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Anderson Video - Gauss's Law Example

Professor Anderson
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So let's say we have the following, we have 3 objects: O1 has a charge of five microcoulombs. O2 has a charge of minus six microcoulombs. Object 3 has a charge of zero. And let's put those three objects right here. Here's object one, here's object two and and here's object 3. And let's ask a few different questions. Let's ask, what is the flux through area 1? Let's ask, what is the flux through area 2? And then let's ask a follow-up question, what is the flux through area 3? Okay, so with a big sort of peanut, we have a sphere here for A2, we have a sphere here for A3. And we want to find the flux through area 1, area 2, and area 3. All right, flux we usually write with a capital phi. I don't know if he does in your textbook or not, but typically you write it with a capital phi. That is just e times the area and Gauss's law tells us it's Q over epsilon naught. This is enclosed in that surface. Enclosed in the gaussian– let's be clear– gaussian surface. Okay what is inside that closed surface? Okay, so area 1. Phi 1 is just going to be how much charge is there inside area 1 divided by epsilon naught. Area 1 covers everything so that's going to be the charge of object 1: 5 microcoulombs. Plus the charge of object 2: minus 6 microcoulombs. Plus the charge of object 3, which we said was 0. All of that over epsilon naught. And so you get minus one micro coulomb which is times 10 to the minus 6 coulombs divided by epsilon naught, which we said was 8.85 10 to the minus 12 and you can put on the unit's there if you want. And so what do we get? I don't know, somebody run that in your calculator and tell me what you get. It's about 1 over 9. A little bit more. 1 over 9 would be point 11 but it'd probably be more like 0.12 and then this is going to be 10 to the minus 6. I have to add 12 so I got a 10 to the 6. So I'm going to say this is minus 1.2 times 10 to the 5, all right. I moved it over one, and so we need to drop one from there. Is that what you guys got in your calculator? Did you try that? 1.1? Okay so 1.1 in the calculator times 10 to the 5 and you can figure out what the the units are on that. Okay, it's going to be E times A, and so the units we said for E are Newton's per coulomb. And A is square meters so it's going to be Newton meter squared per coulomb. Okay that's flux Phi 1 through the whole thing. Let's think about flux Phi 2 through A 2. Phi 2 is gonna be the Q enclosed in its shell divided by epsilon naught. How much Q is enclosed in A2? It's only O3 but O3 has zero charge on it. So this is zero over epsilon naught and so that is zero. There's no flux coming out of this one. What about Phi 3? Phi 3 is also Q over epsilon naught. How much charge is enclosed in A3? Is there any charge in there? No, there's zero. Zero charge enclosed in A3 so that one is also zero. Let's talk about that for a second. And let's ignore some of these other areas. Let's just worry about A3. A3 is sitting right there. Let's draw a little bit bigger. And let's draw O1 which we said was a positive charge. O2, which we said it was a negative charge, and here is our area A3. I have a positive charge sitting right here. Are there electric field lines coming out of that positive charge? Absolutely, right? We know what that's doing. They're pointing radially out from the charge, like so. A few of those are pointing right at A3. And so this field line doesn't just stop, its going to keep going and then it goes into A3, but then it comes back out of A3. And– we don't have to draw that one. Let's think about the negative charge. The negative charge also has field lines, but it's coming into the negative charge, but this line came in from much further out so it had also gone through A3. But what is flux? Flux is field lines in minus field lines out. How many field lines are going in minus how many field lines are going out. Actually, it's the other way I think is the way we typically define it. So we'll change this to field line out minus field lines in and that way it's a positive number if there's a positive charge in there. So how many field lines are going out? Well I've got one going out there. I have another going out there. So that's two going out, but I have two field lines going in. That one came in there, this one came in right here. And so this becomes 2 minus 2 which is of course 0. And so we're back to this idea that it only depends on the charge enclosed. How much charge is enclosed in your Gaussian surface, that determines how much flux is coming out of your Gaussian surface. Now, the entire point of Gauss's law is to find E. It's to help you use mathematical tricks to find the value of E. That's all you're trying to do with it. You're not really interested in the flux, eventually, you're interested in E and Gauss's law helps you find that. All right, questions about that? Everybody fully saturated? all All right, why don't we wrap for today and I will see you guys tomorrow online and Thursday face to face. Have a great day. Cheers!