Electric Fields in Capacitors - Video Tutorials & Practice Problems

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concept

Intro to Capacitors

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Hey, guys. So in this video, we're gonna be talking about these things called capacitors. Alright, We're gonna be working with capacitors much more in depth later on in the course. So let's go ahead and get familiar from right now. All right, let's get started. So if you have two parallel plates and they have equal and opposite charge, this is actually the important part. They're gonna produce an electric field between them. We know that charges produce electric fields. So in this specific arrangement, if you have two plates and they have equal but opposite charges, they produce a uniforms electric field. That uniformed just means that it's the same at all locations. Always has the same magnitude everywhere. Always. It's just the same. Okay. And to see why that is, we just have to remember that electric fields always point from positive charges. Whoops. They point from positive charges or from the positive plate towards the negative plate. And the reason for that, right? Because basically, you can think of this plate of positive charges is just a whole bunch of tiny little mini positive charges. We know these many positive charges have electric fields that point outwards. So what ends up happening is that this electric field that points in this direction right here these guys are going to point downward over here and then these guys right here, they're right next to each other. These electric fields there, pointing in this direction, are gonna group up and basically cancel each other out and point in this direction as well. Now we have the negative charges on the opposite side of the plate that want to produce an electric field that points in words like this. And just like in this case, where basically they symmetrically cancel out the symmetric, the symmetric electric field lines from these negative charges is going to result in an electric field that points downwards. They're just gonna cancel each other out. Just is just like it for the positive charges. In other words, will be end up getting is we just end up getting this uniforms electric field that is always in this direction from the positive plate towards the negative plates. So this is E. And the deal is that this at any point that you decide to measure the electric field e is always gonna be the same so not. No matter if I chose here or here or here anywhere like that. These these are always going to be the same. So I'm gonna write equal signs. So that is gonna be the same. Is this e and this is gonna be the same. So it's always the same at all of these parts. Got it. And so if you have these two parallel plates that have equal and opposite charge, that's what we call a capacitor. So that is the definition of a capacitor equal and opposite charge, and they're separated by some distance. And basically all the capacitors are We're to talk about them a lot more later on in the course is there. There's thes things that store charge and this energy between them. All right, we'll talk about that later. And so we talked about the direction of the electric field in this diagram right here. Now the equation for that electric field is just Q divided by epsilon. Not so. Let me write that a little bit neater. So this Greek letter right here that kind of looks like an E, but it's called Epsilon. The Times A which is the area of each plate. So this new Epsilon not right here is called the vacuum permitted ITI, which isn't a name you probably don't need to know. And the value for that constant is 8.85 times 10 to the minus 12 and it's got some units associated with You don't really need to know that now. This absolute not is actually related to our other constant that we know called columns constant. And the way that these things are related is that K right that 8.99 times 10 to the ninth is actually equal to 1/4 pi times Epsilon knots. Now, this is something we're gonna talk about much later on. We're gonna talk about in greater detail. But it's just a good thing to know for now that these two constants are related and we have the A is just the area of each plate. Whereas Q is the charge on each plate. Now, remember, these things are equal and opposite. So let's say one is gonna be negative. Two columns one is gonna be positive. Two columns. The queues is gonna be too in that case, and then a is the area of each plate. That depends on the geometry. Sometimes they could be square. Sometimes they could be cylinders, Things like that. We'll get into it later. Um, but this is the equation for inside of the, uh, inside of the plates. So, in other words, the space right between the plates Here, this is going to be Q over e que Over epsilon. Not times a but outside of the plates. What if you were actually go at some point over here? What does the electric field in this area? Well, outside of the plates, it's just going to be zero. So anywhere outside of the plates, the electric field to zero anywhere inside of the plates. It's this equation right here. Got it. That's all we really need to know. For now, let's go ahead and start dealing with some problems. So we've got electric field in between. Two parallel plates were given the magnitude of the electric Field 1000 Newtons per column, and we're told with the area of these electric plates are so we just we need to figure out the charge on each plate. In other words, we're looking for Q, so let's go ahead and start out with our equation. So the electric field between two parallel plates Aaron and it's in a capacitor is gonna be Q over Epsilon, not times a. So we have what the electric field is and we have with this area. Is this absolutely not? It's just a constant. So let's go ahead and figure out what this que is. All right, so let's go ahead and move these things over to the other side. We've got Epsilon, not a we're gonna bring over. So that means that Q is equal to Epsilon, not a times e. So we've got 8.85 times 10 to the minus 12. You don't really need to know the units for that. You've got the area which is five centimeters squared. So I wanna be careful here. Remember that five centimeters squared is not 0.5 m squared. You actually have to do the conversion twice because there's an exponents here of two. Alright, so just don't make this mistake. It's a very, very common mistake. So do not do this. It's not this. Instead, five centimeters squared is equal to five times 10 to the minus 4 m squared because again, we have to apply that conversion twice. All right, Just just wanna sort of refresh your memory. Eso We've got five times 10 to the minus four, and then you got to the electric field, which is just 1000. So if you go ahead and work this out, you're gonna get a Q. That's equal to 4.43 times 10 to the minus and that's gonna be in columns. All right, so that's the charge that's built up on each one of these plates. Man, my highlighter is not working. Alright, guys, that's it for this video. Let's go ahead and take a look at some examples.

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example

Example 1

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2m

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Hey guys, let's work this one out together. So we have a capacitor that produces an electric field. We've got these two plates with two charges and we're going to have to double the charge half of the area between you know, of the plates. And we need to figure out what happens to the electric field. This is going to be one of those proportioned reasoning problems. We have to figure out what happens to E. So in other words, e new as we change some variables is gonna be what of e olds, right? So this is how we handle these kinds of problems. So let's go ahead and set up what the electric field is uh between two plates, right? So this electric field is going to be the charge divided by epsilon, not some constant divided by the area as well or times the area, I guess and just make sure this is in the denominator. OK. So then what happens is that this charge, right? This this Q new now has to become two times the original charge and you're not gonna actually double the both charges together because remember this Q just rep represents the charge on each plate. So in other words, you don't have to like multiply it by four or anything like that, right? So it just becomes two Q and then this area. So this a new is just going to become uh one half of a new. So now let's go ahead and using these two new variables to set up our new electric field. So this new electric field, which I'm gonna call E new, it's just going to be Q new divided by epsilon knot times a new. And now all we do is we just substitute these expressions and oh, I'm sorry, I didn't mean to write a new that, that should just be a. So we just have to substitute these expressions in for our uh our new equation right here for a new. So this is just gonna be two Q divided by epsilon knots times and then this is gonna be one half of A. So now what we do is now we've substituted those expressions in, we just have to see if we can pull all of these factors and constants out to the outsides. And then we're gonna basically just gonna extract the original expression. So what happens is this is gonna be two divided by one half. So when you divide by a half, you're gonna multiply it by two. So you're gonna have to multiply it by two again. And then you're going to get Q divided by epsilon not times a now notice that this is just the original expression for e right, this is just the original electric field. So in other words, what happens is that we've gotten four times the original electric field and that's our answer. So if you double the charge and then half of the area between the two plates, so that the, the new electric field is going to be four times stronger than the old electric field. And that's basically the answer. That's what happens to the electric field. It multiplies by four. All right, let me know if you guys have any questions with this stuff.

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Problem

Problem

An electron moves into a capacitor at an initial speed of 150 m/s. If the electron enters exactly halfway between the plates, how far will the electron move horizontally before it strikes one of the plates? Which plate will it strike?