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Cornell University

University of Michigan - Ann Arbor

University of Winnipeg

03:38

Keshav S.

(II) A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 $g$'s. Calculate the force on a 65-kg person accelerating at this rate.What distance is traveled if brought to rest at this rate from 95 km/h?

01:23

Suman Saurav T.

(II) According to a simplified model of a mammalian heart, at each pulse approximately 20 $g$ of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?

0:00

Muhammed S.

(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?

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when we begin our discussion of charge. We started with columns, law and electric force between charges, but we quickly moved on to the electric field. That was because we knew even if we don't have to, charges that could exert forces on each other. We knew that one charge was doing something of the space around it, and that's something turned out to be creating an electric field that other charges could react to the force. So now let's consider that we have this box in space. We don't know what's inside the box. All we know about the boxes that it's there, we can't see inside it. And otherwise it's imaginary. There's no mass. There's nothing else there were told There may or may not be charged with in this box. How do we determine if there is? Well, we could take this test charge in red and bring it to a point on the box, and we expect that if there's charge inside the box, we feel a force on this test charge. However, maybe the charge inside is a charge distribution of positive and negative charges and the net force of where we've placed our test charges. Zero. Well, we can take it to another spot on the box and test the force there, and we could take it to another and another. And if we don't feel any charge, any force, at any point on this box, we'd expect Well, there's probably no charge within the box and we'd be correct. However, we might feel a force and say We're told her is at most one charge in the box. Well, say we bring it here and we feel a repulsive force. Now we know we have a positive charge within the box because the force is repulsive. And if we move the particle over here, well, we know we have one charge and we know it must be positive. So we know we'll feel a force this way and we can move the charge around. And we know we will always feel a repulsive force no matter where we are, because there's only one charge in the box. And if it was positive at one point, if we felt that it was positive at one point on the box, then it's got to be a positive everywhere. And conversely, if the charge is negative, then we'd expect all the forces two point to be attractive to point us into the box. Now let's consider this box and say we're told that there may be charging it and there could be any kind of charge. It could be positive. It could be negative. It could be both. Or we could have positive and negative charges. Um, so we can use our test charge and say we move it here. Maybe we feel a repulsive force, and we move it again and we feel unattractive. Force into the box and we move it again. And maybe over here, we feel no force whatsoever. Well, we know we're feeling the superposition of electric fields from all the charge within the box at any point on this box, and that could point in or out, or it could be zero. And so we kind of begin to develop this concept that there's kind of a flow of the electric field either into as in here are out off, as in here. This box now recall we only use that as an analogy to better understand it. The electric field doesn't flow for fixed charges. The electric field is fixed everywhere in space. But we can use this analogy to help, better understand and eventually determine what kind of charge we have inside. And so to better develop this analogy, let's work with um, or concrete example. And let's say we have some charge off, plus capital que inside our box. Well, we know here where we have our test charge, we get an outward force. And as we move that charge around, well, we know everywhere on the surface that should be an outward force because we have two positive charges and the force between them will be repulsive no matter where we go on this box. And so we think, in our analogy terms that this positive charge creates a flow of electric field outward. But the more, as we said, electric fields don't flow. The more appropriate term is that it creates an electric flux through the surface outward of this box everywhere on this box. And that's the important term here. Electric flux and what it is is kind of, ah, definition of the magnitude of any field through the surface. How can we quantify this magnitude more? We'll say we take this charge out and we put in some new charge with plus two capital Que Well, we expect our force here to double and our force here to double. And wherever we tested again, we should measure double the force we did when we just had a charge of plus capital Q. There. And so we see that the flux through a surface. It's proportional to the charge inside, and we also know it's proportional not only to the magnitude but to the sign of the charge inside. So if we put a negative charge in there, suddenly, all of our forces well, point into the box and we say there's electric flux into the surface of this box. So now let's go back to our original example of where we had a charge of plus Q within the box. And now, instead of doubling it, let's say that we added a second charge of plus Q so not plus to Cuba to different plus Q charges. Well, if we put our test charge over at this surface again, we'd expect to see an increased force on it, compared to just the Central Central plus Q charge here. We don't expect us to double. We might see the original force that we saw in one direction and an additional force in this direction and some net force this way. But the magnitude by a factor edition wouldn't necessarily double, but we would see is an increase of the force everywhere around the box. And so when we add up, those forces are on the box. We would see a double compared to what we measured for just one Q plus charge. So this leads us to believe that well, So now let's consider another situation. We're back to just the charge of plus Q in the box plus capital Q and raising our test charge of plus little Q. And we measure a force here and, of course, really everywhere. So we said we have an outward flux all over the box, which we expect. And let's say that the total flux we're gonna call fee is equal to some value phenomena. Well, now, instead of simply doubling the charge, the single charge inside the box from plus Q two plus two Q. Let's put a second charge somewhere in the box, also, plus Q. So now when we measure the force well we don't necessarily expect the force at a certain point to double from before the component from the original plus Q charges the same, and we know that the new plus Q well give us a force in another direction. And when we add those by superposition, the total force we feel on our test charge will be bigger than before, but not necessarily double because of the nature of Vector Edition. However, as we move this around, we'll see that everywhere we measure the force will increase compared to when we just had the plus Q charge inside the box. We'll get different magnitudes and different directions, depending on where we're measuring mhm. But if we originally measured a fee of plus Q. Now when measuring fever plus two Q all over the box, we'll see it actually doubles. And so the flux, as we said, is proportional to the magnitude of the charge inside. But it's actually a proportion of the magnitude of all the charges inside, which leads us to one other conclusion, say instead of a second plus Q charge, we put a minus que charge in there. Well, we'll measure everywhere and in some points again, we'll see some charges into the some field pointing into the box, some pointing out at some points. It will just be zero. And so our fee for plus or minus que well, actually equal zero when we added up all over the surface of the box. Now we'll cover one last important example, and that is okay, putting this box in a uniform electric field. In other words, let's say there's no charge in the box, but we have on electric field straight to the right. In other words, we have e equals some magnitude in the X direction everywhere. And let's place this box like so so that the top and bottom faces in the front and back, which I'll draw in with dotted lines, are all parallel to the electric field and the front face over here and back. Face over here are perpendicular to the electric field like so now we'll note that the electric field lines on these four surfaces the top bottom, front and back are parallel to these walls. They don't actually go in or out of those walls. And so we say the flux through here equals zero, and we'll know whatever flux comes in at the back comes in at a perpendicular trajectory and exits at a perpendicular projection trajectory. And so all the flux that comes in the back goes out the front. And so the total flux is zero not just through the four surfaces that are parallel to the doctor field but through the other two faces. And so through this entire surface. And so there's no charging clothes, and we've shown there is no flux. More importantly, we've shown that in order for there to be flux, there must be some component of the electric field perpendicular to a surface. So let's explore that concept a little further. Let's say we have some small bit of surface here and some electric field line in that direction. Well, we know we only want the electric field component that's perpendicular to the face. In other words, E co sign of data. And so if we call this surface some DS, then we see the flux is equal to E. D s cosine theta. Or at least this little piece of flux defy. You might recognize this as a dot product or S is just a surface were integrating over. So when we draw a full surface, such as a sphere or a box, we can integrate these terms and see that the total flux is equal to the integral, the electric field perpendicular to a surface over the entirety of that surface. Now let's take a look at the units of this flux. The field as units of Newton sparkle, Um, and as we integrate over a surface area, we get meter squared. And so flux has units of Newton's meter squared per cool. Um, and we know this is proportional to some charge within a surface, which has units of cool arms. And so to get from electric flux two columns we need to have a factor that has units of Newton's meter squared Cool, um, squared. Now you may recognize these as the units of our electric constant K, or also the universe units of our Epsilon. Not, and so we can begin to tie this all together. We'll do it like this. Let's draw some imaginary sphere three dimensional surface and at its center, let's place some point charge with charge. Plus Q. Now we know for a point charge. As we've shown the field field lines on the field vectors point radio the outward, which means they're perpendicular to the surface of the sphere. So when we calculate our fi e. D s over the surface of the sphere, first, we'll know anywhere on the sphere. Since the electric field of a point charge goes as cake over R squared, the electric field is constant in magnitude all around this fear, and so we can pull the first term out, and we also can get rid of the dot product because it's perpendicular to the surface all around the sphere. Now we just integrator on the sphere and we get its surface area four pi r squared and the magnitude of the electric field is cake over r squared, which, when we multiplied by the surface area, simply gives us four pi que, que or que over epsilon. Not so we can see. We've just shown, at least for a sphere, that the flux through a surface is a charge over F. Selena. And more importantly, it's the charge enclosed, and so we see that if we have a point charge to draw sphere around it, the flux through that sphere, it's constant. It's independent of how big we draw that sphere. And it makes sense because, say, we drew a bigger sphere. Well, the electric field lines still passes through the second service service, and you'll know you might note that the electric field is smaller out here at this point than it is on the first sphere. However, we're integrating around a bigger service. And if we do the math again with some are to, we'll note that it doesn't change this calculation. And so we might say, Well, let's forget about a second sphere instead, let's draw some irregular surface, some three dimensional surface and there we go, some three dimensional surface that isn't a sphere. It's not uniformly separated from the point charge in any way. But again, we'll note that all the flux that passes through the first fear must pass through the secondary surface. And so it doesn't matter what type of surface we pick. We've just shown that the flex through a surface is always equal to the charging closed over epsilon not, and this will give us a powerful new tool that will allow us to cook to derive more complex electric fields for more complex charge distributions. And this general equation that we found is called KAOS is long, and it tells us that the electric flux is equal to all the electric field lines, the past perpendicular early through a surface we integrate all over that we get the total charging closed over Epsilon. Not the important thing to note here is it's not limited to the point charges, as we've done in this. In this example, this includes any charge distribution, all charges enclosed. In the next video, we'll see a few examples of how we can put this thio toe work in order to help us solve for more complex charge distributions and the electric fields they create.

Electric Potential

Capacitance and Dielectrics

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Magnetic Field and Magnetic Forces

11:08

09:57

10:50

11:12