:12:42.300 --> 01:12:47.400
What if you expanded it and expanded it until it is infinitely big?
01:12:47.400 --> 01:12:50.800
We almost have the solution to an infinite plane of charge here,
01:12:50.800 --> 01:13:12.600
let us do that by taking the limit as R approaches infinity of our answer which should be Σ/ 2 E₀ × 1 - z/ √z² + √R².
01:13:12.600 --> 01:13:22.400
As I do this, R gets really big we are going to have 1 – 0.
01:13:22.400 --> 01:13:31.900
The R is going to dominate over here so we end up with Σ/ 2 E₀.
01:13:31.900 --> 01:13:40.600
If that disk is infinitely big as it becomes an infinite plane, there is no dependence on how far you are from that plane at all.
01:13:40.600 --> 01:13:49.400
Which makes sense, if it is an infinite plane, that distance from it really have any tangible practical meaning any more.
01:13:49.400 --> 01:13:58.600
We get this nice simple Σ/ 2 E₀, it is all a function of your surface charge area.
01:13:58.600 --> 01:14:00.100
Practice is good, let us keep going.
01:14:00.100 --> 01:14:07.800
Hit the break and by all means do so but comeback in we will finish this up here with one last problem.
01:14:07.800 --> 01:14:15.700
Find the electric field due to a finite uniformly charged rod of length L on its side at some distance D away from the end of the rod.
01:14:15.700 --> 01:14:21.800
Uniformly charged rod we are some distance from it, let us find the electric field here.
01:14:21.800 --> 01:14:26.900
By symmetry arguments, it looks like you are only going to need the x component.
01:14:26.900 --> 01:14:29.800
We will do the same thing, let us take and break this up.
01:14:29.800 --> 01:14:35.900
Let us call this little piece of that Δ Q and close there.
01:14:35.900 --> 01:14:44.300
We will define some x distance and distance from P to our Δ Q, there is our x.
01:14:44.300 --> 01:14:54.900
This Δ Q is going to be = to some linear charge density × whatever that little piece of x is.
01:14:54.900 --> 01:15:06.900
Δ x or we could say that the differential of Q that tiny little piece of charge enclosed is the linear charge density × Dx.
01:15:06.900 --> 01:15:13.800
Where Λ is linear charge density is Q/straight forward L.
01:15:13.800 --> 01:15:31.500
The electric field due to this little tiny I, bit of charged in the x direction is 1/ 4π E₀ Δ Q/ distance².
01:15:31.500 --> 01:15:38.400
With that I think we are set to start doing our math.
01:15:38.400 --> 01:15:49.000
Electric field due to that little bit i in the x direction is 1/ 4π E₀ Δ Q/ x² which implies then
01:15:49.000 --> 01:16:02.800
that the total electric field in the x direction will be the integral of 1/ 4π E₀ R DQ / x².
01:16:02.800 --> 01:16:21.800
Which we already said DQ is Λ Dx so we can write that as pulling 1/ 4π E₀ out of the integral sign, integral of Λ Dx/ x².
01:16:21.800 --> 01:16:33.800
Where our limits of integration, we are going to integrate from x = that distance d to x = D + L to get us that total length of our line of charge.
01:16:33.800 --> 01:16:47.300
This implies then that the electric field in the x direction is Λ/ 4π E₀, our Λ can come out as well, it is a constant.
01:16:47.300 --> 01:17:12.700
The integral from D to D + L of x², x⁻² Dx which is Λ/ 4π E₀ × -1/ x from D to D + L,
01:17:12.700 --> 01:17:39.400
which is Λ/ 4π E₀ substituting in we have -1/ D + L - -1 / D which implies then that the electric field in the x direction is Λ/ 4π E₀.
01:17:39.400 --> 01:17:47.100
And doing a bit of rearranging here, putting a common denominator D × D + L we end up with.
01:17:47.100 --> 01:17:57.500
Let us write that down here first, D × D + L we end up with D - D + D + L.
01:17:57.500 --> 01:18:14.400
It is little bit easier to deal with which is Λ/ 4π E₀ - D + D is going to be 0 that is just going to leave an L / D × D + L.
01:18:14.400 --> 01:18:26.600
We also have said again that Λ = Q/ L so that is going to be = I will replace Λ with Q/ L.
01:18:26.600 --> 01:18:39.600
We have still got our 4π E₀ down here × L/ D × D + L.
01:18:39.600 --> 01:18:45.600
L in the bottom and L to the top make a ratio of 1 electric field in the x direction.
01:18:45.600 --> 01:19:02.100
1/ 4π E₀ × Q/ D × D + L.
01:19:02.100 --> 01:19:04.400
That looks like enough practice problems for now.
01:19:04.400 --> 01:19:10.800
Please take your time, if you need to do these again go back practice some and see if you can do them on your own or see how far you can get,
01:19:10.800 --> 01:19:14.500
then check the video again, pause and rewind our wonderful things.
01:19:14.500 --> 01:19:17.700
Thank you so much for watching www.educator.com.
01:19:17.700 --> 01:19:21.400
We will see you in the next lesson when we talk about Gauss’s law.
01:19:21.400 --> 01:19:22.000
Thanks and make it a great day.