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(a) Find the total electric field at $x=1.00 \mathrm{cm}$ in Figure 18.51$(\mathrm{b})$ given that $q=5.00 \mathrm{nC}$ . (b) Find the total electric field at $x=11.00 \mathrm{cm}$ in Figure 18.51$(\mathrm{b}) .$ (c) If the charges are allowed to move and eventually be brought to rest byfriction, what will the final charge configuration be? (That is, will there be a single charge, double charge, etc. and what will its value(s) be?

a)-\inftyb)203500$\frac{\mathrm{N}}{\mathrm{C}}$c)+q (4 charges)

Physics 102 Electricity and Magnetism

Chapter 18

Electric Charge and Electric Field

Joshua F.

October 6, 2020

Good day Mr. Kevin, I would like to ask why did you consider using small positive numbers in item letter a. Why not negative?

Hope College

University of Sheffield

University of Winnipeg

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Welcome back to New Murad seven truck. Let's take a look here where we have a whole bunch of charges that are all put into a line and we're being asked to figure out what the some of the electric field is first at the position X is equal to one centimeter, but we can recognize from the picture that X is equal to one centimeter is gonna put us right here on top of first charge. So even if we were to consider just trying to sum up all of the electric fields, the very 1st 1 that we're going to sum up is going to be this Kate Times Q. Where he was going to be this negative to you. So actually, write that and negative to queue. The first charges needed to two divided by r squared and the distance between this charge and the position one is going to be zero. We can't actually divine bias here, so we could consider just a little bit to the left or the right and looking achievement, the right or the left. And then we could just consider what it would have been right before that. And we get to that. That would be a very, very small numbers. We could just replace this in our mind with a very small positive distance. That one squared and having a negative number divided by a small distance is actually just gonna get something really, really big. So this entire term is almost like saying we have a negative infinity here and then anything else that I add. All of these other electric fields caused by these other charges are just going to be numbers because their distances is going to be a reasonable real numbers. We're gonna have negative infinity plus numbers. What's numbers? And that's just going to simply not matter this for charges just too big. It's a number that's huge, so we can say that the electric field at the first point, it's just going to be negative. Infinity are negative, infinite value, so it's going to transfer point for part. Once we could just say it's going in big of infinity, way too large, way too negative. Now let's consider what would happen instead, if we looked at the position at X equals 11 centimeters, that's gonna get somewhere out here, and we can again go ahead and calculate this, but summing up all of the electric fields. But sometimes I find it helpful to draw in what the electric fields are going to look like first. So if we consider the first charge, the electric field lines are going to be coming into this object from both sides, whereas from the 2nd 1 I'm going to be having electric field lines that come out from the 3rd 1 I'm also gonna have electric field lines that come out because it, too, is positive. And this 3rd 1 is going to have something go in because it's negative. So now that when we add up all of these lifer fields, we're gonna attach the appropriate signs of the appropriate magnitudes, the 1st 1 is going to pay times to you. Where against me? That negative to okay, times Q. One over R one squared R one squared is gonna be the difference between that one centimeter and that 11 centimeters so sometimes also helpful to write this in using the best most to keep us immediately 0.1 meters 0.5 meters 0.8 meters, 11 meters in 110.14 meters. When we do this difference on the bottom. It's 11 centimeters minus one centimeter gives us 10 centimeters, 4.1 square. Let's not do this for a second term since BK times you to work here too. It's just gonna be a positive one. You in the difference between five centimeters and 11 centimeters is going to be 67 years 670.6 And let's make sure that square next we'll dio plus k times the third charge which is going to reach you and divide that by the difference between eight centimeter position of the 11 centimeter position with the point rate centimeters 0.3 meters, which were then square in the last one is going to be a negative. You work in a positive on top. Even with the charges negative, we're gonna have signed the chart. We're gonna sign a positive motivation to it, because again, take a look at this. These arrows, These negative for these red arrows we're gonna be pulling to the right. So we're actually making that's positive. Keep the magnitude of few over the difference, which is going to be that 0.3 squared. So just a recap. When we're putting in all of the signs. The signs don't have to do with the signs that would see in the picture here directly they have to do with the direction of the arrows were really thinking of them or inspectors. Now we do have the value for Q. We also have a value for Tae, which is just the elect, the constant that goes along with electric fields in winter constant. And so we're just gonna plug all this in north back later. And when you dio, you'll get that the answer comes out to be about 203,500. Her cool 203,000 500 Newtons per kula result from plugging in this charge here of five times 10 to the negative night Ghoulam's or five Nano, Colom's and putting in the Electric constant. The last question we have. If the electric charges are allowed to move and eventually brought to arrest by friction, what will the final charge configuration be? That is what will be the result of all these charges. Well, this was actually the easiest. All the questions We just need to add up all of our charge it and see what the overall result would be if I have negative to you, Rick, you another positive feel and another negative. You. If we add all that together, we're gonna be left with a positive you. And so after all of these kind of settle on a common location, the result is going to be just a positive value. So even say that Positive Q one positive two is one of the answer for part C. All right. I hope you enjoyed this video. If he did, just take a moment.

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