In an alternate universe, the smallest negatively charged particle, analogous to our electron, is called a blorvek. To determine the charge on a single blorvek, an experiment like Millikan's with charged oil droplets was carried out, and the following results were recorded : (b) Further experiments found a droplet with a charge of 5.81 * 10-16 C. Does this new result change your answer to part (a)? If so, what is the new largest value for the blorvek's charge?

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Hello everyone today. We are being given the following problem in a hypothetical universe, the charge of the negatively charged Zargar was determined via method similar to Millikan oil drop. Experiment with the following results using the tabulated results, calculate the minimum charge of as arg for the first portion and for the second portion it says an additional droplet was determined to have a charge of 3.26 times 10 to the negative 15th columns. Does this affect the previously calculated charge for a Sarg? If yes, calculate the new minimum charge of as Argh. So we're going to handle answer choice, a first or answer choice a but not answer choice A We're going to handle part A first and that's to calculate the charge difference. So essentially we're just going to take these droplets and subtract them from the next one in line. So what does that mean? That essentially means we would take, for example, Droplet three and subtract droplet four from that. And so if we did that, that would be 2.45 times 10 to the negative 15th subtracted by 1.63 times 10 to the negative 15 And that would give us 0.8, two times 10 to the negative 15th columns. We would then take droplet to And subject that subtract droplet three from that. And so that would look like 4.89 times 10. So then they get a 15th columns -2.45 times 10 to the negative 15th columns and that would give us 2.44 times 10 to the negative 15th columns. And lastly we would take the droplet one and subtract Droplet two from that. And so that would look like 7.34 times 10 to the negative 15th columns, subtracted By 4.89 times 10 to the negative 15th columns. And that would give us 2.4, 5 times 10 to the negative 15th columns. And so the approximate charge of hazard Is going to be, or the minimum charge is going to be 0.82 times 10 to the -15 columns. And so using that, we can calculate the total charge as well as the number of Zardes to get a more accurate value. And so that's essentially going to look like taking each of these droplets and dividing them by our approximate charge of as argh, so we're going to do that in purple And as follows. So for Droplet one we're going to have or droplet four first, we're going to have one 0.63 times 10 to the negative 15th columns, divided by that. 0.82 times 10 to the negative 15th columns. And that is going to give us two, two sharks to be exact, We're gonna do the same for each of these droplets for droplet three, That would be 2.45 times 10 to the negative 15th columns. We would divide that by the 0.82 times 10 to the negative 15th columns to give us three Zardes. And we're essentially going to repeat this process for the first or the second and the first droplet. So for the second we took our 4.89 times 10 to the negative 15th columns, divide that by 100.82 times 10 to the negative 15th columns to give us six Zargar. And lastly we'll do one over here. That's going to be 7.34 times 10 to the negative 15th columns, divided by 0.82 times 10 to the negative 15th columns to give us nine Zardes. And so now we can calculate in green below the questions how much is our would be? So ones are good, we equal our droplet 17.34 times 10 to the negative 15th. Cool Homes plus are second droplet Plus our 2nd droplet Which is 2.45 times 10 to the negative 15th columns. And then lastly Our First Droplet which is 1.6, 3 times negative 15th columns. And we're essentially going to take those values and divide those by the number of Zurich's total 79 plus six plus three plus two. And that would give us 8.16 times 10. That's a negative 16 columns as our minimum charge of a czar. Now we can move on to part B And that involves calculating the charge difference between each charge values that includes that additional droplet. That additional droplet charge was 3.26 times 10 to the - columns. And so essentially we're gonna go through each one of these and we're gonna subtract them. Buy the new one. So that is going to include taking that 2. times 10 to the negative 15th columns. And subtracting that by 1.63 times 10 to the negative 15th. That's going to give us 0.8, two times 10 to the negative 15th columns. And we're essentially going to do what we did in part a to this new part. And so as you could guess this new droplet goes in between here. So the process would be the same. We're going to take that new 3. Times 10 to the -15 Columns. And subject that by the 2.45 times 10 to the negative 15th columns and get 0.81 times 10 to the negative 15th. We're gonna do the same thing for the rest of these. So that's gonna be 4.89 times 10 to the negative 15th columns minus 3.26 times 10 to the negative 15th columns. And that's going to give us 1.63 times 10 to the negative 15th columns. And lastly we have 7.34 times 10 to the negative 15th columns minus our 4.89 times 10 to the negative 15th columns to give us 2.45 times 10 to the negative 15th columns. We have to check and see if this 1.63 value here Is a multiple of that. 8.16 times 10 to the -16 as we calculated earlier as a minimum charge. And so to do that, we're simply going to take that 1.63 times 10 to the negative 15th columns And subtract that by the 8.1, 6 times 10 to the negative 16 columns and we get 1.9975 to which we can round 22. And so our final answer if it affects the previously calculated charge, it's going to be no it does not. And with that we have answered this question. I hope this helped and until next time.

Verified Solution

Key Concepts

Charge Quantization

Millikan's Oil Drop Experiment

Coulomb's Law