Problem

(a) Two point charges totaling 8.00$\mu C$ exert …

10:16

Problem 37 Hard Difficulty

A certain five cent coin contains 5.00 g of nickel. What fraction of the nickel atoms' electrons, removed and placed 1.00 m above it, would support the weight of this coin? The atomic mass of nickel is $58.7,$ and each nickel atom contains 28 electrons and 28 protons.

Answer

$1.02 \cdot 10^{-8}-\left(11.02 \cdot 10^{-6}\right) \%$

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Physics 102 Electricity and Magnetism

College Physics for AP® Courses

Chapter 18

Electric Charge and Electric Field

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Video Transcript

Welcome back. My name is Kevin. Trucks were being asked Teoh, remove some charges from a nickel that weighs five grams and placed these charges one meter above it so we could see if we could support the weight. What would be the fraction of the nickel atoms electrons it would have to remove? Well, we could imagine all of this since we're suspending, suspending something with weight, we could imagine that this is going to have a free by a diagram that goes along with it, where we have the electric force pointing up and we have the gravitational force pointing down off this Nicholas being suspended. So we know from Newton's second Law that we have the sum of forces equal Teoh zero in this situation, if it's going to the stationary, which means that the force of the electric or the electric force and minus gravitational force has to be zero all of this to say that we can just simply said the electric forest in the gravitational force equal All right, well, let's not begin to break that down. They grab it there. The electric force that's going to be calculated by K Times Q one Q two. Where the queues right? Moved a certain amount of charge in this I have it. So spending above and a deficit of it within the people just call that cues. Get cake. You square over our Where are we know is eventually going to be that one meter. So case Q squared over r squared and that's going to be equal Teoh the gravitational force which is just going to be Mng. And I'm not gonna worry about affecting the mass of the Nicholas removed the charges because there's just so little maps that would get removed. It all you're doing is taking away civil life tries. Okay, Great. Let's now begin to expand this up. So if we're trying Teoh solve out for what? These these cues are over here. Then we can do that by simply saying that Q is equal. Teoh um times g times R squared, divided by tae square root. What's going to the square root 0.5 times G, which is 9.8 times are squared, which is just going to be one divided by tape, which is 8.9 times 10 to the night. This is all just going to be equal to 2.3. Far if I was 10 to the negative six. Great. That's gonna be our charge. Overall, we can get the charge of each electron where we didn't divide that by the charge of each electron to get the number of electrons. So if I don't divide that by 1.6 times 10 to the negative 19th then we're going to get one point for seven times 10 to the 13th electrons. And that's how many I have suspended. So now let's figure out how many I started with. Well, I did start with 0.5 kilograms of the nickel. I do know from the molar Mass that if I divide that by 58.7 kilograms, I will be left with the equivalent one whole. I know that one mole is Allah. God Rose number of Adam's 6.22 times 10 to the 23rd Adams in each one of those atoms. I know that I have 28 electrons, 28 electrons for everyone. If you go ahead and multiply all of this through, where you're going to get is that you have 5.5 times 10 to the negative 12. You see they are number. There were one point for three times 10 to the 21st total electrons that were within the substance. So if we've taken out 1.47 from that, we can simply ratio the to to see that what the amount we've taken out here's 1.2 times 10 to the negative. It's Or you could think of this as being 1.2 times 10 to the negative AIDS are negative. Six her set if you want. I don't enjoy this video. If he didn't please take a moment. Click that part about that screen to let me know as well let me retract more students. Thank you.

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