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Chapter 22, Problem 063spherical water drop 1.3 pm in diameter is suspended in calm air due to downward-directed atmospheric electric field of magnitude E 542 N/C. ...

Question

Chapter 22, Problem 063spherical water drop 1.3 pm in diameter is suspended in calm air due to downward-directed atmospheric electric field of magnitude E 542 N/C. (a) What is the magnitude of the gravitationa force on the drop? (b) How many excess ectrons does it have?(a) NumberUnits(b) NumberUnits

Chapter 22, Problem 063 spherical water drop 1.3 pm in diameter is suspended in calm air due to downward-directed atmospheric electric field of magnitude E 542 N/C. (a) What is the magnitude of the gravitationa force on the drop? (b) How many excess ectrons does it have? (a) Number Units (b) Number Units



Answers

A spherical water drop 1.20$\mu \mathrm{m}$ in diameter is suspended in calm air due to a downward-directed atmospheric electric field of magnitude $E=462 \mathrm{N} / \mathrm{C}$ (a) What is the magnitude of the gravitational
force on the drop? (b) How many excess electrons does it have?

So over here to Sprinkle particle jobs with charges of -1 in the 2016. And I was centered to send the suppression of funds. And what is the mind beautiful study for accepting between him and how much excess charge are on the edge of the job? Uh It it starts imbalance giving it. It started imbalance. So basically you know that charge each of them is equal to 1- of one into 10. to the power of -16 columns. I and Supposing we were sent here to send us a present. Between them is nothing but one send them there. And since they are very particle then we can just consider them each a point charge. Right? So force between them in the first part. That will be nothing. But have you called to if you want you to upon our square. It's okay. It's nine into 10. To the power of nine into Cuban and U. Two. It seems to minus of one into 10. To the power minus 16 square For the distance between the mid-1. Send him to just point like particles into the power of -2 square. All right. Yeah. So it will be it becomes anything. But So it becomes nothing about nine into 10 to the power of nine into this minus and minus plus. So this will return to the four minutes 32 upon 10 to the power of -4. So it will be 10 to the power of -28. And this will be 9:10 to the power of -19. Tells. So the force between them that comes out to be nine into Into the power of -19. Newton. Okay, let's move to the next part. We have to calculate exist charge what it exists, charge, what injury? That how much excess electrons are on drop giving it is in balance or imbalance because they are the negative charge. So that means we need to find the number of electrons. So that means we know that uh, E equals two, I think what you sow. And into each manners of 1.6 into 10 to the power of minus 19 equals two E. That is 1- of 1.20 government 16 -1. In addition to the power of -16. So and over here that comes out to be 6- 5. 6- five electrons. Okay, thank you.

Okay, So for the Watcher, drop it to re minister story The weight off the water that his energies should be balanced by the electric field force to directly plans should be opposite. So empty should be called minus times queuing to eat No mg easygoing with jars is equal to the number off a grown stand Charles off one electron, my necessities And he so the number of electrons that require is equal to minus off mg by charts off one electron into T No mass off the water ball is the volume inter density So for every by our cue for what? Three fire two times Roll the density of water times G back Citizen, you grabbed you on Hee Hee. Now hear the radius are is equal to 0.18 millimeters Suit into the bar minus three meters. The density off water. You know very well that is for the 1000 kisses for me. Jerk you on. You know the value off the G. That is 9.82 meters per second squared. You know the value off in a chrome jobs electoral and that is the manager. So 1.6 entered into the power minus 19. This minus and minus in. The new editor gives canceled out. Andi, the electric field e is equal to how much? What is the great fear? 1 50 So 1 50 Nugent park with them. So if you clog all this panic here, this gifts and is equal to 9.98 into chain to the power six horns, right, so almost equal to came to the power seven across. So this is the solution to this question. Thank you.

Okay, So for this problem, you know, this corresponds to a famous for six problem. Maybe the Millikan problem. Yeah, Um, a droplet. And it's suspended in the air because it has some charges. Um, so it's in an electric field, Given that it's an electric field, it has, ah electric force up due to the electrons, given that the electric fields pointed down. And ah, it has a weight for us of gravity. And so those balance each other. So, um, so we can figure out how many charges by setting the electric force equal to the weight and then solving for the charges. So the weight is mg. The electric force is, uh, the total amount of charge times the electric field so we can get that Q is equal thio g over, um, the electric field. And then if we want to convert it, um, into the number of charges we need, Thio multiply by a conversion factor, which is one charge is, um, 1.6 times 10 to the minus 19 cool homes. So, like this, I'm g over. You will be in Coolum. So we multiply by one charged about it. I won't 10.6 times. Hello, my stinking columns to get it to the number of charges. One more step is we need to get the mass. So the problem doesn't actually give us the mass of the droplet. Only the radius, which you can imagine that's but would be measured. And then they were given that the radius is 0.18 millimeters. It's really small. And then so to get the massive it you can use mass is equal a density times volume. Ah, and the density of water is 1000 kilograms per meter cubed. And then the volume is gonna be this 4/3 pi times this r cubed in the mind. Plug that into a calculator. I got, um, 2.44 times 10 to the minus 11 kilograms. Just like very small. It's like micro is not on the level of my program. So it would be, um, kind of minus eight Graham. So I guess I'll be on the level of micrograms, and I've definitely seen ah, things that are micrograms. So suppose that's not too off the mark. Um, okay, so that's what we're putting in for mass for E. Let's go ahead and write that down. The problem gives us that the electric field is 150 Newtons per cool. Um, and then I think we have everything. Gravity's 9.1, uh, on and yeah, When I plugged all this into a calculator, I ended up getting 10 to the, uh, seven charges.

Here for the solution here. Giving dead water drop is stationary in the air so the magnitude of the droplet must be the same as the weight of the droplet. Water droplet is spherical in shape, so the volume of droplet is four by three pi r. Q. Here r is the radius of the drop. Mhm. The density of water drop is p the radius of water. It is article 20.18 or 1.8 multiplied by 10 to the power minus 5 m. Uh, sorry. Electric capital equal to 1 50 Percy intensity of water p equal to 1.0 multiplied by 10 to the power three deep as easy acceleration due to gravity is equal to 9.8 m per second squared. The magnitude of electric force on the droplet is F equal to QE and we consider it as a question one. Hair charge on the water droplet that is Q and E is the electric field. The weight of the water droplet double equal to mg air mass of droplet am equal to four by three pi r cubed B Therefore the weight of the droplet W equal to four by three by R Q p multiplied by G and we consider it as a question to like, number of the access electrons on the water droplet is equal to end the charge on the droplet that is critical to any hair charge of the electron that is small equal to 1.6, multiplied by 10 to the power minus 19 column Further droplet to remain stationary, the magnitude of the electric force on the droplet must be same as weight of the droplet. Therefore, the from Equation one and two Q is equal to mg here by substituting the value of you and I am so we get an small E capital equal to fall by three pi r. Q p g. So in terms of an an equal to fall by argue PG Divide by three Smalley Capitally Now substitute the values we get Access electrons on the droplets here by substituting the values of variables. We get an equal to 451.8, multiplied by 10 to the power minus five m l to the power 31.0 multiplied by 10 to the power 3 kg per M Q. 9.8 m per seconds, divided by three. Multiplied by 1.60 to multiply by 10 triple minus 19. Coolum in Baghdad, 1 50 Percy Electrons. So here, By solving this, we get an equal to nine point nine six, multiplied by 10 to the power six electron or through the round of 1.0, multiplied by 10 to the power seven electron. So this is a compared solution. Step by step in detail with the explanation. Please go through this. Thank you.


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