Two small aluminum spheres, each having mass 0.0250 kg, are separ...

Question

Two small aluminum spheres, each having mass 0.0250 kg, are separated by 80.0 cm. (c) What fraction of all the electrons in each sphere does this represent?

Accepted Answer

Welcome back everybody. We are taking a look at two charged copper balls here, we are told that they are cm apart or .55m apart. We are also told that electrons are transferred from one ball to another until an attractive force is established. Now this attractive force is going to have a magnitude of 7500 newtons and the mass of each ball is .2 kg. And we are tasked with finding two things here. We are tasked with finding first the number of electrons that are being transferred back and forth. And then to expressing that as a fraction Of the total electrons in one ball. So let's go ahead and start with part one here. What? We are going to need to use our two separate formulas. We're going to need to use that. The magnitude of charge for one of the balls is just going to be an E. And then we are also going to have to use Cool. Um law that states F is equal to K U one U two all over R squared. So looking at this first formula here, we want to find the number of transferred electrons. So we need to isolate that term. I'm going to divide both sides by E. And then these terms will cancel out leaving us with that. The number of transferred electrons is equal to Q over E. But what is this Q part here? That's where cool. Um law comes in. So first I do wish to establish something, although we are told that the magnitude of charge for each one is Q. You may be wondering? Well one's positive and one's negative. Keep in mind we're looking at the magnitude of charge. Right? So say we have Q. One and Q two. The absolute value of Q. One equals the absolute value of Q. Two equals Q. So this queue is what we are looking for. Let's go ahead and find it with Coolum. Coolum Law states that F is equal to a constant times the magnitude of charge of the first ball, times the magnitude of charge of the second ball. Both of them are Q. So it's just gonna be Q squared divided by the distance between them squared. Now we need to isolate this Q term. So I'm going to multiply both sides by R squared over K. Also put R squared over K. On this side. You'll see that these terms will cancel out over here and we are left with U squared is equal to F. R squared over K. Then finally isolate that cute term. We're going to take the square root of both sides and we get that Q is equal to the square root of F. R squared over K. Now let's go ahead and plug in our numbers for this. We have that Q is equal to the square root of 7500 newtons times the distance between them of 75000.55 m squared. All over nine times 10 to the ninth. This gives us that our few Is equal to actually going to move this over here are Q is equal to 5.02 times 10 to the -4 columns. Now that we have that we're go ahead and ready to find our number of transferred electrons here which once again is just Q over the charge or the magnitude of charge of an electron. We have 5.02 times 10 to the negative fourth columns, divided by 1.6 times 10 to the negative 19th columns. Which gives us that there were 3.14 times 10 to the 15th electrons that were transferred between each of the charged copper balls. So now we are ready to find this ratio right here. But first we need to find the total of mount electrons on one ball and here's how we are going to do it. Let me scroll down here. Just a little bit. We are told that each of the balls are made of copper and the atomic number or copper is equal to 29. And the molar mass of copper is 63.55 g per mole. We're going to use these facts in conjunction with the fact that we are told that the mass of a single copper ball is 0.2 g. And we are to convert all the way to the total number of electrons per ball depending on the mass. So let's go ahead and begin that process here. We are told that the mass of one of the balls. Oh, apologies not .02 kg, but rather .2 kg. And there are in one kg 1000 g. So these units are going to cancel out. We're told that Uh in one mole of copper there is 63.55g. So these units will now cancel out In one mole of anything we have That. There are 6.02, 2 times 10 to the 23rd atoms. So the mole units will cancel out and then in one atom of copper there are 29 electrons. So then the units of Adam will cancel out and we are left with our desired unit of electrons. So if you multiply straight across, you find that the total number of electrons on one copper bowl is going to be 5.49 times 10 to the 24th electrons. So now we will just go ahead and express the number of transferred electrons as a fraction of the total electrons. This is equal to our end of 3.14 times 10 to the 15th, divided by total number of electrons which is 5. times 10 to the 24th. And we find that this fraction is equal to 5.72 times 10 to the negative 10th electrons which corresponds to our answer choice of a thank you all so much for watching. Hope this video helped. We will see you all in the next one.

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