Case over this problem. You're given a sphere, and we want to know with a conducting shell surrounding it. So here's this fear. And then here is the show. No, it's not really important of the whole thing in there. Let's just do that. And, uh, you the radius of the inner spheres, eh? This is B. This is C and but write out all the variables. There's Q one on the inside, and that's five um, fundo columns. So 5.0 times. 10 to the minus 15 Crooms Um, que tu So that's Q one Q twos that total charge on the outer shell and that's minus Q one a is two centimeters. So 20.2 meters never too early to convert something two meters, in my opinion, be used to a so I'll just read it out. It's quaint. 04 meters and see is two point for a Oh, I'm just gonna leave it in terms of a MME. And then for completeness soldiers, right? This is to pay. Great. So, yeah, we want to find the electric field at different points. So finding it in the center in each case you're gonna, um, the idea is to draw a galaxy and surface. And in the case of the sphere of the galaxy and surface is always a spherical shell. Um, around where? And you wanted the galaxy and surface to intersect where you're concerned with the field. So for, um, for are our equals zero your gals, Ian's here would be a point, Really? And so it wouldn't encapsulate any charge. So the field there is, ah, groups for a night A equals but e sorry, it's an ugly e cool zero. But now, as you go further out, you're gonna start encapsulating more charge. So I'm just gonna go ahead and draw pictures. It's just sort of one of the earlier one. So here is that close up of that sphere, and then we want we're concerned with the radius, which is, um, half of the radius of the sphere. So it over to so are is a over two point. 00 and so here's your Gaussian surface. Um, so this is like a radius are. And, um, now we wanna apply calluses law, which says he and a girl of ee dot d a equals Q and closed over epsilon. That's an l That's just randomly high over Absalon night. Um, and the this integral in the case of the choice of this fear, um, spherical shell of radius are it's just you times the surface area of that fear. So it's for pie. Well, the prize struggling. It's another leg. Four pi r squared and then that's equal to the human closed over Absalon. Not so how much charges enclosed here? So it's how much charge would be sort of within this red circle. And so it would be the percent of charge this enclosed or it's a percent of the total charge, um, which is proportional to the percent volume, um, within this fear. So that's just gonna be q one, not keep the Epsilon. Not, um, come on. 10 times the ratio of the volumes. The volume of a sphere is 4/3 pi r cubed and so and we want to set that we wantto divided by the volume of the total sphere to get the proportion charging closed so we can cancel the poor thirds and the pies and we get caught. So too next I do some algebra to determine what ah formula for ye at that particular are. And I would say it's Q one over four Pi upsilon nine. A queue of times are so the field just gets bigger, bigger as you, um, go further out as long as you stay within the the sphere. And then if you plug in numbers to that. So you're looking like you're ours too, eh? Well, I'm your end. The two centimeters, um, maybe I get when I go. It which was 0.563 Newtons per Coolum. Okay. And then for see, at this point, we're going to the edge of the circle, right? And saying our equals a Yeah, so are equals a And so we're still kind of technically within the sphere so we can apply this same formula. Um, and so you can say that he is just If our we could just say that our is equal to a So hee is Kyu won over for pie El Salon. I, um a squared right e did our equals saying canceled out the days. Um so if you put that into computer, you'll get what I got, which is 0.112 noons per cool. It makes sense. It just a bit bigger than, uh, the the a bit further end where there's less charged contributing to the field. Great. And then for D, it's, um, you want to know at 1.5 s? Oh, that's still not up to be. So it's between A and B. Ah, and then again we would draw our galaxy and surface. This time we would draw lit I R equals 1.5 a, and the only charge that would be enclosed would be Q one. And so you can say that E equals Q. If you go too fast, you will be punished, says my tablet Kyu won over for pie. Absolute night r squared again on DSO this time where this R is the 1.5 a so plugging that into a computer. I got 0.5 moons for Coolum, so it kind of starts to like sink back down again like it's a mazar gets bigger, it goes up, then six pack down as you get further away. Hey, almost done. So Hee is asking at 2.38 So that's beyond because B is to a and so at that point you could draw your calc and surface and since, um, and you ask how much charges enclosed within my surface? Andi, you would count the charge from the inner, too, in the charge from the outer, and that total is zero charge. And so there's no charging clothes. Therefore, the electric field is zero. And for F, it asks, Um, at 3.5 a. Oh, my bad. I, um that that shot should be the reasoning for F. So 2.3 a is not, um, it's It's in between B and C because, see, is 2.4 a. And then we were concerned it's 2.3 a. So it's between B and C. But the electric field is still zero because the the shell is a conductor. And as a rule, uh, the electric field is always zero in a perfect conductor. So the answer is no change, uh, Fergie and asks what the charges on the inner surface. So, um, I've done this really someone before I think a few questions back. So I'm just gonna quote a result that, um, you have the charge in the middle, and it's going to draw the exact opposite charge on that inner shell. So that's minus Q. One. And this is always the case because you can draw the galaxy and surface within the conductor, and then the A total charge. The electric field field a zero and therefore the total charging clothes must be zero, So the minus Q. Must balance the positive cue in the center and for each. How much are just left on the outside. So in principle, you could have trudged left on the outside. But it's the problem, states. It was only minus Q one total on the outside. We already have minus Q one on the inner surface, so there's none left for the outer, so that zero.