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QuesIlon 7/12 Uchorde TU MCI-laculed W LE &tiq acond chaldc VC plnced Detenine the point (other than infinity) at which the electric field (5 Zero?Iho / Wlal=M...

Question

QuesIlon 7/12 Uchorde TU MCI-laculed W LE &tiq acond chaldc VC plnced Detenine the point (other than infinity) at which the electric field (5 Zero?Iho / Wlal=M10 WI-1Wl 1550) mi1-10 9 1-137 WM

QuesIlon 7/12 U chorde TU MCI-laculed W LE &tiq acond chaldc VC plnced Detenine the point (other than infinity) at which the electric field (5 Zero? Iho / Wlal=M10 W I-1Wl 1550) mi 1-10 9 1-137 WM



Answers

$\mathrm{P} 15.27$ determine the point (other than infinity) at which the total electric field is zero.

In this question, we are given that there are two charges over here minus negative, 2.5 micrograms. Any other six micrograms positive and in the distance off, 1 m apart. So what you find is a point right along this axis in which we have electric few. There s a close to zero. No, Teoh. More or less likely case would be a point that is on the left off. This two charges with your and a reason why is because we have a larger magnitude off, uh, church for this but girl. But go over here. Which means that the point in which the electric for your zero must be further away from this point then from a weaker church, right, so that they will be able to Palin stopped. And so you cannot be on the right hand side where it is closer to this higher magnitude off charge. Right? There definitely would have no zero value. And it can't be in between right in between the two charges. Because at this point, the electric theories also non zero because it is both pointing in the same way rights for the negative charges point you tow us he left and for the positive charge, also pointing toe a left so they will definitely add up to be non zero. So assuming that ISS as some distance d away from, uh, negative charge, right to be candid and try to find this point using the equation for electric for you to be K times Q. For are square. So we have to, charges said. We have to take care off so we have first tst negative 2.5 Michael Groom's at a distance off de and then we have six Michael Grooms resistance off the plus 1 m equating this to zero. Since we won the electric field to be zero and uh, can rearrange this bit right, we bring this negative term to the Ryan site 2.5 gloves times deep. Thus one squid, because to six clones times D square, so we can do is we divide 2.5 over to the Ryan site to find my cloakrooms and your left. This is 2.4 de scratch, So opening out the Ricketts on the left we have the square plus two d plus one. It's keeps us one point for the square minus to D minus one equals zero. So it's a quadratic equation you can soft to find the close to 1.82 uh, the close to negative 0.3 nights. So the disciple unions to once again, uh, as mentioned before, it cannot be in the center between these two chargers for negative value immune states. The point they're looking at is in between this two charges and there is not possible. So you have to reject the negative answer and only tiki positive answer. So therefore the position must be 1.82 m towards the left, off the negative 2.5 Michael Crooms, judge

For this problem on the topic of electric fields, we have a minus 2.5 micro column charge and a six micro column charge separated by a meter. And we want to find the point At which the resultant electric field due to these two charges is zero. Now the electric, the point designated Is designated in the sketch a distance d from the -2.5 Micro column charge since it has a smaller magnitude. And because the two charges of opposite sign, the point cannot be anywhere between the two charges. So the electric field due to the -2.5 macro column charge will call anyone and the one is equal to K. E times a charge Q over R squared. This is the electric constant 8.99 times 10 to the nine newton meters squared. Cool. Um squared times the charge, 2.5 Times 10 to the minus six columns divided by D squared the electric field at this point p. Due to the second charge again is K E times Q over R squared. And so this is equal to eight 0.99 Terms 10 to the main newton meter squared. Coolum squared Times the magnitude of this charge, which is six Micro Columns, or six times 10 to the -6 columns divided by its distance two point P, which is D Plus one m, all squared. And so we have two equations one and 2. And if we equate the right hand side of these equations, we get the plus one m squared is equal to two 0.4. The square they're creating these equations means the electric field is the same at point P button opens directions which produces a net field of zero. So from here we can see that D plus a meter is equal to plus or minus one 0.55 deep. And so we get the distance D. To be one point eight two m. And we also get a negative value, Which is -0.392 m. Now, the negative value for D is unsatisfactory because it locates a point between the charges where both fields are in the same direction, and so therefore the position of point P Is equal to 1.8 m to the left of the minus 2.5 Micro column charge.

In this problem were asked to take a look at the figure 19.17 and determined the point other than Infinity were. The electric field is equal to zero. So to begin, to get a handle on this, um, I think the best thing to do is to examine what the direction of the electric field would be due to each of these point charges separately, and then think about how they're going to add up or superimposed on one another. So, first of all the positive charge, the electric field from the positive charge will always be pointing away from the positive charge. And so that means that it's pointing to the right on the right and to the left, on the left of the charge and even, you know, over here by the negative charge as well there's gonna be an electric field pointing to the left, then from the negative charge the negative. The electric field always points towards the negative charge. So near the charge, it's going to be pointing to the left. Um, we're sorry to the right on the left hand side and to the left, on the right hand side, and then we can think about how the electric field from the negative charge is gonna look over in the region of the positive charge. And it's gonna be pointing towards the left as well over here so we can see automatically that in between the charges, there's no way that the electric field is gonna be zero because they are both the electric fields from both of the charges air pointing towards the left and they're gonna add up to create, you know, a total that is pointed towards the left as well. On the right hand side of the positive charge, perhaps the ah, perhaps we have an opportunity here for the electric fields to cancel out and the same thing on the left hand side of the negative charge because they point in opposite directions, the electric fields. Then there's an opportunity for them to cancel out. So basically what we're doing as we're looking for the point where the electric field from the negative charge, the magnitude is equal to the electric field of the positive charge, and we're only going to be accepting solutions to the left hand side of the negative charge or the right hand side of the positive church. We know that we any solutions that we get in the middle there not going to be valid just because of their directions. So basically, from this, we can create a little equation. Ah, what I'd like to do before we do that it was just set up a coordinate system. So what I'm gonna do is I'm going to take X equals zero to be the point where the negative charge sits. And then that means that the positive charge is going to sit at X equals negative one meter because they are a distance one meter separated from each other. So let's get on with the math here. So the electric field from the negative charge is going to be K. Whatever the negative charge is Ha divided by X squared, right, because X equals zero is the point where the negative charge sits. So the distance from the negative charge, too. Whatever. Ah, the point is that we're considering is gonna be X and then the electric field from the positive charge is gonna be k Q plus, divided by X plus one squared. And that's because let's say you know, we're considering a point over here. How far away is the positive charge? From that point, it's X plus that extra distance of one meter away. So what we can do to make our lives easier is cancelled out the K That's the same on both sides. And then I'm gonna cross multiply. So what do I get here? So I get Q plus X squared equals Q minus X plus one squared in the next step, I'm gonna substitute in the charges. So Q plus is six Micro Coombs Q minus is 2.5 and I am just using the man to because I'm just considering the magnitude of the electric field here. And then I'm going to expand out the X Plus one squared. So I get X squared plus two x plus one. If I expand out that bracket so it's make lives easier again. We're gonna cross out the micro columns and we're gonna multiply. Everything's we've got six x squared, 2.5 X squared, class five rate, 2.5 times two So five x plus 2.5 And then I'm gonna pull this six x squared over to the other side. So I have zero on the left hand side and negative 3.5 x squared on the right hand side. Everything else stays the same. And now I just have a quadratic equation that I need to solve. You can use your favorite method for solving, um, quadratic equations. You can use your calculator. Um, it has a nice built in function. Most calculators thes days, Or you can use, you know, the quadratic equation, Which would be, ah, something like this. Since this is not really the point of the question, I'm not gonna go through subbing in into the quadratic equation to find the solutions. I'm just going to use my calculator on my calculator says that X is either 1.82 or X is equal Teoh negative 0.39 and one or both of these solutions could be correct. Could be a point where the electric feels equals zero. But we've already said that any solutions that we get in the middle of the charges, they're not gonna be valid. In particular, Knake X equals and negative. Um, negative 0.39 is somewhere in between them. So we know that this is not actually a real solution. Okay? The only solution here that makes sense is X equals 1.82 And just to refer back to the picture, that means that we are 1.82 meters to the left of the negative charge. The way that we have to find our coordinate system. So this is our final answer here.

So this is the graph drew for his portion. So negative charge. And there's one positive charge. Neighbor Charles has a modern times off negative negative 2.50 micro calls and puzzle charge has amounts of six point old micro cola and the distance between never charging other charges. 1.0 meter. So in order to find the points that I can even outside the EU, that review between the charge and the charge that trying to put the points on left out off the mega charge okay, is because it posits heart has a larger magnitude Glenda name in charge. Okay, so in order to even out the letter if you when you make sure, yeah, that's a few from the positive charge become lesser. So the only way we can do this is increase the distance between the positive charge, Any less of you. OK, so that's why I would pick the left side. Is this closer to winning? A trash buys murder upon the puzzle charge and they say the distance on their points to a Navy chances are Okay, so now can these such inclusion here, which is that's a you want is he going to eat too? So you want to the that review around the neighbor Charge E two is a few phone calls. Charge, OK. And his chemicals. Who? Okay, have a society Q one over. Art Square is equal to que absolute value. Q two over R plus acts square. Okay, so cute. Wise the charge wrong. Nega charge here and accuse used to charge for the charge here. The reason why we absolutely are here because we are looking for back. Okay, so no half we know gang cake and because all year Okay, so we'll have two point by zero. My girl columns over our square is equal to 6.0, micro cooler over out plus acts square. Okay, so that's utilize the our value, which is one I'm sorry that you realize the expat of years. Okay, which is 1.0 meter. Okay, so if we plug in will have to a point by zero Micro Coolum over our square, You see? Go to six point. Hello. Over. Look are plus 1.0 meter and then square. Okay, as against help microclimate program, can we can sew up. Okay, so for the next step That's trying to ignore the unit meter here. Okay, because it's for the convenience off calculation. We were eventually pulling doing this for the answer. Okay, so for the next day, we have 2.50 over. Our square is equal to 6.0, over R plus one square name will have our plus ones. Where is equal to six point. Hello. Divided by two point file. Times are square. They will have Remember our plus wants. We're just simply are square plus to our and then plus warm. Where does he go to, uh, to a going for our squatter. If we move everything from the left side to arise that they will have zero. You see, co two to a point for are square minus are square plus to R plus one. No que us Switch this. I would have to going forward. Our square minus are square plus two are plus one is equal to zero. Therefore F 1.4. Our square minus to our minus. Warren is equal zero and we can use the form of that here, which is negative. B plus and minus square. Root bees were minus Voisey over It's a way to determine the are fatty. Okay, they will have if you're plugging where that could be, which is to plus and minus square root to square. Plus, uh, 1.4 times for Okay, you go. This is, uh, minus one times 1.4 and then it's negative before the four. So we have positive about it. Okay? And over two times, one point for they'll have the first solution. A one is equal to 1.82 meter. And the second solution Arts, who is a well, negative 0.4 meter. We know that that this is can never be connected. Okay, so the solution here is impossible. Therefore, the honest lotion that is possible is our one was which would have the body for ours. People go our which is even one point. It's one meter. So this is the answer for this question. So if so, the points at which the last few zero is 1.8 meter away from the neighbor charge on the left side. Ok, so now we have I said here R is equal to 1.2 meter. Okay, so this is the answer for this question.


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