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Find the total Coulomb force on the charge $q$ in Figure 18.52 given that $q=1.00 \mu C, q_{a}=2.00 \mu C$ $q_{b}=-3.00 \mu C, q_{c}=-4.00 \mu C,$ and $q_{d}=+1.00 \mu C$ The square is 50.0 $\mathrm{cm}$ on a side.

$F=0.102 \mathrm{N}$ downward

Physics 102 Electricity and Magnetism

Chapter 18

Electric Charge and Electric Field

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11:53

In physics, a charge is a …

10:30

The electric force is a ph…

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(a) Find the total Coulomb…

10:17

Using the symmetry of the …

07:42

What is the force on the c…

02:03

What is the force on the …

11:10

(a) Find the electric fiel…

07:36

(a) Using the symmetry of …

10:29

Four identical charged par…

05:52

(a) Find the total electri…

11:14

Three charged particles ar…

06:58

Two $2.00-\mu \mathrm{C}$ …

08:51

06:38

06:00

Two equal charges $\quad(Q…

15:03

Find the electric field at…

10:58

Four identical point charg…

11:05

Three charges are at the c…

02:46

Identical point charges $q…

03:54

Point charges $q_{1}=50 \m…

0:00

Identical charges $q = +$5…

Kevin Chirac. Let's consider here where we have a charge in the center of a square, which we're gonna call Q. And then we have. Q A Q B. We also have Q C and Q D. And we're gonna be trying to figure out what the Net forces here just on this middle, que we're gonna just call que well, we know that forced the sum of all the forces is going to be equal. Teoh the force from each of the ad together as a factors we can say that forced from a plus the force from B plus the force from C plus the force from D. And we'll just make note of the fact that these air each defectors. That means we're going to have to break this into a sum of forces in the X direction. We're going to break this into it. Some of forces in the Y direction now, as far as the magnitude of each of these forces, remember that it's going to be of the equation structure. K Q one Q two over R squared and the ours for each of these are going to be the same because they're just going to be the shortest distance along this diagonal down here in this square, the case, we're all going to be the same. And for at least one of these cues, let's take you to That's also going to be the same for all of them. So the only thing we need to consider is the angle, the direction and in the varying charges of a Q eight Q B kids seeing Judy. So, for example, let's consider the charge that the force that results from Huet onto the centre since you is a positive charge and the center is also a positive charge, we know that's going to be a result into charge that's going to be going down into the right now. The interaction between the center and B is going to be an attractive charge, so that one's gonna be going up into the right for C is going to be going down into the left and for D. We're going to be going up into the left, so we're gonna be using each of those pieces of information as we add together all of our forces. Another thing to keep in mind is that because this is a square, and the angle here is 45 degrees. We can use our knowledge of unit circles to say that really, that just means that we're breaking this into a ratio of where the up component or the sideways component, and similarly, the down or sideways components, depending on the direction or it's gonna be a factor of routes you over to. So all of this together is going to allow us to write the following for the X direction. We're going to have k Hugh over R Squared times route to over to, and then we're going to be multiplying this by each of the signed magnitudes of these charges. So, for example, Q. A is going to result, like we said in pushing down into the right. So for our extraction risk when it caused Call that to be a positive direction that's going to be too micro coolants to 10 to 6 for the next one. B is going to be pushing it or excuse me, pulling up into the right so again that's going to be plus in this time is gonna be plus three times 10 to the native six for the next one see, this is going to be pulling down into the left, so that left part is going to give us the site of negative. So it's gonna be negative four times 10 to the age of six. It's more room here, and then the last one is going to be the is going to be up into the left. So again it's gonna be the laughs we're gonna say, minus one times 10 to the English days. So just to reiterate, if you were to expand all this back out or try to figure out where this whole came from, we're taking this equation for each one of these terms. And I've just made the calculation simpler about recognizing that K is constant. All of them, the center charge and the radius squared, or it's gonna be constant. There's going to be a a resulting scaling that's going to occur because of the direction of doing co sign of the angles between our side of the angle between respectively. And that's just always gonna watch what routes over two and then all that's left and unique to each of the terms is this Q one or the size of each of these cues. All right, Now you go to the exact same thing for the sum of forces in the X direction far exceeded some force in the Y direction again. K is gonna be constant for each one of those terms. The center charges constant. The radius is going to be constant. There's going to be a route to over two factor that's going to occur. But now again, we're going to include the sign for each one of these interruptions. So the interaction again between A and Q is going to be down into lots of this is gonna be a negative two times 10 to the negative six down part of down into down into the right is gonna be the negative that's gonna pop up here. The next one is going to be up into the right, so that's going to be a plus three times 10 to the pagan sites. The next interaction between the negative center charge and the the positive center charge and the negative see charge is going to be a pull down into the left so that down is B negative four times 10 to the negative six in the last one is gonna be an erection between charge D, which is positive, and you were just positive it's going be pushing up into the left. But that up part is gonna give us a positive. So it will be up one times 10 to the negative six. So when you add all this together, you're gonna get some of the forces in the Y direction You're gonna get the symbol of reporters in the X direction and by simply squaring those and then adding them together and doing the square root, we can get the total result in forces like we did back in our course, chapter writes. We could say that the total force is equal to the square root that symbol, the ports in the extraction squared, plus that some of all the forces in the Y direction squared. You can just plug those right on. When you do, you get the answer being that the magnitude of the force is going to be 0.102 news, and what you're going to recognize is that this is going to occur because the force in the Y direction is going for the force in the X direction is actually all going Teoh zero out and you can see that just because here we have two times 10 of them get six returns. 10 8665 times 10 to the naked 60 out together. Here we subtract five. Well, so actually, this is zero. All of this is zero. And this answer right over here become 0.102 news we get. The magnitude is 0.10 to Newton's and we get that is facing down. I hope you enjoyed this video. If he did this, take a moment. Click that part department scream to let me. Now it's well to help me reach more students.

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