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Question 10 part 1 of 1 10 points particle with charge 5 pC is located on the T-axis at the point 10 cm and a second particle with charge 6 pC is placed On the T-ax...

Question

Question 10 part 1 of 1 10 points particle with charge 5 pC is located on the T-axis at the point 10 cm and a second particle with charge 6 pC is placed On the T-axis at 4 cm The value of the Coulomb constant 8.9875 X 109 N. m? /C22~10-8 ~6 ~4 ~210 (cm) What is the magnitude of the total clec- trostatic force 0n a third particle with charge ~5 pC placed on the €-axis at ~2 cm Answer in units of N_

Question 10 part 1 of 1 10 points particle with charge 5 pC is located on the T-axis at the point 10 cm and a second particle with charge 6 pC is placed On the T-axis at 4 cm The value of the Coulomb constant 8.9875 X 109 N. m? /C2 2 ~10-8 ~6 ~4 ~2 10 (cm) What is the magnitude of the total clec- trostatic force 0n a third particle with charge ~5 pC placed on the €-axis at ~2 cm Answer in units of N_



Answers

Ten positively charged particles are kept fixed on the $x$ -axis at points $x=10 \mathrm{~cm}, 20 \mathrm{~cm}, 30 \mathrm{~cm}, \ldots, 100 \mathrm{~cm}$. The first particle has a charge $1.0 \times 10^{-8} \mathrm{C}$, the second $8 \times 10^{-8} \mathrm{C}$, the third $27 \times 10^{-8} \mathrm{C}$ and so on. The tenth particle has a charge $1000 \times 10^{-8} \mathrm{C}$. Find the magnitude of the electric force acting on a $1 \mathrm{C}$ charge placed at the origin.

Okay, So in this problem, you have to particles. They each have a charge of five times 10 to the minus six. Cool. OEMs. Right down and their distance between them. The distance between them is 24 centimeters. Let's convert that two meters 20.24 meters, and we want to get the force exerted. 1/3 particle. So I'll call this. I don't know. Thank you. And that has a charge minus 2.5. Micro cool ums. Um, and that's 13 centimeters from each of them. Um, so I'll call that. I don't know. Let's see Big D, I guess to keep along with the theme. Not that That's Ah, convention. Okay, so, um, our goal is to get that the forest that they exert on this particle. So, um, actually, I would need to know how it's configure it, so let me really the problem. I guess it had It almost has to be a triangle. Right? So if the two cues are like, um so this they're this far from each other, right? This is our d. Here's our two cues in order for them for, uh, this que to be 0.13 from both of them. It kind of has to be up here or down here, um, sort of. And so So, Yeah, that's probably important to acknowledge that this is the picture we need tohave based on the description. So, um so here we go, And then our goal is to get the force on here, so I'm just gonna draw free body diagram on cue. So it's gonna feel a attractive for us to both of the charges. So something like this, And, um uh, but the horizontal components are gonna cancel, so we're just gonna be left with the vertical, some of these vertical components. So, basically like this, that would be the vector sum of these two. Probably a little bit shorter. Um, but, um yeah, so it's just and then these air the same magnitude vectors. So we're just gonna take the vertical component and double it. So you basically, this problem boils down to finding the vertical component of the force of, um, one of these charges on one of these little cues on the big Q. So let's write that down, then. Um, that's right down after you Q first. So it's just gonna be a k q q over, um, Big D squared. Um, and then now we need to get the components, and so we need to figure out the ankle. Um, so it's gonna be pointing along this line that connects the two. Um, this line has a magnitude of d. And then, um, we know that the full line is off is, um de so this side's gonna be d over two. Alright, so I've just been trying to get this angle, so I'm drawling this triangle to represent with crayon. So, um, and then this is going to be our data that we and we can take the vertical component by, um, looking at the length of of this poor sorry by, um, let me just remind a little bit to make sure I'm making full sense. Yeah, I just I drew the triangle that was connecting all of these three charges and to in order to get the angle that this vector is putting down because I know the vector is gonna be pointing along the side of the triangle and so that I can identify this data and then, um, I can if I multiply this forest by the coastline of this data that I'm getting the vertical part of the forest. So it's gonna be so I'm and need to do coastline data and then to get the net forest, I need to double this. Right? As I was saying before, so then I get that, um f net equals this times two. So now it's I just need to plug in numbers co sign of data. It's adjacent over high pot news. Um, and so I could if I wanted, I could find that I using tangent here. I could say tan data is your little do or two, um divided by big D But I can also use Pythagorean two to determine this side, and then I could just to directly do this side over D so the site is gonna be my Pythagorean theorem D squared minus B squared over four. So then the coastline, if data is just gonna be this divided by d. So actually, I'll just go ahead and said that in there. And, um, I'll put my two in the front so I don't forget about it. Is that, too, that I raced? And yet now I'm gonna do a Jason over iPod news. So it's d squared minus d squared over four, divided by D. So that's gonna make this d cubed. Okay, great. Now I can start plugging in numbers. Um, you pull up a tab. Okay, So we're gonna do two times eight point. How many Sig figs? I think it was to about 32.92 times 8.99 times 10 of the nine. And then I'm just gonna use magnitudes of Q. Um, because I already took into account for direction. Um, in my drawing, which sign is the only time would affect the direction. And I already took into account the direction of drawing. Um, that's personally how I like to approach these problems. So then I multiplied the two charges, and then I'm gonna look a climber. The square root of big B squared big. These points 313 squared minus little D is 0.24 square divided by four square. That whole thing you're not going to arrive at by 40.13 to the third power. So let's see what I got. Um, I got 50 and ice. It comes out to just know ah, times 10 to the anything. So 5.11 Newtons is, um the what I get for the net force.

So from Sim entry, we see that the net force um come on the net force component along. Why access is zero now, the net force component along X axis will point right word Now, given Fada is 60 degrees. So if on the force would be twice off cue three times Kyu won plus component of Theodora or ah, prediction of fate Our protection off the component along, um X axis. That's why we have cool science data here. And then we have four pi unsettled, not discord. So that's the force Company lets a spread Excuse me. So that's the production of force along X axis. And since we know that co sign 60 degree is half so we can use that number on DA, we see that F three will be okay. Q three Q one divided by two. Sorry, uh, you wouldn't be here. It's a squid because we can get rid of this half with the two multiplied in front of the force. So this is equal to 8.99 times 10 to the bar. Nine Newton meters squared are Coolum SCA. Red Q three is five times into the bar. Native 12 Coolum and Q one is given us two times 10 to the power negative 12 column and then for its 0.9 five meter and then prescribed there, which finally gives us the force as 9.96 times tend to the bar negative 12 years. Thanks for watching as you

24.56. So we have this configuration of charges here. One and two are fixed and there's a distance via four centimeters separating them. And then three could be moved along the X axis and were shown the potential energy as a function of its position. And so we want to figure out what the charge on particle three years. So if you note that when this is a 10 centimeters, which let's call X nod, the potential energy is zero. So we just write down the total potential energy of our system, which is going to be equal to zero we have you want. And you too, um actually will have an easier time in life if we just take the four pi upsilon. Not that we know we're gonna have in every term out in front of everything. So the way if you won you too over D plus you want Q three? Yeah, over plus X not because their distance be plus X. We're looking at a specific value. X not here. Plus you too. You three over x mas. So, uh, if we just do some algebra, we find Q three. I was kyu won over deep us X Not plus you too Over X not is equal to negative. You want cute too Over. We just divide both sides by this little sum here and we find out that you three as a charge of negative 5.7 micro cool homes.

So we have given to charges three microfilm and minus one point from my curriculum and the distance between the mistrust continued US. We want to find the force between that force Coolum forces given by K you want you do. Why are square so that is will simply substrate our values. You know that Kay is 8.99 Thanks. 10 part in part nine Newton meters square. But Coolum Square you would thank you too will simply substitute. So that's three times 1.5. So we have a £10 minus six and 10 bar my six. So for total often bar minus 12 Coolum squares divided by 12 centimeters, which is 0.12 meters squared. We can cancel the meter square on this Cool, um squared and our units will be Newtons, which is exactly what we expect on the number comes out to be to a 0.81 The force feeding the mystery find it when unions


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