This problem are given a capacitor first without an electric and we are dielectric in and the end goal is to find what the surface charge will be on the dielectric. And before we get into the calculation of that, let's talk about how to arises. So here is the situation where there's no doubt that electric and is charged up to a charge Q one. I've used four electric field lines to represent the electric field between the plates. Remember dr field is tangent at any point following the general sense of the arab, now we put it to electric in between. What happens if I remember dialectics are insulators, you don't have free movement of charge, but they can move slightly in their general location. What you get is a polarization where the pluses and minuses would be on top of each other. Their at least their distributions would be. Now they will be slightly off. I'm making this larger than then. In reality they'd be slightly off. So the miners would be closer to the plus plate and the plus I was close to the honest plate. Now, let me just put in a couple, you know, I won't put in the big thing I just put in the end results decide Factory, this surface will seem to have a negative charge, and likewise, this surface will seem to have a positive charge. We're not charging the dielectric. It still it still has no access charge. We're just causing this polarization effect. And it gives us this this charging of the, you know, this appearance of charge surface. Yeah. And in that respect, operationally, now, let's talk about the electric field aspect of this. We know there's electric field, original electric field. Now, with these charges been in induce the factory on the surfaces. Think about what goes on. This would make it look like. So we would have the original field this way that's from here and that looks like this would give me a field back to the left. And so I'm going to get this is my now new field between the place have reduced electric field and that is certainly the case. And how do we represent that? We left a few lines well, remember they begin on positives and and on negative. So there's one line, here's another line you can think of these guys as knocking each other out. You know, if you think of them as separate. A little bit, there's a little, there's a little bit of a arrow from the plus, the minus, but it's all very, very, very small. Uh, in terms of that does not really impact anything. So, notice two lines is before that visually tells me lesser feel. That's what the field lines are. Therefore in addition to giving me direction. So we have a lesser field. Now, let's talk about that. That's the key ingredient here. We have a lesser electric field. Now, this is hooked up to a battery in this problem. 12 votes. Everything's hooked up to a battery. The same situation just you know that we're just going from one step to the other hooked to a battery. Um, so that says between these places be 12V. But I just told you the electric field drops, electric field drops. That means the potential difference between the plates drops, But that can't be. We're hooked up to something that says you're gonna be 12V between those planes. So what has to happen? Well, more charge must flow to bring up the electric field back to the way it used to be. So that we have the same potential difference of 12 votes across the plates. So let me put six. I still got my, Still got my two. So there is our situation charge grew but in combining these together because remember it's all the surface Combining together. It's got to give me the same as Q1 which gave me the original field because we're hooked up to a battery. If you're not hooked up to the battery, that's a different problem. That's not what we're dealing worried about here. Now, what we're looking for in reality is this QUE que Q cap, I should say. People say Q cap capital for the electric counselor, looking for the for this come out and obviously this is just minus Q. K. Q. Capa we should say. Um so getting using that symbol. Let's look at this play here. That tells me what that que tu minus Q. K. I'm thinking cute kappa Q coppa as as a positive number. We always we can pass is always thinking about the talk about the positives. Que tu minus Q. K. Is equal to Q. One. That's how we get the same electric field that we had before. That gives me the same potential difference. So now we can solve for Q Kepa. You kappa you too ask you what now? This is given by the compassionate when we have the dielectric. I see mine is the casualties capacities when we don't have no cause C zero transfer fee. Now what is the capacities when you have a dielectric? Well that's very simply kappa times. See not if you remember see not just see not was epson zero for parallel plate San zero A over D. So this is just going to be kappa times that. And so we can factor out, see not see kappa minus one. And we can put in our numbers now. Mhm 32 times 10 to -6 ferrets 12 votes 4.5 -1. billion. The dielectric constant. And this works out to be 1.3 times 10 to manage for cool apps so that that is a positive charge. That is amount of charge on the electric surface here or minus status, The amount of charge that is seen on the dielectric surface on the left, whichever, wherever you're looking. So that is that is our answer and that is the whole problem.