A spherical, thin conducting shell of radius 8cm has a charge of –6C. If a 4C charge were placed at the center of the shell, what is the electric field 4 cm from the center? At 12 cm?

E_{4cm} = 2.25 × $$10^{13}$$ outwards;E_{12cm} = 1.25 × $$10^{12}$$ inwards

A spherical, thin conducting shell of radius 8cm has a charge of –6C. If a 4C charge were placed at the center of the shell, what is the electric field 4 cm from the center? At 12 cm?

Hey, guys, hopefully be able to take a look at this one your own using the other videos that we've seen before on how to calculate electric fields. Using gas is law. You're able to try this one on your own. So we have a thin spherical conducting shell and it has this charge on it. Negative six columns three idea is that you're gonna place a four Coolum charge at the center of this spherical conducting shell, and you have to figure out what the electric field is at certain points. So we know we're trying to figure out the electric field and we have charges we can use. KAOS is law, and God's Law basically says that if you're trying to figure out what the electric field is, at a certain point, the first thing you need to do is construct a Gaussian surface at that location. So what we're gonna dio is if we want to figure out what the electric field is at four centimeters from the center, then we're gonna have to construct a Gaussian surface which we know is a sphere by now. And we know that this Gaussian surface here is going to be at four centimeters because then that's going to tell us what the strength of the electric field is at that specific location. So that's sort of like that little our distance that you plug into your gas is law. It's basically gonna be wherever you're told to evaluate that electric field at and we know we're gonna We're gonna have a sphere because the electric field points outwards and the area vector also points out. So that's E. And that's a So kAOS is, Law says that the total amount of flux that's traveling through that's basically going out from the surface here, which is E. A. Times Co. Sign of theta is related to the Q enclosed over Epsilon not, and we're trying to figure out what the electric field is now. The only thing we have to dio is figure out what we can cancel out. So, for instance, this co sign of theta always equals one, because again this data is equal to zero. Any point along the surface, the electric field and the area vector will always point in the same direction, which means that the coastline is always just gonna be zero. So now we just have to figure out what the area of this Gaussian surface is. That's just four pi times R squared where r is the radius of our Gaussian surface. That is this four centimeters. I'm gonna highlight that this four centimeters is actually what this are is so now all we have to dio is just move this a term to the other side and then figure out what the with the electric field is. So the electric field is just going to be the queuing closed. Now the what's the charge that's enclosed by our Gaussian surface? It's just at four columns. So we've got four columns divided by 8.85 times 10 to the minus 12 which is our epsilon not and now we just have to do one over for pie and then we have our squared. That's our distance. Is this four centimeters? Remember? So we have to convert that 2 m. First we have to use this 20.4 and then we have to square that. So basically, this is what the electric field is. It's sort of like, you know, if you're to just figure out the electric field due to this point charge, but we're just using gas is law. So the electric field, if you actually go ahead and plug all of this stuff in, is equal to 25 times 10 to the 13th. That's Newtons per column. That's the answer. Which, by the way, if you hadn't used houses law and you were just told to figure out what the electric field is due to this point, charge and just use cake you over r squared, This would be the sorry. Yeah, cake over R squared. You would actually get the exact same answer, right? Because that's what Houses Long tells us. Cool. So just want to make that point. So then B is now we have to figure out what he is at this 12 centimeter distance. So now we have to dio. So we have to construct another galaxy in surface. That's gonna be all the way out here, right? And God's law says that the amount of flux that passes through the surface is gonna be related to the amount of charge that's enclosed. So again, we're just gonna start off with. If we're trying to find E, then we have to use fine nets on. We know that e a cosine theta is just gonna b e a right. But the same exact reason the cosine theta is gonna cancel. And now all we have to do is just relate this to the amount of enclosed charge. Okay, well, are Gaussian surface in encloses this negative six Coolum charge. That's all distributed around basically the outer walls of this little conducting surface right here. And it also includes the four Coolum charge that's at the center. So that means that the total amount of in enclosed charge here is just equal to negative to cool OEMs, right. You're just gonna add those charges together? So that means that this is going to be negative to cool. OEMs. So you have negative to cool ums divided by epsilon knots. Right, So we go ahead and move that now the area of this surface is not the same as it was. Here it is. Four pi r squared. But now this four pi r squared is so if this distance r equals 12 centimeters, then the are that we're gonna plug in for this equation. Is this are not this are so you gotta you gotta remember. It's basically wherever you're told to evaluate that electric field, as that's where you plug in for your our distance. Okay, so that means that the electric field is going to be negative to cool. OEMs divided by 8.85 times 10 to the minus and then we're gonna have one divided by four pi, and then we have to use 12 centimeters in the right units, so 120.12 squared. So that means the electric field is going to be negative 1.12 times 10 to the 13th. That's gonna be Newtons per cool. Um, what I wanna do right now is just briefly talk about the direction. So what happens is we got a positive number over here. So in other words, the electric field points outward in this case, and that's because the electric field at this place at this specific location or the Gaussian surface that we've constructed couldn't only enclosed a positive charge, which means the electric field should actually point outward. That should be right. But over here, this Gaussian surface that we construct on the outside basically acts as if there was just a negative to cool, um, charge on the inside. So that means that the electric field lines actually point in word for this. And that's what we would expect for a negative charge. All right, so I just wanna point out the fact that this is gonna be inward on the outside of this, and this is gonna be outward on the inside of the show. All right. Let me know if you guys have any questions with that.

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A spherical, thin conducting shell of radius 8cm has a charge of –6C. If a 4C charge were placed at the center of the shell, what is the electric field 4 cm from the center? At 12 cm?

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