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Negatively charged il drop of mass m distance is between apan: two horizontal parallel metal platesoil drop mass mMgdhorizontal metal platesWhen the potential diffe...

Question

Negatively charged il drop of mass m distance is between apan: two horizontal parallel metal platesoil drop mass mMgdhorizontal metal platesWhen the potential difference (p.d:) between the speed. When the p.d_ plates is Vi IS decreased t0 Ihe oil drop value Vz the il rises drop constant falls at the same constant speed Air resislance acts on the drop When it is moving The upthrust on (he drop is negligible, The acceleration of free fall is g. What is lhe charge on the oil drop? mdg mdg 2mdg V-V2

negatively charged il drop of mass m distance is between apan: two horizontal parallel metal plates oil drop mass m Mgd horizontal metal plates When the potential difference (p.d:) between the speed. When the p.d_ plates is Vi IS decreased t0 Ihe oil drop value Vz the il rises drop constant falls at the same constant speed Air resislance acts on the drop When it is moving The upthrust on (he drop is negligible, The acceleration of free fall is g. What is lhe charge on the oil drop? mdg mdg 2mdg V-V2 2mdg Vi+Vz Vi-V2 Va+Vz



Answers

A positively charged oil drop is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced $16 \mathrm{cm}$ apart. If the mass of the drop is $1.0 \times 10^{-15} \mathrm{kg}$ and it remains stationary when the potential difference between the plates is $9.76 \mathrm{kV}$, what is the magnitude of the charge on the drop? (Ignore the small buoyant force on the drop.)

The water particles. Oh, are, ah, the good drop that has got a sweat in the downward direction who use MGI, but it is being pushed up by a force of Q E. According to new. Those seconds are, if that it was zero. So with that, Q E equals M. G. So the charge will be m D over E. It was now the electric field because we are given the potential difference. It will be Get the V over day. It was I am G B over Delta V now putting the numbers in. We have God 10 to the negative 15 kilogram times. Gravity 9.8 meters per second. Squared displacement is 0.16 meter. Put in jeopardy frenzies. 9.76 times tend to that three world. All together we come. We find the charge to be 1.6 times 10 to the native 19. Could, um

This is Problem 38 of Chapter 17. This problem has two charged plates in a charge. Drop of oil is being put in the center, and this problem wants you to find the charge of the drop of oil. So to start off, we need to find the electric field that it's in. The electric field between two large parallel plates is relatively constant. We learned that back in the gases Law chapter, you can go prove it yourself if you want to also recall that the electric field is the negative radiant of electric potential. And what that means in this context is the distance derivative of voltage will be equal to the magnitude of electric field. Using this and also using that the field between the two plates is very constant. We confined electric field, so the distance derivative of voltage would be the change in voltage over the change in distance. And that's equal to 3000 Newtons per cool. So that's our electric field. Given that electric field, we could do the Coolum Force equation, F equals Q e. We have e we have f Now we confined que f is 9.6 times 10 to the minus 16 Newtons. That's in the problem. In the field we found, was 3000 Newtons per Coolum. Plug it in self. You find that the magnitude of charge is equal to 3.2 times 10 to the minus 19 columns. But the problem asks for this in terms of E electron charges. If you divide this by the electron charge, you find that Q is about to e look.

For Mexicans. Oil drop experiment The radius of the oil drop is given by AIG Was squared it off. Nine. Enter V over to row G. Putting in the numbers, we would have acceleration. It was 1.618 times 10 to the minus six media. So that is the radius and mass of the oil drop. These density times for third by a cube equals 1.4. Do times 10 to the minus 14 you know Van.

Okay, so we know the potential difference is even an actual few times that distance and therefore have left you see, with the potential difference Dubai by the distance and was another forces people to charge tax the Latapy. So therefore, charge Q will be going to the forced you have out in that review ever will have charged que. It's also equal to the force you have. I don't be over the okay because lots of your easy going down that way about the eventual have charge. Q. Is the Goto ft over data Be so we know the forces Givens 1.2 times 10 to apartment 14 you tops and the distance given a 0.64 centimeter, which is 0.64 times 10 to apartment to Amir and the voters. I mean, the potential difference is give us 240 boats. Okay, so we unplugging all these values back into the equation. Here, therefore can determine the charge, which is Q is equal to 1.2 times 10 to the power off when they were 14. New times Times 0.64 well times 10 to the power meaning to meter and over 240 votes. It isn't. Give us. The charge on the drop is about 3.2 times 10 to the power off Nega six. Next in 19. Cool arms. Okay. And in order to determine how many that strong will miss it well, we need to Eustachy question, Jer, They say the number of electrons and and this is equal to the charge from the drop Dubai body charge of individual natural, which is this A Q B. And this will give us three point you times 10 to the power of the late 19. Cool arms over 1.6 times potential apartment igniting cool arms. Okay, because the charge on the electron is 1.6 times tens of power in a 19 Koloss, and this would give us the number of electrons they were missing is about to and is on my answers for this question.


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