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Question 4 (6 points) A tiny ball (mass 2.125 ke) carries a charge of -81.00 UC What electric field (magnitude and direction) is needed to cause the ball to float a...

Question

Question 4 (6 points) A tiny ball (mass 2.125 ke) carries a charge of -81.00 UC What electric field (magnitude and direction) is needed to cause the ball to float above the ground?Prevlous PageNext PagePage 1 of 4

Question 4 (6 points) A tiny ball (mass 2.125 ke) carries a charge of -81.00 UC What electric field (magnitude and direction) is needed to cause the ball to float above the ground? Prevlous Page Next Page Page 1 of 4



Answers

A tiny ball (mass $=0.012 \mathrm{kg}$ ) carries a charge of $-18 \mu \mathrm{C}$. What electric field (magnitude and direction) is needed to cause the ball to float above the ground?

So to find the electric field 19 direction, you need to cause a boat afloat. Uh, basically what you want to do. You want to balance out a few forces. So there's first the force of gravity OMG on then the other force, which that that's the fourth pointing downwards. So I have to forces to force that this would be energy. Then you have the other one pointing upwards, which is the electric force. And you find that with the equation you charge times electric field, so simply just want to balance out the forces. So you want Teoh. So since if they're equal, the a levitating the trucker levity So that's another force. You said that you go to each other. So what we're looking when the problem is the electric field to let this value so we simply want to do is isolated. And since we know the values for everything else the mass 0.12 the gravity constant is 9.8 a second squared and then the charges given the problem is negative. 18 10 10 to the negative six Coombs meadowland your calculator and you get around negative 6.5 tens, 10 to the third new me cools. That should be the electric field. And as for the direction, it should be upwards since the

Hi, welcome to this edition of this problem. Now imagine this is the object that has masks 3.8 g. It is suspended in this region. So just imagine that this whole region of this white body Sir region that contains directed feeling ideally this should fall down because after MG falls downwards but it is being suspended. It means that any equal falls is being apply it on it in the opposite direction. That force is due to the electric field. So will he be you? Mhm. Okay. So this means that the magnitude of Q. E. Should be equal to the magnitude of. Okay, so we can right it S. Q. E. The equal to M. G. Now in the problem, The M. Mars has given us 3.8 g. The charge skills given us -18 new c. And it would be very easy for us to find the electric field by an equation is equal to MG. By keep by an equation. He is equal to MG By okay applying the values we get 3.8 multiplied weight and raised in my industry. Because we have to convert Graham two kg as kilogram mr S a unit and 9.8 m per second squared divided by -18 in dude. And raise 2 -6. But here we have to use it at plus because we are using that uh magnitude only it injured invoice to minus six. And we get a value of 2.07 in didn't raise to three to find uh 2.07 into 10. raise too -3. Newton Per. Cool, it's a dangerous two plus three. Sorry, 10 ways to three. in September is 2, 3 Newton college. This is the magnitude of the electric field. Now, what will be the direction of the electric field? Now we have to remember that this is a negative judge this this up negative selves. So for a negative chance, the direction of the electric field will be opposite to the direction of their electric force. So earlier, we had already seen that the direction of the electric forces upward, so here its direction will be no one will. So The magnitude of the electric field is 2.07 injured and raised to three Newton per column and the direction Yes downwards. Thank you. Happy learning.

Hello everyone. This problem we are asked to find the electric field magnitude for the follow scenario. We have a five times into the minus six column charts or or five micro problem charge of mass, 0.500 g. So five times 10 to the -4 kg, moving at a height observed 1 60 m with initial velocity of two m per second east. And it's entering a uniform electric field of some sort and has a final uh speed 5m 3 seconds as it hits the ground. So we know that the acceleration due to gravity is 9.8 m/s squared. And we're looking for once again the electric field magnitude. So actually there is an assumption that you have to make here because you know, just telling you that the final speed is five m per seconds. There are many ways that you could engineer the electric field depending on its direction and its magnitude. Such that that would be true. So this five m per second squared is the combination is the sum of the X and Y components of the final velocity. And so those components can be pretty much arbitrary unless you make an assumption. So here, I'm going to assume that the electric field is horizontal and then we'll see where it gets us. So with the assumption we can start by finding the time it takes for the charge to hit the ground. And that of course, is unaffected by the electric fields. This is pointing in the horizontal direction. So it doesn't have it's not providing any force in the in the vertical direction for their for the charge. So we know that the initial uh y. Speed or velocity is near zero m per second squared meters per second. So we can just work out using the height, what the time it takes for it is charged to fall to the ground is. So we're just going to use our basic mathematic equation of the height or why is equal to you know, initial, my position plus or final position is equal to initial Y position plus initial y velocity. Or write speed times time plus a half times acceleration in the Y direction times T squared of course the first two terms of zero. And so we just know that it falls through a height of each under purely the action of the acceleration due to gravity. So the acceleration is cheap. I mean just have that tea were except to be the square root Of two times h over G. Which if you calculated that 0.35 seconds now then we can find what the final speed in the Y direction is or what the white component of the final velocity is. And then of course because the charge starts with a zero y velocity or high velocity component, it's just going to be, this final component is just going to be the velocity gained for the acceleration due to gravity. So it's just equal to G times. T can also work out to be 3.43 m per second. Okay, so then now we know what the why component or what the magnitude of the Y component of the final velocity is, but we still don't know what the X. Component of the final velocity is But we do know what their combination in terms of the magnitude of the final velocity is, which was if we knew it would be 5m/s. So what we can do is we can rearrange or we can say that you know, the final speed is equal to by pythagoras is through the square of its components, of its of its X and Y components. And rearrange this equation for the X component's magnitude, which we find to be, we have squared minus to b f y squared square rooted. And if you calculate that, that there is that to be 3.64 m per seconds. So this new speed, we started with an initial Y, our initial X speed of two m per second and we've moved on to gain 1.64 m per seconds. So then we can use this information to find out what the acceleration was because we also know the time it took to attain this new speed. So we just use that the final X speed is the initial explained. Plus the acceleration in the X direction times time. Re arrange that for the acceleration. And we get that the situation is 4.67 m/s squared. So this is the acceleration in the exploration, do you purely due to the electric field? Now, we also know that the acceleration in the X. Direction is the sum of the forces. Um So, net force in the X. Direction divided by the mass of the particle in the mass of the charge. Because this is what you do. The second law tells us. So, knowing that there's only the electric field acting on the particle. We know that the force that the electric field exerts is given by cute types. He recused charge and is the magnitude of the electric field as we're divided by m to get the acceleration. So we can finally rearrange this to find the electrical magnitude as the Mass of the charge, times the acceleration in the installation, divided by the charge of the charge. So divided by Q. Down here. And so if you put in all the values that we have calculated so far, you find that the electric field magnitude is 467 mutants for coolant, and this would be by the way, pointing to the right, so it would also be in the easterly direction. But we're only really asked to find the magnitude.

Hi in this given problem there is a uniform electric field in a horizontal direction from left. So right and as we know elective it always goes away from positive to negative. So this is positive and this is negative and in this electric field there is have all suspended with the help of a threat and the ball is given a negative charge. So under the influence of this electric field the ball is real estate from the vertical such that the thread makes an angle of going to disagree with the vertical. In this situation weight of the wall will be acting vertically downward but electrostatic force acting over it being negative, it will be attracted towards positivity. So this is electrostatic force which is given by charge. You time selected few. This will be the tension in that thread. So if this angle is 20 degree here this angle will also be 20 degree. So this tension will be resolved into two parts vertically upward. This will be T post 20 degree horizontally right word. This is p sign. 20 degrees charge given to the ball is negative. We have to find it now as the ball is in equilibrium. So there should be no net horizontal force and no net vertical force sitting on the ball. So it's creating the horizontal components He signed 20° is equal to Q into E. And he cause 20 degree vertically upwards. Could be equal to wait acting vertically downward. Now if you divide these two equations canceling tension we get and 20° is equal to U. E. Bye MG. We get an expression for that charge. Unknown charge which is U. Is equal to M. G. Times of 10 20 degree divided by electric field. So plugging in all the non values here, this charge will be given by 0.60 graham was the mass of the ball, so this is 0.60 into 10 days, par minus three kg, Multiplied by 9.8 acceleration to the gravity, multiplied by 1020° and divided by Electric field which has been given as 700 Newton park colon. So it comes out to be 3.1 in 2 10 for minus six colon. Or we can say this is 3.1 Micropal. Um and that is negative, which is the answer for this Given problem here. Thank you.


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