(b) What is the change in potential energy if the distance
separa...

Question

(b) What is the change in potential energy if the distance
separating the two electrons is increased to 1.0 nm?

Accepted Answer

Hi everyone for this problem we are asked to calculate the electrostatic force of attraction in newton's between a proton and electron that are separated by 1.32 nanometers. Okay, so for us to calculate this, we need to recall columns law, which is force is equal to K times charge one times charge to over distance squared. Okay, so we're told we have a proton and an electron. So that's going to represent our cues. Okay, So we need to know what is the charge of each. So let's go ahead and make a note of that are charge of a proton is 1.60 times 10 to the negative 19 Kalume. Our charge of an electron is the same thing, but it's negative. So negative 1.60 times 10 to the negative 19 Kalume and RK Colom's constant is 8.99 times 10 to the ninth jewel meters over Kaloum squared. And they tell us our distance is 1.32 nanometers. So we're going to need to convert this nanometers two m. So let's go ahead and do that now. And one nanometer, There is 10 to the -9 m. So are nanometers cancel. And we're left with 1.32 times 10 to the -9 m. Okay, so now we have all of our variables to calculate our electro status electrostatic force of attraction. So let's go ahead and plug in what we just noted. Okay, so our force is going to equal 8.99 times 10 to the ninth jewel meters over kaloum squared times K q one Q two. So we have 1.60 times 10 to the negative 19 Kalume And negative 1.60 times 10 to the -19 Kalu. And this is all over our distance squared. Remember we have to square it. So 1.32 times 10 to the negative 9th meters. And this is squared. So our final answer, once we do this calculation, we're going to get an electrostatic force of negative 1.3, 2 times 10 To the negative 10th jewels over meter. And this is our final answer. This is the electrostatic force of attraction Between a proton and electron that are separated by 1.32 nm. That's the end of this problem. I hope this was helpful.

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