Understanding Electric Fields and Potentials

So if you've got a charge Q, that charge is going to create an electric field and that charge is gonna create an electric potential. What I mean is, here's a charge Q, and it's orange, because this charge here is orange because this charge here is creating, this is important. This is the charge in this formula that creates that electric field. Creates that electric field, creates it everywhere around it, radially outward if it's a positive charge, and if it were a negative charge, Swap the direction of the arrows, it would go radially inward. Similarly, this charge creates an electric potential everywhere, everywhere around it, so creates that V. So if you plug Q into this formula for V, that is the Q creating the V that you're talking about, that you're solving for. Now these formulas aren't that hard to like, solve for mathematically. You just plug in the R you want, so if this E is 10 meters away, you'd plug in 10 here. plug in the Q that's creating the E at that point, and it gives you the electric field. Similarly for electric potential, you don't even have to square the R, or worry about the vector nature, because electric field is a vector. This has a direction. Electric potential doesn't even have a direction, it is simply a number. Could be positive or negative, but it's simply a number at some point in space. And so this Q and every other Q is creating an electric field, an electric potential everywhere around it. You find it with these formulas, fine. Easy. However, what if your electric charge isn't a point? What if it's spread out over space? Say you took that same charge cube, but you distribute it uniformly across a, like, rod right here of length L. And let's say this time you want to find what's the electric field at point X, distance, you know, X away from the right side of the rod. Well, you can't just blindly be like, okay, 1 over 4 pi epsilon naught total Q over what? X squared? What are you going to put? X? because that's how far the e at this point, like we know if this is positively charged, the e would point radially to the right because it points away from a positive charge, but what are you gonna plug in? X, that's how far this end of the charge is from that point. What are you gonna plug in? X plus L, that's how far this end is. You can do the average. You can't do any of that. You have to actually solve this using an integral, so you have to be clever. If your charge is distributed over some like, you know, Line or area or something like that a volume you're gonna have to do an integral to find the electric field So what do you do you use this formula? But you're gonna have to be clever the way you're clever about it is you you say look this formula gives me the electric field From a point charge well you turn this into a point charge Specifically you turn into an infinite amount of point charges so imagine you break this up now. This is not infinitesimal, but imagine This is an infinitesimally thin, right here, little slice of Q. In fact, that would be a differential amount of Q. That's what we mean by dQ, a differential amount infinitesimally small. Well, that you can tell me how far away it is. So, this is exactly that far, whoops, too far. This is exactly that far away from that point where we want to find the electric field. Like, let's say we wanted to find the electric field over here. This has an exact location. So I could say, alright, that little dq, that infinitesimal charge creates, it's gonna be a really little e, you know, it's gonna be a really tiny differential amount of e. I can't even draw it differentially small, but it'd be a infinitesimally small amount of electric field because this is hardly any charge at all. So you turn this into d e. You say, well, if a big charge creates electric field e, you say that it... differentially small amount of charge is going to create a differentially small amount of electric field over here and so this formula is true it's just r you have to be careful r is from that charge to this point here so we need you know uh we could label this in a certain way we can call this r right here so that's going to be r from the dq to that point where we want to find the electric field okay but that's just the electric field from one little infinitesimal amount of charge we need the electric field from all of them Those that are closer will contribute more because they're closer to that point. Those little differential amounts that are far away, like this one's gonna create the least amount of electric field, they're all contributing. And they're all gonna contribute with electric field to the right, that's important. Positives all create fields radially out. So they're all pointing rightward, we can just add them all up to get the total. How do you add up an infinitesimal amount of things? You integrate, so we'd have to integrate both sides. Integrate both sides, the left-hand side just becomes the total E. So this side's gonna become E, and that's gonna be equal to, I can pull out all this extra stuff, I can pull out the one over four pi epsilon naught. I cannot bring out the dq. Q, obviously, that's the differential that's gonna be integrated. I also can't pull out the R squared. This R's gonna be different for each little charge. Sometimes you get to pull out R squared. In this case, we do not get to pull out R squared, so this is gonna be DQ, differential amount, over R squared. Well, I've got DQ, and I've got R down here. So what you're gonna have to do is change DQ into something else. This is the first order of business. What you're typically gonna do is just say that, all right, we have to talk about charge per length. So lambda. Lambda, is there a B? I'm gonna put a B. Lambda is gonna be charge per length. That's what we mean by lambda, it's the amount of charge per length. And let's just say this is uniformly distributed. You could have some crazy lambda that depends on X, it gets more and more dense as you go. Fine, whatever, you'd plug that in. But all we need here is to say that, well, Q is gonna be lambda times L. Okay, so how about dQ? I mean, I want a differential amount of charge. Well, I'm gonna put a B. If you take the charge per length and you multiply by a differential amount of length, you'll get a differential amount of charge. So all this formula's saying is, if I know how much charge I have per length, coulombs per meter, that's what this is gonna be, coulombs per meter. If I know how many coulombs per meter I have, and I multiply by this little infinitesimal amount of meters, I'll get the infinitesimal amount of charge that's in there. So this is what we're gonna use for dq, at least if you have a line, like a 1d example. So dq's gonna become that. So we're gonna have Equals in fact. I'm just going to write it in here. Let's just erase it you're going to have Lambda Dl Lambda Dl I'm going to pull it the lambda out so lambda is going to come out here That's a constant if it wasn't a constant you'd leave it in the integral and integrate whatever it was as a function Maybe it's like x x squared who knows but lambda comes out here. You have Dl. Well. I'm still like crap was L That's R. I need the same variable here, but But what is dl referring to? It's referring to this little bit of length here. So that's dl right there. An infinitesimal amount of length. That's the same as dr. Look, r goes in the same direction here. I mean, if you want to be very careful about it, you could define l. Define l as like this distance here from the one end. And then you could say, all right, you could say that l plus x, right, l. plus x, so every one of these gets a little address. This one would have an L of zero, this has an L of L, this over here has an L of capital L, L gives you, little L gives you the location of the charge. You could say that that L, that location from the right end, plus x has to equal r, but look, if I take a differential of both sides, x we're just keeping fixed, so that means dl just equals dr in this case. So I'm gonna replace this with dr. DL, DR, like it's the same thing. You can call these DX and write this R as X. We're talking about the same direction here. These are all along the X direction, so we call this DR. And this is what we're gonna use to solve. Now it's not too bad, I just need to get my limits right. I'm integrating R, not L. So R is gonna go from not zero. None of these little differential amounts of charge were zero away from this point over here. Remember, r is from your infinitesimal amount of charge to the point where you want to find the electric field. So the closest I get is this little dq right there. That is the closest I get, and that is x away. So the r value, which is from charge to the point where you want to find the field, the closest I get is x. The furthest I get is over here, and that is going to be x plus l. So the largest r value I get is x plus l. So you solve this, you're going to get... Lambda over four pi epsilon naught integral of one over R squared is negative negative one over R Oops negative one over R you evaluate that between between X and X plus L I'm just gonna write it up here. You're gonna get I'll pull the negative out lambda over four pi Epsilon not Evaluated between X plus L so you get one over X plus L, I pulled the negative out front, so I don't have to deal with it, minus one over X. If I don't want that negative out here, I can bring it inside if I want, and just reverse the order of these. So I can bring that negative inside, and I can say, actually, I'm gonna do this term here, minus that term there, just swap the order, and you bring the negative inside. And so that's our formula for the electric field at this point. x away from the right hand side of this bar. Like this is the e. E is this, the total e. Total e equals that. What would you do if you want to find the electric potential over at this point? Well, it's pretty much the same thing. Like, you got this formula. What's the difference? Well, it's not a vector, but that didn't even really matter here. All these E's went in the same direction, so we didn't even really have to worry about the vector nature of this E. We just added them up blindly. V doesn't even have a vector, so we wouldn't have to worry about that anyway. and it's R instead of R squared. So let's just look. Let's just try to identify. What would be different? Well, this would obviously be dv. This would just be R instead of R squared. Oops. So this would be... Oop, it's not even letting me erase that. This would be r instead of r squared. dq still becomes lambda dr. r still goes from x to x plus l, right? This would be v. So when you integrate this, the only difference is that instead of getting negative one over r, you'll get that v is lambda over four pi epsilon naught. The integral of one over r is log, so you're gonna get log of, and there's no negative now. So integral of one over r squared gives you a negative. The integral of one over r doesn't it doesn't give you a negative. It's gonna be log of x plus l. Minus log of x because it's gonna be evaluated integral of 1 over r. Sorry, integral of 1 over r is log r. And so log of x plus l minus log of x. And you could simplify this because you can write that as, you can write the difference of logs as the log of a ratio. So you could say that v total at that point would be lambda over 4 pi epsilon dot log of x plus l over x. There we go. You can simplify it. You can write it as 1 plus l over x if you wanted to. But that's v total at this point. The total v created by all of these dqs added up. And it's going to be... This is going to be the E at that point due to all the Q's added up. Notice it's not as simple as just taking, like, this original formula and, like, plugging an X or plugging an L. You have to do an interval to get the value. Recapping, you break this up into a bunch of infinite. in testable amounts of charge. Those reach at a given distance and then you add them all up by taking the integral. So this was the linear case. What do you do if you get this case? You're like, no, not the curvy. The curvy's the worst. It's actually not that bad. Let's say you wanted to find E at this point, right in the center. This time you got a curve of charge. Oh, wait, I didn't finish this one. Technically, here's what you'd have to do. I wrote this in terms of lambda, but let's say you weren't given lambda. Like I wasn't given lambda, I was given Q and L. So the At the very end, instead of writing lambda here, you'd write it as q over the length of your object. So everywhere I had a lambda, I'd write it as q over l. So this would have to be q over l right here for lambda at the very end, and I can write this as q over l for lambda right here. If you're not given lambda, sometimes you're given lambda, sometimes you're given q. Use whichever one you're given, obviously. Okay, so what if you had this case? So, in that Q, you distributed over half of a circle, a semicircle of radius R. We want to know the electric field right in the center. Before we do that one, let's do the electric field, or sorry, the electric potential. This one's gonna be easier. Let's say we want to know the electric potential in the center. Well, what do we do? We again break this into infinitesimal little pieces. So I break this into an infinitesimal amount of charge, dQ, right there. So an infinitesimal amount of charge, and that's gonna create an infinitesimal amount of V. So an infinitesimal amount of V at that point. Basically nothing but not nothing. So dV gets created by an infinitesimal amount of dQ. How much does it make? This much, because like you plug in the Q you got, you plug in the V. Let's integrate both sides. What are we gonna get? The side becomes V equals, alright, so I pull out my constants, one over four pi epsilon naught. Now here's the magical thing. Look at every single DQ that you can consider. This one here, this one down here. There's something similar about all these, and that is that they are all the exact same radius away from this point. That wasn't true for our line charge. Our line charge had points that were close, points that were far. For this loop, the nice thing about it is that they're all the same distance away. So you're like, what? That means I could pull out r? Yeah, you could pull out r. r is the same for all of those. And you're like, hold on, that just means I get integral dq? Yes you do, and what's the integral of dq? It's just q. So you get v is q over four pi epsilon naught r. Now you might be like, wait a minute, that's just the original formula up here. You know what I mean? Like if I got rid of this integral nonsense, if I had just gotten rid of all this and plugged in my total Q, I get the exact same thing. That's because V is... It's just a, you know, not a vector, it's a scalar. You can just add up all the values here. This dq here at the top contributes just as much v as this point does, as this point does, as that point does. So having it in this loop is no better than just concentrating all of this charge right there. Say you just took all that positive q and put it right there, it'd still make a green v value of this value at that point, same as if you spread it around, doesn't matter, as long as it's all the same distance away, from a point, it's gonna contribute the same v at that point. So you might be like, alright, well then e should be just this, right? Well, no, unfortunately. And the reason is, electric field's a vector. This piece right here is gonna create a dE that goes this ways. Let me draw it straight. That goes that way. So that dE from this piece of dQ goes radially out, cause it's positive. That's the little bit of dE from this piece. What about the DE from this piece over here? Well, it would create an E that goes radially out from it. And you're like, uh-oh, those ain't gonna add up easy. I can't just add up these two numbers. Like if this one was two and that one was two, the total field isn't four. Look, the vertical components of these are canceling, and that's true for any of these points. This top one creates a field to go straight down. This bottom one creates a field to go straight up. All the vertical components cancel. If you had this one here, it'd be perfect. It goes right along the horizontal. The only part that's gonna matter is the horizontal components. You have to add all those up. So we don't actually wanna find the total E, I mean we do, it's just the vertical parts are gonna cancel. These verticals cancel those verticals, these cancel these verticals. The only part that survives is the horizontal piece. So what I have to do is, like, okay, shoot, this is a vector. I need to make sure I find E in the X direction. So I'm gonna take this, I'm gonna call it DQ, fine, and that makes this DE. But if I only want the piece in the x direction, you know, if I only want that piece right there, so that's gonna be DE, little component in the x direction. This is DE in the x direction. I need to multiply by, you know, a sine or a cosine. And it depends on how you want to define this. I'm gonna define my angle. You have to do this. You have to draw this out. This is important. Draw where you want to measure your angle from. So if you're taking sine or cosine, there has to be an angle measured from somewhere. I'm just gonna choose this center line as where I'm gonna measure theta from, so I'm gonna call that theta. So I'm gonna say that this dq is located at that theta right there. Well, that means this is also theta. These are interior angles here. So if I wanna find dex instead of de, I would take my total de, which is the hypotenuse here. This de was the total field that that dq up here created, and then I multiply by cosine theta because this is the adjacent piece. piece. So I multiply by cosine of theta. Now I'm ready. These I can add up. You can add up all the DEXs because those all go the same direction. There's no point in adding up all the DEYs. Those are going to cancel. You can do it if you want. They're all going to be zero. Take my word for it. Or try it if you want. It doesn't matter. So now you can integrate. Integrate both sides. Left-hand side becomes E in the X direction total, which is just E total because the vertical is going to cancel. Equals, bring out your constants, 1. Over four pi epsilon naught. R squared, you might be like, wait, is R squared still constant? Yeah, this is just as far away as that is, as far as that, as far as that. These are all the same distance from this point where we want to find the electric field. So you can still bring R squared out. This is the great news. R squared comes out. I write it as capital. And I get integral dq. Let me find my yellow here. Integral of dq just times cosine theta. So we're not quite as lucky. Over here we just got dq. Alright, well if you get integral dq, that's just q. But integral of dq times cosine theta is not just q. You might be like, can I bring the cosine theta out? You can't. Look, these are all at different angles. They're all at the same r, but this one's at this much of an angle. This one up here would be at that much of an angle, you know what I mean? So they're all at different angles. I can't just bring my theta out of the integral. It's varying. So what do I do? Same thing we always do. You write dq as something else. So you're like, shoot, I actually have to do the integral. What do I do? Well, I know Q is gonna be lambda times L. So just like we did before, DQ is gonna be lambda times DL. Where DL is this little bit right here. So this little length right there is DL, that infinitesimal length right there. Width is DL. And if I multiply that little length by the charge density, linear charge density. Lambda times that little dl gives me the infinitesimal amount of dq in that region. So this is what I replaced dq with down here. So this dq is gonna be lambda times dl. Does that help at all? It does. Lambda gets to come out of the integral because it's a constant, again. Maybe lambda's not a constant. If your lambda's not a constant, fine. Again, as always, leave it in the integral. You'll have to be told what that lambda is, and if you know what the lambda is, you can just integrate it according to whatever the integral is. But in this case, I'm gonna assume that this lambda is constant. We can bring it out. We still have a problem. This is theta, that's L. They need to be in terms of the same variable. That's not too hard to fix. The way we're gonna fix that is we're gonna say that this dl is a little bit of an arc length. Look at, so hopefully you learned about arc length at some point, s. s is r times theta. Well, I should say delta theta. A little bit of an arc length is gonna be s is r times delta theta. Well, how about an infinitesimal amount of arc length? That would be r times d theta. you know, a differential amount of theta. So the theta between this line here and that line there, I mean it's the slice of all slices. This angle in here would be d theta. The ultimate slice right here, the thinnest slice you can imagine. If you took that thin slice of theta, multiplied by the r, that just equals arc length. So this is the formula for arc length. You need to know that to do these problems. But that's ds, well that's just what I'm calling dl. dl is the same as ds. So dl here is gonna be rd theta. You might object and be like, wait, ds is curved. dl is straight, not if it's infinitesimal. If it's infinitesimal, ds is straight. There's no time for it to curve. I know that's weird, sorry, that's math. And so ds is the same as dl. Plug this in for dl down here, and you're gonna get r times d theta. So dl becomes r times d theta. Now you can integrate, you just need some limits. So what are we gonna integrate from? Well, it depends on how you defined your theta. So I defined theta from here, so I'm gonna come all the way from negative that. Like if this is theta equals zero right there, I need to come from negative pi over two, because that's where this point is. My dq with the smallest possible theta would be negative pi over two, based on where I made my zero. And my biggest one would be pi over two, so I gotta go from negative pi over two. to pi over two, the r gets to come out. So like r I can just bring out, in fact I can cancel this r with one of these r's. So that's not harming anybody. I would take the integral of cosine, and that's not too bad either. The integral of cosine is just gonna be sine. You might be like, is it negative? No, integral of cosine is positive sine. So we're gonna get lambda over four pi epsilon naught times r, integral of cosine is just sine theta evaluated between Negative pi over two to pi over two. What does that become? You're gonna get lambda over four pi epsilon naught r. Sine of pi over two is just one. Minus sine of negative pi over two. This is negative pi over two. Sine of negative pi over two is negative one. So that's just two. One minus a negative one is just two. And so you get that the electric field, I'll write it over here, the electric field's gonna be two times lambda over four pi epsilon naught r. Now, again, I didn't give you lambda, so you have to plug that in. What is lambda? Lambda's always just defined to be the charge per length. Again, if you're given lambda, fine, leave it like this. If you're not given lambda, you gotta put it in terms of stuff you know. So you put this in as charge per length, but you're still not done. You're still not done because I didn't even give you the length. You might be like, do I put r? No. Q isn't spread over one radius, Q is spread over one semicircle. This is the length that the charge is spread over. What is that length? Well, it's half of a full circle. A full circle is, you know, a full circle's circumference is two pi r. That'd be the length of a full circle. This is a half circle. Half circle's just pi r. So you divide this by pi r and you can clean this up. You can write this as Q over, I guess you'd have two. Pi squared, epsilon naught r squared would be the electric field, it's in the x direction, the total electric field in that x direction that this little loop creates. That would be the electric field in the x direction. The vertical component would cancel. How do you know that? Well, look, we took cosine theta to get this horizontal piece. You do sine theta to get the vertical piece. And if you integrated sine theta, you'd get cosine theta, but cosine of pi over 2 and cosine of negative pi over 2 are zero. So it's all going to cancel out. So the vertical parts cancel. You only get the horizontal part. For v, it's easy. You just use the formula. Why did we get to just use the formula? It's because v doesn't have any vector directions. We had to multiply by this cosine in order to pick just that piece we wanted in the x direction, that piece that wasn't canceling, so that we add all those up. And if you do that, you do the integral, you solve, you get your electric field. So to recap, what you need to do is use the formula for electric field. but plug in dq instead of q, and that means you have to integrate it. Dq you almost always have to write as lambda dl. Once you have that, pull out anything that's constant. Make sure you're consistent with your limits based on how you defined your angle up here. That'll dictate how you define cosine theta here too. Since I defined theta from this point, I've gotta use cosine theta to find that e. Run your integral, get your limits, solve, you get your value, and then at the very end, if you're given lambda, leave it as lambda. If you're not given lambda, write lambda as what you're given.

Creates that electric field, creates it everywhere around it, radially outward if it's a positive charge, and if it were a negative charge, Swap the direction of the arrows, it would go radially inward. Similarly, this charge creates an electric potential everywhere, everywhere around it, so creates that V. So if you plug Q into this formula for V, that is the Q creating the V that you're talking about, that you're solving for. Now these formulas aren't that hard to like, solve for mathematically.

You just plug in the R you want, so if this E is 10 meters away, you'd plug in 10 here. plug in the Q that's creating the E at that point, and it gives you the electric field. Similarly for electric potential, you don't even have to square the R, or worry about the vector nature, because electric field is a vector.

This has a direction. Electric potential doesn't even have a direction, it is simply a number. Could be positive or negative, but it's simply a number at some point in space.

And so this Q and every other Q is creating an electric field, an electric potential everywhere around it. You find it with these formulas, fine. Easy. However, what if your electric charge isn't a point?

What if it's spread out over space? Say you took that same charge cube, but you distribute it uniformly across a, like, rod right here of length L. And let's say this time you want to find what's the electric field at point X, distance, you know, X away from the right side of the rod. Well, you can't just blindly be like, okay, 1 over 4 pi epsilon naught total Q over what?

X squared? What are you going to put? X? because that's how far the e at this point, like we know if this is positively charged, the e would point radially to the right because it points away from a positive charge, but what are you gonna plug in? X, that's how far this end of the charge is from that point.

What are you gonna plug in? X plus L, that's how far this end is. You can do the average.

You can't do any of that. You have to actually solve this using an integral, so you have to be clever. If your charge is distributed over some like, you know, Line or area or something like that a volume you're gonna have to do an integral to find the electric field So what do you do you use this formula?

But you're gonna have to be clever the way you're clever about it is you you say look this formula gives me the electric field From a point charge well you turn this into a point charge Specifically you turn into an infinite amount of point charges so imagine you break this up now. This is not infinitesimal, but imagine This is an infinitesimally thin, right here, little slice of Q. In fact, that would be a differential amount of Q. That's what we mean by dQ, a differential amount infinitesimally small. Well, that you can tell me how far away it is.

So, this is exactly that far, whoops, too far. This is exactly that far away from that point where we want to find the electric field. Like, let's say we wanted to find the electric field over here. This has an exact location. So I could say, alright, that little dq, that infinitesimal charge creates, it's gonna be a really little e, you know, it's gonna be a really tiny differential amount of e.

I can't even draw it differentially small, but it'd be a infinitesimally small amount of electric field because this is hardly any charge at all. So you turn this into d e. You say, well, if a big charge creates electric field e, you say that it... differentially small amount of charge is going to create a differentially small amount of electric field over here and so this formula is true it's just r you have to be careful r is from that charge to this point here so we need you know uh we could label this in a certain way we can call this r right here so that's going to be r from the dq to that point where we want to find the electric field okay but that's just the electric field from one little infinitesimal amount of charge we need the electric field from all of them Those that are closer will contribute more because they're closer to that point.

Those little differential amounts that are far away, like this one's gonna create the least amount of electric field, they're all contributing. And they're all gonna contribute with electric field to the right, that's important. Positives all create fields radially out. So they're all pointing rightward, we can just add them all up to get the total. How do you add up an infinitesimal amount of things?

You integrate, so we'd have to integrate both sides. Integrate both sides, the left-hand side just becomes the total E. So this side's gonna become E, and that's gonna be equal to, I can pull out all this extra stuff, I can pull out the one over four pi epsilon naught. I cannot bring out the dq. Q, obviously, that's the differential that's gonna be integrated.

I also can't pull out the R squared. This R's gonna be different for each little charge. Sometimes you get to pull out R squared. In this case, we do not get to pull out R squared, so this is gonna be DQ, differential amount, over R squared. Well, I've got DQ, and I've got R down here.

So what you're gonna have to do is change DQ into something else. This is the first order of business. What you're typically gonna do is just say that, all right, we have to talk about charge per length.

So lambda. Lambda, is there a B? I'm gonna put a B. Lambda is gonna be charge per length.

That's what we mean by lambda, it's the amount of charge per length. And let's just say this is uniformly distributed. You could have some crazy lambda that depends on X, it gets more and more dense as you go. Fine, whatever, you'd plug that in.

But all we need here is to say that, well, Q is gonna be lambda times L. Okay, so how about dQ? I mean, I want a differential amount of charge. Well, I'm gonna put a B.

If you take the charge per length and you multiply by a differential amount of length, you'll get a differential amount of charge. So all this formula's saying is, if I know how much charge I have per length, coulombs per meter, that's what this is gonna be, coulombs per meter. If I know how many coulombs per meter I have, and I multiply by this little infinitesimal amount of meters, I'll get the infinitesimal amount of charge that's in there.

So this is what we're gonna use for dq, at least if you have a line, like a 1d example. So dq's gonna become that. So we're gonna have Equals in fact.

I'm just going to write it in here. Let's just erase it you're going to have Lambda Dl Lambda Dl I'm going to pull it the lambda out so lambda is going to come out here That's a constant if it wasn't a constant you'd leave it in the integral and integrate whatever it was as a function Maybe it's like x x squared who knows but lambda comes out here. You have Dl.

Well. I'm still like crap was L That's R. I need the same variable here, but But what is dl referring to? It's referring to this little bit of length here.

So that's dl right there. An infinitesimal amount of length. That's the same as dr. Look, r goes in the same direction here.

I mean, if you want to be very careful about it, you could define l. Define l as like this distance here from the one end. And then you could say, all right, you could say that l plus x, right, l.

plus x, so every one of these gets a little address. This one would have an L of zero, this has an L of L, this over here has an L of capital L, L gives you, little L gives you the location of the charge. You could say that that L, that location from the right end, plus x has to equal r, but look, if I take a differential of both sides, x we're just keeping fixed, so that means dl just equals dr in this case.

So I'm gonna replace this with dr. DL, DR, like it's the same thing. You can call these DX and write this R as X. We're talking about the same direction here. These are all along the X direction, so we call this DR. And this is what we're gonna use to solve.

Now it's not too bad, I just need to get my limits right. I'm integrating R, not L. So R is gonna go from not zero. None of these little differential amounts of charge were zero away from this point over here. Remember, r is from your infinitesimal amount of charge to the point where you want to find the electric field.

So the closest I get is this little dq right there. That is the closest I get, and that is x away. So the r value, which is from charge to the point where you want to find the field, the closest I get is x.

The furthest I get is over here, and that is going to be x plus l. So the largest r value I get is x plus l. So you solve this, you're going to get...

Lambda over four pi epsilon naught integral of one over R squared is negative negative one over R Oops negative one over R you evaluate that between between X and X plus L I'm just gonna write it up here. You're gonna get I'll pull the negative out lambda over four pi Epsilon not Evaluated between X plus L so you get one over X plus L, I pulled the negative out front, so I don't have to deal with it, minus one over X. If I don't want that negative out here, I can bring it inside if I want, and just reverse the order of these.

So I can bring that negative inside, and I can say, actually, I'm gonna do this term here, minus that term there, just swap the order, and you bring the negative inside. And so that's our formula for the electric field at this point. x away from the right hand side of this bar.

Like this is the e. E is this, the total e. Total e equals that.

What would you do if you want to find the electric potential over at this point? Well, it's pretty much the same thing. Like, you got this formula. What's the difference?

Well, it's not a vector, but that didn't even really matter here. All these E's went in the same direction, so we didn't even really have to worry about the vector nature of this E. We just added them up blindly. V doesn't even have a vector, so we wouldn't have to worry about that anyway.

and it's R instead of R squared. So let's just look. Let's just try to identify.

What would be different? Well, this would obviously be dv. This would just be R instead of R squared.

Oops. So this would be... Oop, it's not even letting me erase that.

This would be r instead of r squared. dq still becomes lambda dr. r still goes from x to x plus l, right? This would be v. So when you integrate this, the only difference is that instead of getting negative one over r, you'll get that v is lambda over four pi epsilon naught. The integral of one over r is log, so you're gonna get log of, and there's no negative now. So integral of one over r squared gives you a negative.

The integral of one over r doesn't it doesn't give you a negative. It's gonna be log of x plus l. Minus log of x because it's gonna be evaluated integral of 1 over r.

Sorry, integral of 1 over r is log r. And so log of x plus l minus log of x. And you could simplify this because you can write that as, you can write the difference of logs as the log of a ratio.

So you could say that v total at that point would be lambda over 4 pi epsilon dot log of x plus l over x. There we go. You can simplify it.

You can write it as 1 plus l over x if you wanted to. But that's v total at this point. The total v created by all of these dqs added up. And it's going to be...

This is going to be the E at that point due to all the Q's added up. Notice it's not as simple as just taking, like, this original formula and, like, plugging an X or plugging an L. You have to do an interval to get the value.

Recapping, you break this up into a bunch of infinite. in testable amounts of charge. Those reach at a given distance and then you add them all up by taking the integral. So this was the linear case.

What do you do if you get this case? You're like, no, not the curvy. The curvy's the worst.

It's actually not that bad. Let's say you wanted to find E at this point, right in the center. This time you got a curve of charge.

Oh, wait, I didn't finish this one. Technically, here's what you'd have to do. I wrote this in terms of lambda, but let's say you weren't given lambda. Like I wasn't given lambda, I was given Q and L. So the At the very end, instead of writing lambda here, you'd write it as q over the length of your object.

So everywhere I had a lambda, I'd write it as q over l. So this would have to be q over l right here for lambda at the very end, and I can write this as q over l for lambda right here. If you're not given lambda, sometimes you're given lambda, sometimes you're given q. Use whichever one you're given, obviously.

Okay, so what if you had this case? So, in that Q, you distributed over half of a circle, a semicircle of radius R. We want to know the electric field right in the center. Before we do that one, let's do the electric field, or sorry, the electric potential.

This one's gonna be easier. Let's say we want to know the electric potential in the center. Well, what do we do? We again break this into infinitesimal little pieces. So I break this into an infinitesimal amount of charge, dQ, right there.

So an infinitesimal amount of charge, and that's gonna create an infinitesimal amount of V. So an infinitesimal amount of V at that point. Basically nothing but not nothing. So dV gets created by an infinitesimal amount of dQ. How much does it make? This much, because like you plug in the Q you got, you plug in the V. Let's integrate both sides.

What are we gonna get? The side becomes V equals, alright, so I pull out my constants, one over four pi epsilon naught. Now here's the magical thing.

Look at every single DQ that you can consider. This one here, this one down here. There's something similar about all these, and that is that they are all the exact same radius away from this point. That wasn't true for our line charge. Our line charge had points that were close, points that were far.

For this loop, the nice thing about it is that they're all the same distance away. So you're like, what? That means I could pull out r? Yeah, you could pull out r. r is the same for all of those.

And you're like, hold on, that just means I get integral dq? Yes you do, and what's the integral of dq? It's just q. So you get v is q over four pi epsilon naught r.

Now you might be like, wait a minute, that's just the original formula up here. You know what I mean? Like if I got rid of this integral nonsense, if I had just gotten rid of all this and plugged in my total Q, I get the exact same thing.

That's because V is... It's just a, you know, not a vector, it's a scalar. You can just add up all the values here.

This dq here at the top contributes just as much v as this point does, as this point does, as that point does. So having it in this loop is no better than just concentrating all of this charge right there. Say you just took all that positive q and put it right there, it'd still make a green v value of this value at that point, same as if you spread it around, doesn't matter, as long as it's all the same distance away, from a point, it's gonna contribute the same v at that point.

So you might be like, alright, well then e should be just this, right? Well, no, unfortunately. And the reason is, electric field's a vector. This piece right here is gonna create a dE that goes this ways.

Let me draw it straight. That goes that way. So that dE from this piece of dQ goes radially out, cause it's positive.

That's the little bit of dE from this piece. What about the DE from this piece over here? Well, it would create an E that goes radially out from it. And you're like, uh-oh, those ain't gonna add up easy.

I can't just add up these two numbers. Like if this one was two and that one was two, the total field isn't four. Look, the vertical components of these are canceling, and that's true for any of these points.

This top one creates a field to go straight down. This bottom one creates a field to go straight up. All the vertical components cancel. If you had this one here, it'd be perfect.

It goes right along the horizontal. The only part that's gonna matter is the horizontal components. You have to add all those up.

So we don't actually wanna find the total E, I mean we do, it's just the vertical parts are gonna cancel. These verticals cancel those verticals, these cancel these verticals. The only part that survives is the horizontal piece.

So what I have to do is, like, okay, shoot, this is a vector. I need to make sure I find E in the X direction. So I'm gonna take this, I'm gonna call it DQ, fine, and that makes this DE. But if I only want the piece in the x direction, you know, if I only want that piece right there, so that's gonna be DE, little component in the x direction.

This is DE in the x direction. I need to multiply by, you know, a sine or a cosine. And it depends on how you want to define this.

I'm gonna define my angle. You have to do this. You have to draw this out.

This is important. Draw where you want to measure your angle from. So if you're taking sine or cosine, there has to be an angle measured from somewhere.

I'm just gonna choose this center line as where I'm gonna measure theta from, so I'm gonna call that theta. So I'm gonna say that this dq is located at that theta right there. Well, that means this is also theta. These are interior angles here.

So if I wanna find dex instead of de, I would take my total de, which is the hypotenuse here. This de was the total field that that dq up here created, and then I multiply by cosine theta because this is the adjacent piece. piece.

So I multiply by cosine of theta. Now I'm ready. These I can add up.

You can add up all the DEXs because those all go the same direction. There's no point in adding up all the DEYs. Those are going to cancel. You can do it if you want. They're all going to be zero.

Take my word for it. Or try it if you want. It doesn't matter. So now you can integrate.

Integrate both sides. Left-hand side becomes E in the X direction total, which is just E total because the vertical is going to cancel. Equals, bring out your constants, 1. Over four pi epsilon naught. R squared, you might be like, wait, is R squared still constant?

Yeah, this is just as far away as that is, as far as that, as far as that. These are all the same distance from this point where we want to find the electric field. So you can still bring R squared out.

This is the great news. R squared comes out. I write it as capital. And I get integral dq. Let me find my yellow here.

Integral of dq just times cosine theta. So we're not quite as lucky. Over here we just got dq.

Alright, well if you get integral dq, that's just q. But integral of dq times cosine theta is not just q. You might be like, can I bring the cosine theta out?

You can't. Look, these are all at different angles. They're all at the same r, but this one's at this much of an angle.

This one up here would be at that much of an angle, you know what I mean? So they're all at different angles. I can't just bring my theta out of the integral.

It's varying. So what do I do? Same thing we always do.

You write dq as something else. So you're like, shoot, I actually have to do the integral. What do I do? Well, I know Q is gonna be lambda times L. So just like we did before, DQ is gonna be lambda times DL.

Where DL is this little bit right here. So this little length right there is DL, that infinitesimal length right there. Width is DL.

And if I multiply that little length by the charge density, linear charge density. Lambda times that little dl gives me the infinitesimal amount of dq in that region. So this is what I replaced dq with down here.

So this dq is gonna be lambda times dl. Does that help at all? It does. Lambda gets to come out of the integral because it's a constant, again.

Maybe lambda's not a constant. If your lambda's not a constant, fine. Again, as always, leave it in the integral.

You'll have to be told what that lambda is, and if you know what the lambda is, you can just integrate it according to whatever the integral is. But in this case, I'm gonna assume that this lambda is constant. We can bring it out. We still have a problem.

This is theta, that's L. They need to be in terms of the same variable. That's not too hard to fix. The way we're gonna fix that is we're gonna say that this dl is a little bit of an arc length. Look at, so hopefully you learned about arc length at some point, s.

s is r times theta. Well, I should say delta theta. A little bit of an arc length is gonna be s is r times delta theta. Well, how about an infinitesimal amount of arc length? That would be r times d theta.

you know, a differential amount of theta. So the theta between this line here and that line there, I mean it's the slice of all slices. This angle in here would be d theta.

The ultimate slice right here, the thinnest slice you can imagine. If you took that thin slice of theta, multiplied by the r, that just equals arc length. So this is the formula for arc length.

You need to know that to do these problems. But that's ds, well that's just what I'm calling dl. dl is the same as ds.

So dl here is gonna be rd theta. You might object and be like, wait, ds is curved. dl is straight, not if it's infinitesimal. If it's infinitesimal, ds is straight.

There's no time for it to curve. I know that's weird, sorry, that's math. And so ds is the same as dl. Plug this in for dl down here, and you're gonna get r times d theta.

So dl becomes r times d theta. Now you can integrate, you just need some limits. So what are we gonna integrate from? Well, it depends on how you defined your theta.

So I defined theta from here, so I'm gonna come all the way from negative that. Like if this is theta equals zero right there, I need to come from negative pi over two, because that's where this point is. My dq with the smallest possible theta would be negative pi over two, based on where I made my zero.

And my biggest one would be pi over two, so I gotta go from negative pi over two. to pi over two, the r gets to come out. So like r I can just bring out, in fact I can cancel this r with one of these r's.

So that's not harming anybody. I would take the integral of cosine, and that's not too bad either. The integral of cosine is just gonna be sine. You might be like, is it negative?

No, integral of cosine is positive sine. So we're gonna get lambda over four pi epsilon naught times r, integral of cosine is just sine theta evaluated between Negative pi over two to pi over two. What does that become? You're gonna get lambda over four pi epsilon naught r.

Sine of pi over two is just one. Minus sine of negative pi over two. This is negative pi over two. Sine of negative pi over two is negative one. So that's just two.

One minus a negative one is just two. And so you get that the electric field, I'll write it over here, the electric field's gonna be two times lambda over four pi epsilon naught r. Now, again, I didn't give you lambda, so you have to plug that in. What is lambda? Lambda's always just defined to be the charge per length.

Again, if you're given lambda, fine, leave it like this. If you're not given lambda, you gotta put it in terms of stuff you know. So you put this in as charge per length, but you're still not done.

You're still not done because I didn't even give you the length. You might be like, do I put r? No.

Q isn't spread over one radius, Q is spread over one semicircle. This is the length that the charge is spread over. What is that length?

Well, it's half of a full circle. A full circle is, you know, a full circle's circumference is two pi r. That'd be the length of a full circle. This is a half circle.

Half circle's just pi r. So you divide this by pi r and you can clean this up. You can write this as Q over, I guess you'd have two.

Pi squared, epsilon naught r squared would be the electric field, it's in the x direction, the total electric field in that x direction that this little loop creates. That would be the electric field in the x direction. The vertical component would cancel. How do you know that? Well, look, we took cosine theta to get this horizontal piece.

You do sine theta to get the vertical piece. And if you integrated sine theta, you'd get cosine theta, but cosine of pi over 2 and cosine of negative pi over 2 are zero. So it's all going to cancel out. So the vertical parts cancel.

You only get the horizontal part. For v, it's easy. You just use the formula.

Why did we get to just use the formula? It's because v doesn't have any vector directions. We had to multiply by this cosine in order to pick just that piece we wanted in the x direction, that piece that wasn't canceling, so that we add all those up.

And if you do that, you do the integral, you solve, you get your electric field. So to recap, what you need to do is use the formula for electric field. but plug in dq instead of q, and that means you have to integrate it.

Dq you almost always have to write as lambda dl. Once you have that, pull out anything that's constant. Make sure you're consistent with your limits based on how you defined your angle up here.

That'll dictate how you define cosine theta here too. Since I defined theta from this point, I've gotta use cosine theta to find that e. Run your integral, get your limits, solve, you get your value, and then at the very end, if you're given lambda, leave it as lambda. If you're not given lambda, write lambda as what you're given.

Transcript for:Understanding Electric Fields and Potentials

Transcript for:
Understanding Electric Fields and Potentials