A 0.10 g honeybee acquires a charge of +23 pC while flying.b. Wha...

Question

A 0.10 g honeybee acquires a charge of +23 pC while flying.b. What electric field (strength and direction) would allow the bee to hang suspended in the air?

Accepted Answer

Welcome back. Everyone in this problem. A 0.15 kg sparrow accumulates a positive nano Coolum charge while perching on a power line, determine the electric field magnitude and the direction required for the sparrow to maintain its position in mid air without flapping its wings. For our answer choices. A is 1.22 times 10 to the eighth power ne newtons per colum. And that's going upward. B is 9.6 times 10 to the seventh newtons per colum downward. C is 1.06 times 10 to the eighth newtons per colum upward. And the D is 1.3 times 10 to the seventh new times per Coolum downward. Now, let's draw a diagram here to make sense of what's really going on. So we have a spiral that's perched on a power line. So let me put my power line on the, on the right here. OK. And wow, this is a great day. Not a lot of days you get to draw birds in physics. So this is pretty good. And I'm gonna try my best to draw a nice bird. OK? It's a bird. There's his beak and ok. Hey, that doesn't look too bad, does it? So he's, he's, it says he is in mid air without flapping its wings. Ok. So let's just imagine that these are its wings and for this bird, what do we know? Well, let's think about it. If it's positioned in mid air, that means it's in equilibrium. So the forces acting downward is equal to the forces acting upward. For the bird, its forces, its force acting downward would have been the force of gravity. And that means the force acting upward would have been, let me put it in a minute here, the force acting upward would have been the force of the electric field. So if we can figure out from those from the, the net forces acting at this point, then we should be able to find the magnitude and direction of the electric field. So first, let's find the sum of the forces in the y direction. So that means the sum of the forces in the y direction equals zero. So fe let me put it in the same color here. The force of the electric field minus the forces of gravity equals zero. Now, what do we know? Let's make a note of the information we have, we know that the mass of the bird is 0.15 kg. We also know that the charge the bird accumulates is 12 nano Coomes, a positive 12 nano Coomes which should be 12 times 10 to the negative nine Coomes. We also know that the gravity acting on the bird, we can assume it to be 9.8 m per second. But what we don't know is the magnitude of the electric field. Neither do we know the direction of the electric field. So since we know that these forces are acting on the bird, how can we relate these here? First of all, let's think about the electric field. What do we know about the electric field? Well, recall that for an electric field, its magnitude would be equal to the force of the electric field divided by the charge per unit. And when we think about it, we have the charge per unit. Therefore, the force of the electric field would be equal to the charge multiplied by the electric field strength. Likewise, for the forces of gravity acting, that force would be equal to the mass of the bird, the bird multiplied by its acceleration due to gravity. So since we have this information, we should be able to substitute it to solve for the ele the magnitude of the electric field. So here we know that the force of the electric field equals the force of gravity since it's in equilibrium. So by substituting these values, then that tells us the charge multiplied. I think, I think I'm going to use the same color here and in the charge multiplied by the magnitude of the electric field equals the mass multiplied by the acceleration of the bird. The charge that the bird accumulates recall that we said it was 12 times to the negative nine Kums. So we're multiplying that by the magnitude of the electric field and that equals the mass of the bird. 0.15 kg multiplied by the acceleration 9.8 m per second squared. So that tells us then that the magnitude of the electric field would be equal to 0.15 kg multiplied by 9.8 m per second squared, all divided by 12 times 10 to the negative nine Coomes. And when we put that into our calculator, when we work that out, we get the magnitude of the electric field to be equal to 1.22 times 10 to the eighth power newtons per Coolum. So this would be the field's magnitude. Now that we have its magnitude, we just need to figure out its direction. Let's go back to our diagram to help us. If we think about it, the electric field required for the spiral to maintain its position in mid air must be directed upward because the spiral has a positive charge. It has that positive charge of 12 nano Coomes. OK. Imagine that this is the spiral here with its charge. So it must be directed upward so that the electric force on the sprow acts in the upward direction, counteracting the gravitational force. If you recall the positive charges would be repelling outwards. So that means the force must be acting upwards. So that tells us then that our answer would have to be a, the magnitude of the field is 1.22 times 10 to the eight newtons per column and it is acting upward. Thanks a lot for watching everyone. I hope this video helped.

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A 0.10 g honeybee acquires a charge of +23 pC while flying.b. What electric field (strength and direction) would allow the bee to hang suspended in the air?

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