Problem

An alpha particle with a kinetic energy of 10.0 M…

05:19

Problem 77 Hard Difficulty

Potential in human cells. Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is $\pm 0.50 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2},$ the cell wall is 5.0 $\mathrm{nm}$ thick, and the cell-wall material is air. (a) Find the magnitude of $\vec{E}$ in the wall between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) A typical cell in the human body has a volume of $10^{-16} \mathrm{m}^{3} .$ Estimate the total electric-field energy stored in the wall of a cell of this size. (Hint: Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of $5.4 .$ Repeat parts (a) and (b) in this case.

Answer

a) $5.647 \times 10^{7} N / C$
b) 0.2823 $\mathrm{V}$
c) $1.47 \times 10^{-11} \mathrm{J}$

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Chapter 18

Electric Potential and Capacitanc

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Video Transcript

So in this problem, we're discussing cells and cell walls and how it could be. Basically, model is a capacitor. So, um, we have a charge. Density of 0.5 times. 10 to the minus. Three problems per meter squared so we can call that Sigma equals. Um, let's just use the positive side when? Five times 10 to the minus three Who moans per meter squared and, um see, what else? And the cell wall is five millimeters thick, so we can say D is equal to five times. Okay, 10 to the minus nine meters. And what else? This cell? Wall materials air. Okay, um, find the potential. Second, my browser's being funny. Okay, Okay. We want to find the electric field between and the wall between the layers of charge. Okay, So the electric field in junk for a capacitor is equal to the ah charged density divided by epsilon. Not so we actually don't even need the distance for this part. So we just want to do 0.5 times 10 to the minus three way. We wanted to buy this Buy excellent on 8.85 times 10 to the minus 12. And for that we got, um, 56 million volts per meter or hits as high as it was in the other by all. Bio probe problem. I was checking at it quite a few times because, I mean, it just seems like such a large electric field, but Okay, so that's what I got for that one. And next, we want to find a potential difference between the inside and the outside of the cell wall and then determine which ones higher. So for Carol Plate capacitor, he is vey over D. And so, um, B is e d the release magnitude wise. So we need to take this number and multiplied by this five times 10 to the minus nine distance. So I got point to a fault, and we want to know, um, which has a higher potential. So the positive side is gonna have a higher potential, and, um, and then there's a negative charge in the inside layer and then on the outside surface, it's positive. So the outside is gonna have a higher be. So I'll just write that down outside higher potential. A few ways to reason that electric field points from high to low because it's the negative derivative of potential. Um, you could get that. Yeah. There's a few different arguments. I think I'll just go with that one for a simplicity. So, for see, we're trying to get so we're given the volume of the cell. So 10 to the minus, 16 meters out. Another be. I'll just use ball 10 time. All right? Not 10. We'll just use 10 to the minus 16 meter is cubed. And then we want to get the total electric field energy, pause the video and think a little bit about this one. Okay, so I wanted to look up the electric field density formula. I guess this set of questions involves other concepts with besides those directly presented in the chapter. So anyway, it's 1/2 epsilon, Not you squared. So this is a density. So to get the total energy, um, you take that and multiply by volume, so, yes, I use you for energy. Um, so I'm gonna take Yeah, I'm just gonna take this electric field this big old number, 56 million squared, and then I'm gonna multiply it by volume. And then Absolutely not. I'm gonna pause while I do that calculation. Okay, When I plug that into a calculator, I got 1.4 times 10 to the minus 12. The him makes me think there is some other way to do this. I'm not exactly sure what it is. I'm just gonna go with this answer. Um, So then we were told that in reality, you know, we there's a di electric constant. So how is that going to change our calculations? So, um, the nigh electric constant is gonna reduce our electric field by a factor of Kappa. So, Kappa, um, is 5.4. I'll go to a new page for this one. So, um, so Kappa is 5.4, and then we're asked to repeat and be a was finding the electric field and then be was finding a potential difference. So, um, for repeating a e goes toc over kappa. So what we want to do is just take this originally that we obtained and divided by Kappa. So the original e waas, 56 million, and we wanted to buy 56 million by 5.4. So that's one over by 10.4. So then I got 10. Basically, just I'm just gonna use to Sig Figs. I think that, too, is the most the least used. So that becomes 10 million volts per meter and then for be the potential difference. So, um, that's just e times d. And so that is also D is going to stay the same, but he is reduced. So that's gonna be a point to a divided by our Kappa, which is 5.4. So 0.28 divided by five points for is 0.52 What?

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