Hey, guys. So in this video, I want to talk about electric fields. It's something that you'll see in your textbooks. You'll need to know how to solve problems with it. But it's also to answer a very important question, which is how to charges exert forces on one another. And the basic answer is that they do them by setting up these things called electric fields. So basically a single charge right which I'll call big Q is gonna produce an electric fields. You might see some confusing definitions in textbooks, but basically the idea is it's kind of like information this big. Q is setting up this thing called an electric fields right that exists in all directions. And then, if you place another charge at some distance, little are away from this produced electric field, then this charge will feel that field and actually have a force on it. So, in other words, it will have a repulsive or attractive force here, which we know from cool arms law. So basically what happens is this thing. Regardless of whether there's another test charge or another secondary charge to feel it, this thing produces an electric field in all directions. And then when you place a charge inside of that field, it feels of force. And that force that it's felt with the electric field is which is set up by big Q is just Q times E in which he has units of Newtons per Coolum, right? You probably don't need to know that, but there it is anyways, So he has units of Newtons per cool. Um, Okay, now which Q are we gonna use in these questions? Well, that Q is always the one that's feeling the force. So the words the secondary charge right here. Now, some of you might be wondering, Whoa, isn't the opposite also happening? Doesn't this Q also produced its own electric field and therefore this one is felt by it, and the answer is yes. So this little Q also produces an electric field that is either felt an attractive or repulsive force. And this is basically how to charges exert forces on one another. You have action reaction because of these things called electric fields. Okay, so again, I just want to reiterate in problems the queue that you're gonna use inside of this f equals qu equation is always gonna be the one that feels the charge or the the electric force. At some distance, little are okay. It's a simple is that Let's go ahead and take a look at an example. We've gotta to cool um, and three Coolum charge. They're separated by some distance. Are So I'm gonna go ahead and set these up. I've got to Cool. OEMs. Uh, got to cool. OEMs. Three columns. They're separated by a distance of little are in such a way that the electric field at this point is 10 Newtons per Coolum. In other words, we have an electric field here at this distance. Oh, sorry, Actually have it backwards. So this is actually gonna be the three Coolum charge, and this is gonna be the too cool in charge. Now, the electric field at this distance is just 10 Newtons per Coolum. And now we have to feel what we have to figure out. What is the force on this to Cool? Um, charge. What does that mean? That means that this is the Q, and this is the big Q. So it all depends on which one is the feeling charge and which one is the producing charge. So this one the too cool in charge is the feeling charge. So in order to figure out the force, we just need Q times E So the force is gonna be Q, which is too cool homes times the electric field at this particular point, which we're told is 10 Newtons per Coolum. And that's just equal to 20 Newtons. Okay, so that's just a simple example. Let's go ahead and get into some or examples involving electric field.

2

Problem

A 1.5μC charge, with a mass of 50g, is in the presence of an electric field that perfectly balances its gravity. What magnitude does the electric field need to be, and in what direction does it need to point?

A

3.27×10^{−7} downwards

B

3.27×10^{2} upwards

C

3.27×10^{4} upwards

D

3.27×10^{5} upwards

3

concept

Electric Field due to a Point Charge

6m

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Hey, guys. So now that we've seen what the electric field is and how charges respond to electric fields, we're gonna look a little bit more closely at the electric fields of point charges. Okay, you don't need to know that. So basically, the idea is we know that charges produce electric fields, and in previous videos, we were sort of like drawing squiggly lines here. But the reality is, instead of squiggly lines, thes positive charges, positive point charge is actually produce an electric field that points radial e outward from it. The way I like to remember this is that positive people are very outward going people. So this is the electric field produced by a positive charge, like one cool, um, or something like that. Whereas negative charges are just the opposite. And we just chose this convention, by the way, hundreds of years ago s Oh, this is just what we established as positive and negative electric field. So these things produce fields that point inwards. That's the main difference between them. We've got outward going and inward e fields like this, but you'll definitely need to know that now We saw how a single charge where it was like this Q. Or this Q over here was basically producing an electric field. And that that when you brought a secondary charge closer to it. So, for instance, if I brought a Q like this, then it would feel a force, whether it's in this direction or this direction, depending on the magnitude of that. Well, the magnitude of the electric field now has an equation, and it's k times big. You divided by r squared. Notice how this kind of looks like Coolum Law, except it's missing one of the little cues. So the main thing is that remember, the electric field is set up on Lee just by one charge. The producing charge Big Q. And the other thing is that this little our distance, instead of being the distance between two charges like we had in columns law. Now it's the distance to the point of interest. So what that means is, that problem will ask you what is the electric field at a certain number, or sometimes you'll be given a diagram of charges, you'll need to figure out an electric field at a specific point. So that's how you're going to get that little our distance anyway, So we have this equation of the electric field generated by a producing charge. Q. And we said that a second charge, if it's moved closer to it that little. Q. It felt a force from that electric field. That equation was given by F equals Q E. What we could do is basically plug in this expression. Now we can actually come up with something interesting. We know that e is just k Q. Divided by r squared. So if you multiply times this Q. If you recognize this formula, all this is is just cool ums law. So this really brings the relationship between the electric field and the electric force. So a producing charge Q produces an electric field. And then when you bring a smaller charge or doesn't have to be smaller, a second charge key. Little. Q. They both feel a force, but the opposite is also true, So you could actually flip little Q. And big Q. One could be the producing. The other one could be the feeling charge, and they're gonna be the same. That's why we have action reaction pairs between two charges because they both exert that mutual forces on each other through their electric fields. So now we get the complete picture between the relationship between Coghlan's law and the electric field. Let's go ahead and take a look at some examples right here. We've got a distance X and the electric field is some number at another distance. Here we have 10 columns and now what is the ratio of between X and Y? So here's what I want to dio. I basically want to draw this little producing charge, which I'll call Q right here Now at some distance away so that I'm gonna call this X. I'm told that the E field here E X is equal to Newtons per Coolum, right, And then at some other distance. So you keep going over here this other distance right here the electric field is going to be equal to 10. By the way, this doesn't necessarily have to be half. It's not to scale or whatever. So this whole entire distance here is equal to why. And now I need to figure out what is the ratio of X and y given those electric field intensities. Okay, so basically what happens is I know that, uh, the electric field at a certain points at, for instance, point x. The equation for that is gonna be K times big Q divided by the distance between X squared. Whereas this guy over here, the electric field at point why is K big Q divided by Y squared. Now what happens is we actually don't know what the producing charge, what the magnitude of the producing charge is. All we need to do is just get the relationship between X and Y. So to do that, you have to divide. We're gonna need a ratio. Right? So what I'm gonna do is I'm gonna divide e X by ey so you're gonna get K big Q over X squared all over K big Q over. Why squared now? The reason this is important is because we can actually cancel out the case and the cues that appear on both sides. Right. If you do copy dot flip. If you manipulate this equation, you're just gonna get the same thing on the numerator and the denominator. So what we end up with when you do this fraction is you're gonna get e X over e y is equal to we have one over x squared, divided by one over y squared. So this is actually gonna go all the way up to the top, And it's gonna be why squared over X squared. OK, so now what we can do is we want to get rid of this X squared over here. So we want to square root both sides. So we're just gonna get this square roots. We make some room for that. We've got the square roots of e X over E Y is equal to the square. Roots are actually when you when you take the square root of these y squares and X squares, they're just gonna turn in tow Y over x. Okay, so now we've got something that kind of looks like our ratio here. We've got y over X. All we need to do is we need to flip it because we need X over y. So all we have to do is just flipped both sides and I'm gonna move this over here, actually, so I get the square root of e y over E X is gonna equal X over y right. So all I did was I just flip that this was the X over e y. Now it's ey over e x, and I would just flip. This is well, right. So you could do that. You could just flip both sides of them cool. So I know I have to plug this stuff in, so I know that e y over here is 10. And I know the X is 20 Newtons per Coolum, and that's just in e equals X over y. And if you plug that and it's just gonna be a ratio, and you should get I get 0.71. So, in other words, that is how far E X is relative T. Y. So this is about 0.71 of the distance between this whole entire line segment here. Okay, So let me know if you guys have any questions, let's keep moving on

4

example

Zero Electric Field due to Two Charges

6m

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Hey, guys, lets you got this problem together. So we've got these two, these two charges which lie on the X axis, and we're supposed to figure out what distance or a point on the X axis Is the electric field going to be zero? But I want you to remember that for this problem, we're gonna be using the fact that again outward or positive charges produce outward going electric field lines, they always wanna go away. Whereas negative charges produced in we're going electric field lines. This is gonna be useful because we actually have a positive charge right here and a negative charge right here. So we're gonna need to remember which directions those electric fields go in. And we've done these kinds of problems before when we were looking with columns law. So in order to figure out basically before we start plugging in numbers, we have to figure out which region is this possible for the electric fields to cancel out. So I've got three basic choices I've got on the left I've got in the middle and I've got on the right. Now, if you look the middle, what's gonna happen is at this point of interest. Some of the words this p right here, the electric field from this positive to cool, um, charge all the electric field lines, we're gonna point away. So that means that at this position right here, the electric field from that positive charge is gonna be towards the right. You see, You see, that always wants to go away. Whereas the electric field from the negative charge always wants to go towards that charge. So that means that at this point, it wants to go to the right as well. So, basically, this is the electric field from the negative charge. And there's never gonna be a situation which these things will cancel out. They're always going to add together, no matter where I am along along this line between them. So that means that it can't be in this area because they never cancel out. So let's take a look at the right region now. So we have the inward going electric field lines, so that's gonna be e minus. And from this positive charge, we have a an outward going electric field line. Now, you might be saying you might be tempted to say Hell we actually might be ableto you know, find a place where these things cancel out. What I want you to notice is that the electric fields for any charges is just k times big Q divided by little r squared. So the words it's the magnitude of the charge divided by the distance between them squared. So this three Coolum charge is always gonna be at a smaller distance from this point of interest remembers to some random point here than this entire distance to the weaker charge. Now, what this means is that at this region here, the magnitude of the net of the negative electric field is always going to be bigger than this positive electric field because it's a larger magnitude of a charge, and it's always going to be a smaller distance because this distance right here is always going to be larger, and it's a weaker charge. So, in other words, the magnitude of this is always gonna be bigger than the magnitude of the electric field. So it's never going to cancel, so it can't be this, and instead we're gonna have to use our last scenario, which is on the right. The magnitude of the electric force. It goes in this direction, their direction of the electric field. Wow. And I've got the inward going electric field from the negative three Coolum charge. Now there is possibly a scenario where I could find a distance in which these will cancel out the weaker charges at a smaller distance. The larger charges that a larger distance so there might be a way than for them to balance out the variable I'm looking for is this distance right here, which is X at what point on the X axis is the electric field equal to zero? And what I'm gonna do is I'm just gonna set the origin over here to be zero. So this is just gonna be X equals zero, which means I'm probably gonna get a negative number or I'm just gonna have to specify to the left of the cuticle, Um, charge something like that. And the condition that I'm trying to solve for is that these things should be equal but opposite in direction to cancel out. So in other words, I should have that the magnitude of the e of the electric field from the positive charge should be equal and opposite to the magnitude of the negative electric field. All right, so we know what the magnitude the electric fields are for This guy over here, it's gonna be K. We've got the producing charge, which is the two Coolum charge. And this is gonna be at a distance of X squared now that has to equal the inward going electric field. That's gonna be K times the three. Cool. OEMs. Notice how I didn't put the negative sign there because I'm just working with the magnitudes of them, right? So just set their magnitudes equal to each other. And now this distance right here is actually not going to be X. But it's gonna be X plus the seven centimeters square, so actually could just write that. So I've got X plus 0.7 square. That's seven centimeters. Just written in s I Right. So now what I can do is I can try to solve for this x variable over here. So I got this equation and I could go ahead and cancel some stuff out. I got the K's. And now, in order to get this X over one side, I have to move this thing over and I have to move the two down, so basically, they just have to trade places. And if I do that, I'm gonna just continue writing all this stuff in black because it's gonna be, ah, pain if I try to switch colors. So I've got X plus 0.7 squared, divided by X squared equals And then I've got three halves. Great. Now, rather than starting to foil all of this this whole, this whole mess out and getting a whole bunch of quadratic terms, things like that, what you can do is realize that both of these terms on the top and bottom r squared, which means that we can take the square root of each side. So we're gonna take the square root, take the square root, and then it just becomes X plus 0.7 over X equals the square root of three halves. Okay, so now all we have to dio is to start isolating for X. I'm just gonna move it over to the other side. This becomes X plus 0.7 equals square root of 3/2 times X. And now all I have to dio is I have to subtract this X to the other side. So by up to minus X, and I have this 0.7 equals the square root of three halves X minus X. Now I can do is I can pull this X out as a common factor because it's like, you know, it appears in both of these terms, so basically have 30.7 is equal to x times the square root of three halves minus one. Notice how if you distribute this X back inside of both of these terms, you're just gonna get back to the expression that you had right above right? So it's just it's a good shortcuts is basically just pull out that greatest common factor. Okay, and now what we could do is we could basically just move that over and divide. So we have 2.7 divided by the square root of three halves minus one, and if you just plug that stuff in, you should get the X distance, and you should get 0 31 meters or that's just 31 centimeters. But remember we said that it's going to be to the left of that too cool in charge. So we just have to specify its to the left off the too cool, um, charge. And this is the answer. So if there was, there was, like, an actual coordinate system here, this would be, like minus 31 centimeters anyway, but this is actually perfectly fine. If you just left it like that, let me know if you guys have any questions. And if not, we're gonna keep going with some more examples.

5

Problem

If two equal charges are separated by some distance, they form an electric dipole. Find the electric field at the center of an electric dipole, given by the point P in the following figure, formed by a 1C and a −1C charge separated by 1 cm.

A

1.80×10^{14} N/C

B

3.60×10^{12} N/C

C

3.60×10^{14} N/C

D

7.19×10^{14} N/C

6

example

Electric Field Above Two Charges (Triangle)

4m

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Hey, guys, let's go ahead and work this one out together. We have these two charges right here, once positive ones negative. And we're supposed to find out what the electric field is at this point of interest over here. All right, so let's go ahead and do that. We've got a positive charge over here. A negative charge over here. So at this point of interest, we just need to figure out without any just calculations where the electric fields going to point. So we know from this, uh, charge right here. We know the electric field is gonna point off in this direction because it's going outwards, right? It wants to point away from that charge. Whereas over here, the electric field is going to point in this direction towards the negative test. Negative charge. And that's gonna be e minus. All right, In order to figure out those electric charges, I need the formula. I've got e equals k que over r squared. So I need to know the distances between the source charge or the producing charge Was the one cool. Oh, and the point of interest right here. Right. So I've got this little our distance over here and I have a triangle. I've got five centimeters and I've got eight centimeters here for the vertical piece. So I configure the high pot news of this triangle using the Pythagorean theorem. So you work this out, you're gonna get 9.4 centimeters. And if you were to do the same thing over here, you would have to do the same exact thing. You know, five and eight. And you have to figure out the iPod news. So this is actually starting to look like we have a lot of symmetry in this problem. So we have five and eight. We basically have this sort of, like isosceles triangle. So let's go ahead and see if we can use symmetry just like we did in the last video. To sort of reduce the amount of work that we have to dio. So we need these same charges. So, in other words, the same magnitude of the charges we got the one cool, um, in the negative one. Cool. Um, at the same distances. Which means they're gonna produce the same electric field, right? And if these things are symmetrically placed around this, these two charges then that means that when we go ahead and break up these electric fields into their components, so e x and the Y, we're gonna have the angle theta That's gonna be the angle. Theta is gonna help us break up these things into the components. But you're also gonna have components in the Y direction in the X direction from the negative. Cool. Um uh, the the negative piece of the electric field. And that's going to be an angle theta as well. So in other words, if these things, if he's test charges in the point of interest, are symmetrically placed, that we're gonna actually have the same exact datas for both of them and what that allows us to do is to eliminate the use of the vertical components. We don't have to. We have to worry about them because they're gonna cancel out. And instead, what that tells us is that the Net electric field is gonna point off in this direction. And it's just gonna be too times the electric field of the X components in which the X component of that electric fields or the X component of anything, is the magnitude times the cosine of the angle as long as your angle is relative to the X axis. So that means that our E nets are the Nets is just going to be too times I have the formula for the electric field over here. So we have cake you over r squared times the cosine of the angle. So I have all of these numbers are have case constant cues is the charge. And are is this distance right here This 9.4 centimeters all have to do is to go ahead and figure out what the co sign of this angle of this triangle is. So I have to go and use the triangle. So instead of using this angle right here, we'll have a whole bunch of stuff written out. E can also say that this angle right here is theta because these lines right here are parallel. So this is the same angle theta one in the triangle and in any triangle, the way we figure out co sign of Fada is if we have the angle, we can just plug it in. But if not, we can use it by relating the sides of the triangle together so we have could So Kyoto, Uh, by using so Kyoto we have co sign is the adjacent over high pot news three. Adjacent side is five. The iPod news is 9.4. Notice how I don't have to convert it to meters or anything like that. I could just plug it in. His meters are centimeters, because this thing is ratio anyways, So I've got 0.53 for my co sign. So that means that I can actually just plug in this number inside for co sign of theta. All right? And so now let's go ahead and just plug in all my numbers. I've got two times 8.99 times 10 to the ninth. Now I've got the queues right for the queues for both of them are just one Cool. I'm just gonna plug in the positive number. This our distance right here is gonna be point 094 because we have to convert m, it's gonna be squared. And now for this co sign of of theta right here, all you have to do is just substitute 0.53. Don't plug in co sign of 0.53. It's just 0.53 if you work this out, the Nets electric field is going to be 1.8 times 10 to the 12th Newtons per Coolum. So that's the final answer, and this net electric field points purely in the X direction. So this is one way that you guys can use symmetry, get very familiar with it. Watch this video a couple times. Just so you understand how I was able to work through this symmetry and I'll see you guys the next one, let me know if you have any questions.

7

Problem

4 charges are arranged as shown in the following figure. Find the magnitude of the electric field at the center of the arrangement, indicated by the point P.

A

8.99×10^{9} N/C

B

1.21×10^{10} N/C

C

3.24×10^{10} N/C

D

8.09×10^{10} N/C

E

1.94×10^{11} N/C

8

example

Balancing a Pendulum in Electric Field

5m

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Hey, guys, let's get more practice involving electric fields. So we have a charge that's at the end of a pendulum. And this pendulum, we're told, is equal in equilibrium and suspended inside of a uniformed electric field With that E over here. Now we're told, with the electric field, Magnitude is and our job is to figure out what is the charge on the end of this pendulum? Que so there's a couple of important things to pick out from this problem. The fact that it's an equilibrium, right? This condition right here, equilibrium means that the sum of all forces in the X and Y directions is equal to zero. Right. All the forces basically cancel out. We're going to see how we can use the forces in order to solve this problem. But if we're solving for the charge. Q. There's basically one of two equations that we can use. We can use cool OEMs law that EFC over our streak of r squared. But the thing is, we don't have another charge, right? There's no second charge over here. So instead, what we're gonna do is we're gonna use the electric force is equal to Q e right, The charge times, the electric field, which we actually have over here. All right, so we're gonna go and use this formula right here to solve for Q. And if we go and rearrange to this, the Q is just getting equal f divided by electric fields. So I have with the electric field is if I can figure out what this electric force is, then I could basically just plug that in and software the charge. All right, so how do I go about solving that force? Well, there's a bunch of forces that are acting on this, So let's go ahead and draw a free body diagram for this. We have an electric field on a point charge over here, so we know that the force is going to be in the same direction. If this is positive now, the reason that thing doesn't just go flying off is because it's held in place partly by this string that it's attached to, which has some tension. And there's also another force due to gravity, right, because this thing has some mass 30 g. So we get We said that this object was an equilibrium, which means that all the forces were gonna cancel out. So in order to set this thing up first we have to break up this tension into its components. So we have an X and A Y component, and we get those components T y and T X by relating the tension with the CO side or sorry, the angle theta that it makes with the X axis. Now what is this angle right here? So right, this over here, this angle is equal to Well, if you sort of, like, project out this sort of horizontal line, this is a 90 degree angle. This is a 15 degree angle, which means this has to make up the difference between 180. So this angle is actually 75 degrees, right? So it has to basically, uh, this angle this angle has to add up to make 90. So we've got the angle right here. So that means that the tension in the X direction is going to be t co sign of data and the T in the Y direction is gonna be t sign of data. So now we have these components. Let's go ahead and set up our equilibrium conditions in the X direction I have that the sum of all forces in the X direction is equal to zero. So what I have to do now is I have to pick a positive and a negative direction. So I'm just gonna go ahead and assume that the top and right directions are positive. If you guys chose to do something else, your answer is still going to come out the same. You're just gonna have some negative numbers first, but the physics still works out. So I'm just gonna choose this to be the positive directions. Now, starting off my forces in the X direction, I have the electric force, which is really what I'm looking for, minus the tension or the X component of the tension. In other words, the electric force here is just equal to the tension times the cosine of the angle. So I'm like, Great. If I have the coastline of the angle and I have the tension, I could just go ahead and figure this out. The problem is, I know what the co sign of data is. I don't know what the tension is. So remember, Think back to Newton's laws. Whenever we had a situation where we needed one variable in the X direction, we usually went to the Y direction to solve for it. So let's go ahead and do that. We're gonna go over to the Y direction. We have the same exact condition we have. The sum of all forces in the Y direction is equal to zero. So you start from the positives, which is gonna be your t y. And then that's gonna equal your minus M G equals zero. In other words, your tea sign of data is equal to M. G once you move this to the other side. So now tea is actually equal to M G, divided by the sine of theta, and I actually have what all of these numbers are. I have the mass I have G and I also have sign of data. So in other words, I get 30 g, which is 300.3 kg, 9.8 for G. We're gonna move down, and then I've got divided by the sine of the angle, which is 75 degrees. If you work this out, you're gonna get attention of zero point through 30 Newton's. And now you can just take this number and then plug it back inside of this formula right here. So I know that the electric force is just equal to 0.30 Newton's times the cosine of the angle, which is 75 degrees. So that means that my electric force is equal to I got 2. 78. So now what I can dio is I can take this electric force and I just have to plug it all the way back into my initial equation right here to solve for the charge. Remember, that's my final variable. So I've got two. Q is equal to this is gonna be 0.78 only highlight that's you guys makes you understand what that connection is. And I've got 0.78 divided by 100 Newtons per cool. Um, so you should get 7.8 times 10 to the minus four columns. And this is, by the way, our final answer. So let me know if you guys think about this problem, let me know if you guys have any questions. I'll happily explain any steps and I'll see you guys the next one

9

Problem

In the following figure, a mass m is balanced such that its tether is perfectly horizontal. If the mass is m and the angle of the electric field is ?, what is the magnitude of the electric field, E?