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Electrostatic potential energy (a) What is the electrostatic potential energy of an electron and a proton at a separation of $1 \AA \equiv 10^{-8} \mathrm{~cm}$ referred to zero potential energy at infinite separation? If charge is expressed in esu, the result will be in ergs. (b) What is the electrostatic potential energy of two protons at the same separation? (Pay special attention to the sign of the answer.)

High. In the given problem, the to charge particles are minus one column and minus to cool. Um The distance between them is given as three meter. So using Ghoulam's law, the electrostatic force of repulsion between these two, like charged particles will be given as K. In two Cuban, into Q two divided by the square of the distance between them. The value of the scale is 9 to 10 december nine. Newton, metarie square park. Coolum square multiplied by Cuban, which is one column in magnitude. And to Colombian magnitude, which is the value of Q two Divided by our square means square off three m. So canceling this school arm and meters squared, the force finally, here will come out to be To into because the square of three which will be nine. That will be cancelled by just nine. So the final answer will be to into 10. To the power nine newton. Hence you can see here of option see is good. Right? Thank you.

Let's use Cool owns law to calculate the energy of interaction for the two scenarios given for a. We can say that we have to attractions and we have radius equal the one times 10 to the negative 10 m, which is equal the 0.1 nanometers. So using two homes Law I this will be 231 times 10 to the negative 19 jewels times nanometers Q one que two It should be the respective charges over our just be 2.31 times 10 to the negative. 19 jewels times nano meters. There are two interactions. Q one Q two over the radius. This would be equal to negative. Uh, let's try the skin. Yeah, yeah, negative. 4.6 times 10 to the negative. 18. Jules, for the interaction in a for be we can see that we have four attractions 0.1, not a meters apart. And we have to repulsion zones across the diagonal. So the diagonal Yeah, we have to calculate. So the length of dag no spirit of a squared plus B squared, which is the square root of one nanometers squared, plus one nanometers squared. This would be equal to right. 0.14 nanometers for the length of the diagonal. Still, calculate the voltage here, which would be 2.31 times 10 to the negative. 19 Jules times, nanometers. We have four attractions. These are 0.1 nanometers apart. Plus, we've got a propulsion of the two positive charges and they're separated by 14 nanometers. And we've got a repulsion of the two negative charges. They're separated by 0.14 nanometers. Yeah. Now we'll have to put this together here. This is equal to 2.31 times 10 to the negative. 19 Jules tens. Another meters. Yeah. So equal to negative. 40 per nanometer. Plus 7.14 our nanometer plus 7.14 per nanometer. We call the 2.31 times 10 to the negative 19 jewels. Purnama meter. This would be equal to maybe 40 Negative. 25.72 per nanometer times culture, 15 minutes. This works so to negative 59 times 10 to the negative 18 jewels. And this would be the energy of the second interaction

Okay, let's do the first approach. That is a energy density is equal to half upset on. Know it. He square so net energy will be volume integrated off energy density. Let's substitute the you in there. Since half absolutely not are constant, we will have e square. Maybe here E is given by electric field Off charge. Cute spherical charge. Cute. So since we are outside of conductor, we can simply choose electric field e equals one divided by full pie. Upset or not Que square divided by our square. No, let's substitute that full pie. Absolutely not square que square integral. One divided by our square. Now we will you just physical coordinates to integrate this question. So this will be a d r times for pirates. I am integrating over all the anvils that is data and by to be full pipe now limits here should be are not to infinity because in spherical conductor capacitor, the other plate is assumed to be at the distance. Infinity. Now, if you perform this integral, you will get Yeah, you equals half, which was already there. You can see that you can cancel this full pie here with the four part here, but it has a square. So 14 vitamins Absalon, Not one of the absolute not denominator cancels with one in numerator. So we are left with one divided by four high. Absolutely not. Que square divided by are not now. Why are not because the integration is negative one divided by our on limits are zero to infinity at infinity that in one divided by infinity zero Oh, sorry. Limit is not zero, but it's are not. It should be are not here so that we have the second term surviving since second term is negative and twice negative is positive. Your answer is what you expect it now In the second approach, let's start with the capacitance of spherical capacity as four pi. Absolutely not. Ah, and then you Question five says you equals half que square divided by sea. So if you substitute value of C here you will get half one divided by full pie. Absolutely not Que square divided by are not I should have I not We're here. No, let's go with the third approach. Third approach asks us to find electric potential energy stole by having small charges off amount dick you from infinity to the spherical conductor. Now to do that, let's consider the work done. Do you? When you bring the charge off cue when you've been bring the charge Oh, thank you. From infinity and toe Find the world done We need to multiply it by the potential difference We No which is Dick You We here will be Q divided by four pi Absolutely not are not Now what I'm assuming here is that this fear here as a charge off. Cute And we are bringing the small charge Dick, you from infinity do this sphere. But to find the net energy all we have to do is integrate, Do you? No. If you integrate, que de qu will get q square by two. So the answer again you will find to be what you find in a and B secure square divided by two

In this problem. On the topic of gases law, we're told that a hydrogen atom can be thought of as a proton at the center of the proton. As a point charge of the center, R is equal to zero and the motion of the electron causes. It's charged to be smeared out into a spherical distribution equivalent to a charge per unit volume, which is given in the problem. We want to find the total amount of the hydrogen atoms charge that is enclosed within a sphere with Radius R and with the center at the Proton. We want to show that our as our approaches infinity enclosed charge goes to zero. We didn't want to find the electric field that is caused by the charge of the hydrogen atom and finally to graph the electric field magnitude as a function of our. So Firstly, we know at the charge, Hugh, as a function of our is equal to the total charge Q minus the integral of the charge density TV row TV. So that's the charge of the proton minus the charge of the electron. So that's Q minus, substituting or expression for role queue times four pi over pie times, a not cubed multiplied by the integral of E to the minus to our over a note multiplied by R squared. They are, so we can write this as Q minus for Q. Over a note. Cubed, I'm the integral from zero to our of our prime squared E to the minus to our prime over a. Not we are prime, which we can simplify to Q minus for Q. Solving the integral that's e to the minus. Alpha ar divided by a not cubed alpha cubed into to E the alpha are minus. Alpha squared R squared minus two alpha are minus two. Simplifying this even further. We get this to be Q mm to the minus to our over a note into two into our over a not all squared plus two into our by a note plus one note that yeah, for the expression of Q as a function of our, which we've now found that as our approaches infinity, then the charge Q. As a function of our we can see approaches zero. So the total net charge of the atom is therefore zero next. For part B of the problem, we want to find the electric field in the magnitude and direction that is caused by the charge of the hydrogen atom. As a function of our now, the electric field is readily outward and gas is. Law tells us that electric field strength E times the circular surface area for pi r squared must equal the total charge. Q. Of our divided by absolute not and so e, the electric field is simply Q overlaps are not divided by four pi r squared. So this is okay. Q. Where K is the electric constant times E to the minus two are over a not directed by r squared. Multiply by two into our over a not squared plus two into are divided by a note plus one. And so we have an expression for the electric field, and that's the magnitude. So not forgetting that the direction of this magnetic field is readily outward. Now, lastly, we want to sketch a graph of the electric field magnitude E as a function of our so we will plot the scaled electric field and the scale electric field. Recently, the electric field e divided by that of a point charge and will plot this against the scale are which is equal to our over a Not so we've taken the electric field of a simple point charge to be K Times Q by r. Squared and we get the craft as follows and this is a graph of the scale electric field against this scaled distance are.