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Problem (10 points) Calculate the total charge of the solid ball z? +4?+2? < 5 (in inches) assuming charge density (in coulombs per cubic inch) of (1,4,2) = (3. ...

Question

Problem (10 points) Calculate the total charge of the solid ball z? +4?+2? < 5 (in inches) assuming charge density (in coulombs per cubic inch) of (1,4,2) = (3. 10-`) (r? +07 +21'2Hint: Consider switching to spherical coordinales

Problem (10 points) Calculate the total charge of the solid ball z? +4?+2? < 5 (in inches) assuming charge density (in coulombs per cubic inch) of (1,4,2) = (3. 10-`) (r? +07 +21'2 Hint: Consider switching to spherical coordinales



Answers

Calculate the total charge of the solid ball $x^{2}+y^{2}+z^{2} \leq 5$ (in centimeters) assuming a charge density (in coulombs per cubic centimeter) of
$$\delta(x, y, z)=\left(3 \cdot 10^{-8}\right)\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}$$

So from the previous exercise, we should have obtain capital Q two b equals two k times R to the power of four times pi. So given that R equals to one and K is equals to 20 capital Q would be equals to 20 pie. And because this is an application question and and the quantity waas initially given to be, um, in micro cool. Um, so we have micro and capital seat.

Once again. Welcome to new problem. This time we are given us for Rickles office with the radius and a total charge of Q. So we have a total charge of Q. And uh the electric the electric field strength. The electric field electric feel strength. Uh huh. Inside the ball, R. Is less than our is given by is given by mm We are equals two. Uh imax hard to the fourth power of capital are default. So this is the variable radius and this is the maximum radius. So there are three things we want to figure out in this problem, determined timing. Uh huh. The maximum electric field strength becoming the imax based on total charge and maximum radius. Uh huh. Sure. Or rather or rather determined determined on expression from the um volume charge density determinant expression for the volume charge density sigma are inside the ball. And this simply means that sigma is function is a function of our sigma as a function of art. And then uh this is but being that we're looking at what C. Is saying mm Sure that the um the charge density sigma arm views a total charge of cube. Show that the charge densities sigma are, gives a total charge. Uh Off the key off. So we're just gonna jump into it. We know that E. R. equals two. Imacs are full of capital are and then by ghosts is gross's law. He lost his law sigma mm The electric Phil strength times the infinitesimal distance is human caused. All of you know you notice the connectivity. Um So at a distance at a distance R. Equals two R. Mhm. The electric field, the electric field here's marksman and so is the Q enclosed. So we want to do this formula. Yes, we have human caused uh you know, but then also imax times four by R squared equals two Q. Of you know what? This is human closed his maximum. So that's why we're gonna use Q at that point. So then imax is the same as a few, you note times four pi R squared since we divided both sides by four pi R squared like that. And so the max is the same as wonderful full pipe. You know what time skew over R squared. That's our imax. And then in the body mm we know that the field strength, the electric field strength has given us the variable radius in terms of the marks. But we've already solved for remarks. So we're just gonna go ahead and plug in the values that we got for imax from the previous step. And so now we do have a new relationship for the electric Uh field strength, which is just Q. four pi We're not we're combining these two. So we have our six And then outside we have are to the 4th power. So electric field strength, mm hmm Inside the ball, the electric field strength. Inside the ball simply mm times the S. Is Q. And closed over the note. And the volume charge density. Remember we were finding a want to find an expression for the volume charge density. The expression for the ruling charge density. We know that Cuban closed. It's gonna be the same as Going from 0 to buy sigma are four pi R. Square our that's our cue. And close. So we go back and connect the two formulas the previous one yard deeds which was um Cuban caused a in art then Q. And closed one of our in our sigma. We bring this one down because this is the same as Cuban closed. Four pi R squared the R. With that mm. And so E. D. S. His E. R. Times four pi squared equals to one of you know, charge density. And this one goes from zero to are we don't want to forget that for by R squared D. R. I'm all right. And we already have a formula. We already have a formula for E. R. If you go back you see that E. R. Is this formula. So we're going to plug it in right here. I want to plug in that formula from E. R. Right there. And this is in the process is going to help us simplify the process. Zero to our sigma. Are four by us. Where are these two cancel out? And then you combine those two. So we have a Q. r. six of six equals two. Don't forget to cancel out the notes. So we just have a simplified version where we can isolate the volume charge density. So we'll go back to this section and we um want to isolate so we have CDMA from zero to our, I just want to make sure we have zero Rds so flip them and we have sigma are for pie our square the R. And then on the right side we have a previous equation. Um Then we do differentials with respect to our that's going to cancel out this. So we have sigma are for parts past where they are when we do the differential for that we get six Q. Are to the 5th power all over mm six to our over our six because our six is a constant. We only do differential with respect to our so sigma are signals are Becomes equal to six q. Are to the 5th power. six Q. R. to the 5th power. Uh Let's see. We want to simplify the six. Cute six Q arch. The fifth bottle, all of four pi R squared on six. So we divided both sides by four pi R squared. All these D. R. Disappeared. So this one after we did the into into uh differential. It disappeared. So we don't really need it normal. These two are going to cancel out. So we have our three. So our final final part as an answer is six cube R. to the power of four pi r. six. This is an expression for volume charge density. And then the final part, part C. We want to verify that um the total charge is equivalent to our based on the charge density. So the charge density uh he's just total charge equal to Q. And so we're gonna take the integral on both sides. Remember we have the charge density right there. This is the charge density. So I want to do that replacement. So from zero to our And we have 6q. Three fought by are to the sixth. Power times four pi R squared the R. I want to see if that's gonna give us Q. So let's pull out the constance uh 60 to our who left with Are the 5th power times they are because these two combined these two council. So it's hard to the feast Powell. Then we go ahead and have 6q. Our six. When you do the integral for that. We have a large the sixth power of six because you're adding one and adding like that to give you the integral From 0 to our. So this is six Q over arch. The six. We plug in the are we have our six. All over six. These two council these to cancel and we have a total charge of cute. So hope you enjoy the problem. Feel free to send any questions or comments and have a wonderful day

Yeah. In this problem we have a charge here. Got a positive charge on it. We know the radius. So that's yours, Capital are and it's uniformly distributed the charge. So we can just calculate the density. This is a this is a charge per unit volume density rho Charge Q. over the volume of a sphere. 4/3 pi are cute. We're gonna need that in a few minutes now. The problem like this is it causes all problems. It has symmetry to it. So let's draw Garcia and surface which looks like a circle. But it's really the sphere radius. Our our goal is to find the electric field. Thank you at any point. That is a radius little R from the center. Now, from the symmetry of this problem, The electric field at all. Points on that here, Guardians here. Yeah. Gonna be E. And the fact that the charge is positive. That's our pointing back to field. Now Gaza's law involves the fuck's let's talk about the fuck's now. This is the left hand side. The total flux is a some nation. Yeah. E. I. Well sorry, data I delta A. I. What does this mean? This This is a summation over all when we break up the surface calcium service into patches, flat, flat patches where the electric field is uniform on that patch. Now you say flat, how can you get flat with the sphere? Well, what you have to do with this fear is you must actually break it up into an infinite very tiny tiny almost zero area patches. An infinite number of those. That's technically where calculus would come in. But don't let that bother you at all. Don't let the infinity or hearing the word calculus. We can do it, we can do the same problem here without having to worry about any technical details, mathematical details. So let me So let me look at four distinct patches of the infinite collection. There's gonna be one here, right here, one here and one here. I have four patches I'm going to look at and they will tell me everything. I need to know. What's important is that I can factor out E. And the co sign turn as you're going to see and I'll just be able to get the area of the calcium sphere. That's the whole goal of this. So, let me look at each of the patches. In turn here is going to be a spot for number one, number two, number three. Number four, Let's learn about these these individual pieces what they tell me and what they give me. So, in patch number one, which is the top here is my area vector, normal to the patch and I have my electric field vector. And the world will be the angle between the two. Trivially, you can see 510 now remember something. Area of actors, normal vectors on a closed service which Gaussian surfaces are outward pointing at inward pointing. So that's our situation in Patch one. Patch too. Here is our area doctor. Here is our electric field vector. The angle between the two. Still. So that's good. Patch number three. Here is our area vector. Here is our electric field vector again. Patch and catch three 50 again theory. But the angle between the electric field and the normal vector and four. There's the area vector here is electric field doctor still 54 Cyril. And that's what we were looking for. We know we'll be able to factor out E one, E 2, in terms of the magnitude all the same. I've already drawn that now that co sign can be factored out and that's important. So let me do that. So let me write out the first two terms just to make everything explicit. Mhm. And here represents the rest of the infinite number of terms. But as I said we can factor out. Yeah The angle in this case is 0°.. And for all and this leaves us with delta A. One Plus Delta A. two and all the rest of the infinite collection. Yeah but what is dealt a one? Remember the magnitude of these area of actors? It's just the area of the patch. So if I add up all the patches that gives me the area of the calcium sphere. So this is the area mhm. Of the Gaussian sphere. So this just becomes echoes and zero is 1 eight times the surface area of the calcium sphere which is four pi r square. Be careful in these problems to maintain the distinction between capital our little. So that's the left hand side of gaza islam. Remember God's law, total flux sequel to the charge and closed within the Gaussian surface over epsilon zero. Uh huh. Now let's calculate the charge enclosed before we put everything together. Charge and closed. It's going to be our density which we have from before. Roe times four thirds pi little are cute. Let's put in our expression from before. For bro, Cute. Over 4/3 pi capital are cute Times 4/3 pi little are cute. So The 4/3 are gone. The pie is gone. So we are left with Q. R. Cubed over Capital are cute. So that that is the charging closed. So now writing gas is low. We have E. For pie are square is equal to don't forget the absence. Here we got one over Epson's hero times. Q. There are acute over capital R cubed. And then we're solving for for E. You can already see that the hours this R squared is going to take away and leave just one are. So I'm going to get from this. E. is equal to one over. We've got a four pi still on the left, 1/4 pi absolute zero. And then we have cube and we still have a little are left and we got our cubed and the battle. That is the magnitude of the electric field at a radius little R from the center of the charge here. And that is the that is the whole problem.

Daughters George is given by integration Off differentials jog on the shells, which he's density as a function of our times Volume in the shell, which is for by our square d r interpreting from zero to our we would have we will substitute the value off road now, which is Renard comes outside the integral for by comes outside the integral zero to our one minus our solar are Yeah time this r squared makes it R and R cube, D r because for by donor heart Times article over two minus one over R Times I did, uh, for over full now wouldn't that limit for our this would be zero and this will be capital are so there's just put it in here That was our square that r squared. So the C's are you the 3/3 salting it we get for by you're going art times our cube over trends. So Q equals this. So we have road not equals three cubed over by are killed. Now if we use Gaza's law so on our this and are we have surface, it goes off. Eat our d a. It was you and close over F salon. This area is for by our spread. So we have e given by 1/4 by excellent not R squared dives Cuba and those which we have to our valiant integration. Now, once again and the volume integration has already been done previously, we just have to substitute a different value now. So this will come out to eat for by row, not times The organs are cube over three minus. No one guess are to the fore over four properties are now No, not we saw skewed the value of flown art Luigi's three Q over by our cube. So we find that the electric field comes out to be you are Yorkis are over for by Ebsen not our cube times four minus three Dorcas are over up. Our guests are and at r equals uppercase are we find that e comes out to be You are over four by s or not our cube dives four minus three because this is a parkas are they cancel out So it comes out to be Q r. How about this are over four by Epson, not our cube So canceling the arts on the numerator and the denominator We have got r squared


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