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The early 1900s, Robert Nmnan used small chargod @ropiets oil, $ uspended clectri Ield, mako Iho Iirat quantitaiive Measurements Iho oloctron s charge. 6d-MIn-diamc...

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The early 1900s, Robert Nmnan used small chargod @ropiets oil, $ uspended clectri Ield, mako Iho Iirat quantitaiive Measurements Iho oloctron s charge. 6d-MIn-diamcin droplat ol cil, hawng cnumc Suspandec mida botroen ncnzonini plates parallel-plate capacltor. Tho upward 2 @Cinc Unncm droplet exactly balanced by Ino downward Voict gravty: The oll has densily kg/m" Cha cIlc plales are 4 0 mI apanPartmustta polanlu cdla roncohotvccnplates Le hold Iho droplet Gquionumt ExDrobe Your answcr wit

the early 1900s, Robert Nmnan used small chargod @ropiets oil, $ uspended clectri Ield, mako Iho Iirat quantitaiive Measurements Iho oloctron s charge. 6d-MIn-diamcin droplat ol cil, hawng cnumc Suspandec mida botroen ncnzonini plates parallel-plate capacltor. Tho upward 2 @Cinc Unncm droplet exactly balanced by Ino downward Voict gravty: The oll has densily kg/m" Cha cIlc plales are 4 0 mI apan Part mustta polanlu cdla roncohotvccn plates Le hold Iho droplet Gquionumt ExDrobe Your answcr with Ihc @ppropriate Unlla AV 26.29 Submll pievioumanta Ho q ucal anuwgt Incorroct; Try Aqoin; utampte romaIning Froudu Fuadbuca MacBook Pro



Answers

In the early 1900 s, Robert Millikan used small charged droplets of oil, suspended in an electric field, to make the first quantitative measurements of the electron's charge. A 0.70- $\mu \mathrm{m}$ -diameter droplet of oil, having a charge of $+e$, is suspended in midair between two horizontal plates of a parallel-plate capacitor. The upward electric force on the droplet is exactly balanced by the downward force of gravity. The oil has a density of $860 \mathrm{kg} / \mathrm{m}^{3},$ and the capacitor plates are $5.0 \mathrm{mm}$ apart. What must the potential difference between the plates be to hold the droplet in equilibrium?

Hello, Strengths. Let's start or discussion. Millikan, Drop experiment. Americans drop experiment. Okay. Suppose you have to metal players off a poser. Charges I draw off electrons is exactly balanced by its weight. It's means that if this is a draw, then a downward force I'm she will acting here. Okay. And two balances downward. Fourth, there should be an electrostatic force is applying on this part. A language. Ah, for direction. It means that this is a port city. Went This is the negative plate. Okay, Now we have to find out that video's off this drone, which is filled with electrons. Okay, so firstly, this'll drove I m g s banished by electrostatic force curie. So from here, mass will come. So, Toby, you eat by muscle drop will be doing by G. Okay, now the dance city or brought really mass by volume and from hair volume comes out to be must buy density and volume over drop is because it's very kinship. Fall upon three ir que mass weight density. So from here, value off our Q will comes. Toby must buy density three up on full by. Okay, So if we put the values off mask. That is this cube i g. So it will be three upon for by you even by G and one upon density and gilded route. So by this formula, we will get the Nadya's off this drug. No charge on electrons are 1.6 inter tenderness. Par minus 19. Coolum. Okay. Electric field that is applied here is three entertained this part for you to Bakula and the density off the drop is 8 58 k g four meter cubed bip and value of G s. 9.8. Meet up a second square. So five putting all the values here. We will get a radius off Jiro point Geo 0.5 to interrupt. Endless bomb minus six meter. It means that we will get rid our radius. All Jiro, point five to my group meter. This is the radius off drop in our Milken's experiment. Well, this is all from me for test video. I hope you will like the video. Thank you.

So to calculate the charge on this oil drop, all we have to do is quit. The force is due to gravity, and you two electric field forced you to gravity equals forced you to electric field. That's why oil drop can suspend in air by electric field. So forced you to gravity is mass times the gravity on electric forces. Cute times, electric field. Now we argument quantities such as mass. In this case, which is 1.1 times 10 power never to 14 on dhe. Gravity is 9.8. Charge is something we supposed to find. An electric field is 1.68 times 10 power five. So from this you can find Cuba's. So let me perform this calculation. 6.41 seven times 10 power. This is 14 negative. 19. Cool. So that's the charge on this oil job. Also, the next part that's be part asks us to find how many extra electrons are there on this oil drop. So Elektronik charge is 1.6 times 10 power negative. 19 Cool. So number of electrons on this oil drop will be cute. That's the charge on the oil drop divided by electron charge. So that's 6.417 divided by 1.6. Because both contain the same factor, they will be cancelled. So the answer here is for so we have full electrons on this oil drop.

Hello, Prince. This problem is based not Milken's oil drop experiment. Milken was a scientist who had first time measured up charge on the electron. He had allowed the Why trump passing through the usual off atomizer So that don't affection office it will. That charge due to transfer of left, what on it is allowed to bus, you know, charge, conducting plates, having that uniformly creepy. And by finding the condition off breast or motion with velocity, he had calculated charge on the electron in the question we have to find the chart on the drop off really? Is. Our brought is at rest between the plates brought will be addressed if it is in equilibrium, that is date of the drop is challenged by forced you to electrically so at equilibrium Que e will be equal to empty the charge that the trump people get and he appoint Yeah, I m is the mass of the drop, which is familiar boardroom of the drop into its propensity to be divided by electrically electric. Please between tell the difference between the plates and that it spins between tablets. The chart on the dropped people get would bite. Are you ruin to d be divided by three to put in Italy. So this is the Antero part eight off this problem, not in the party. The release of the drop is meter by majoring the memorability it a left victory just to stop, then drop it Movie under gravity under gravity and requests Got a minute reality If I drop acquire start of Nebula City the net force on it becomes you know they off. The drop is challenged by viscous force acting on it because six fight Peter are beating, he'll be ab neglected the effect of trust. So trust me, are neglected because it will be very less so that being human to be zero. So for 10 more velocity this first force, we expect the general city below display you are making they volume off the drop For what? Deep ir que going to the So By solving it, you will find the release of the drop movie nine. You job grand multiplicity a point, don't you? No. If we substitute this value said this is the question. But I said this when we substitute this value in question, But we will get the judge. He's done chopped and nobility puberty. I would quite row G in today upon 3. 80. Put any Democrats place and to argue. Argue me nine. Eat up Really upon Rocchi. You're what did you on solving it? The doctor get buried in it and didn't bite gate. So this is the underdog. Be park in the sea part We have two major the charged under drop on releasing the top for the given bellies. See part B appraisal for God given bad news up. Um velocity density response it eight But I tell the prince and the streets between that list the first being tempering the chart manager off actually north a site editing by 3.14 distance between operated one millimeter, 10 to the power My last fight What do you tell the difference is a place 9.161 indoor will be sent off Viscosity off here 1.81 and you can do the par minus pipe. But you all these values argument in the problem That number density it is giving one in tow Country depart by industry apart. We happen wording in meet up our ticket. Thank you. Dance studio right they just given a 24 and G 9.8 on solving it. Chart on that drop even make it 4.8 into 10 to the father off my nuts. 19. Pull up. Which is three times Opie. So, having three access, elect? Not so. This is another answer off this problem last. We have to calculate the radius of the drop. What really is the problem we have to use? So if you have a new here solving it, they didn't want to be white. One dilutive it and so tend to before money. 7 m. So this is the answer for this problem. Thanks for watching it.

So this is the world famous experiment. So I'm glad that I can share the solution of this problem here. The magnitude of the gravitational falls, the equal to the magnitude of the electro static force. So M G equals Q E. Where amazon mas G is the acceleration due to gravity cues the charge and E is the electric field. Now from these situations it's clear that mass is equal to Q E divided by G. Now we can right mass in terms of uh okay, mass in terms of density and volume. So we get volume. We is equal to mass by density so far so good. We have raised to her very comfortable position to answer ignition. Now what is the war limb of a sphere? It is four by three pi. R Q four by three by our cue. So applying that here we get four by three. Fi are killed equals mass. Bye density. Now the masses already given here. So our cube equals three x 4 pi now masses Q E by G and the salt. And finally r is equal to the cube root of three way for pie, Q E by G role. I hope it's clear so far. Now it's already given that the charge here is the charge of the electrons accuse equal, do 1.6 times can raise 2 -9. Cool. And And we have the electric field magnitude of the electric field to be three times 10 raised to fall. Newton for And we know G is the acceleration due to gravity 9.8 m/s squared. So we have listed all the numerical values to solve the problem. Now, putting in the values we get R is equal to r radius, which is the christian radius is equal to. And one more thing that we have to note down is raw. The density, which has given 8 58 kg per meter cubed. 8 58 Killer grandpa. It'll keep so substituting all these values we get R is equal to zero five, two Times 10 raised to -6 m. We can write it in the standard form to be ah 5.2 times can raise 2 -7 m. 5.2 times 10 race to -7 meter. So the radius of the oil drop that could be balanced by the given electric field. The force of the group were given electric field is 5.2 times 10 ways to minus seven m. Thank you. Happy


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