A very large, horizontal, nonconducting sheet of charge has unifo...

Question

A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area σ = 5.00×10−6 C/m2. (a) A small sphere of mass m = 8.00×10−6 kg and charge q is placed 3.00 cm above the sheet of charge and then released from rest. (a) If the sphere is to remain motionless when it is released, what must be the value of q?

Accepted Answer

Hey everyone. So this problem is working with electric fields. Let's see what they are asking us. We know that there is a tiny drop of oil with a given mass given charge and it is released from rest above an infinite non conducting plate. The plate has a uniform charge density that's also given and we need to determine that charge if the oil drop remains stationary when it is released. So that last piece is kind of the hint here where the next step is to, or the first step is to draw a free body diagram. So we know that the force acting in an upwards direction on the drop of oil is the electric force. And then the force in the downward direction is just the weight because it is released from rest and remains stationary. That means that the sum of the forces equals zero. So we need to recall that our electric force is given by Q. E. And weight is just mass times gravity. and so the sum of our forces equal zero means in upwards direction, we have our electric force and that's equal to our weight. So we have Q. E equals M G. We are asked to determine the charge Q. So we'll solve this for Q. So that's M. G over E. So let's look at each of these terms. M is given to us in the problem G is a constant and then we have our E. Which is our electric field latitude. Now we don't have that, but we do have a uniform charge density. So let's recall that the equation for electric field magnitude in terms of uniform charge density, sigma is sigma equals equals sigma over two epsilon. Not epsilon. Naught is our electric constant. So we can plug that in to our equation for Q. And we have Q equals M G. Over sigma two epsilon not. And so from there it's just a plug in chuck. Mass was given to us in the problem and that's 1.5 times 10. So it was given to us as 10 to the minus five g. I'm going to rewrite that as 10 to the minus eight kg to get that into a standard unit gravity we can recall is 9.8 m per second squared two and then epsilon Not our electric constant. We can recall that. That is 8.854 times 10 to the minus 12. Cool m squared per meter squared And all of that over over sigma charge density which was given to us in the problem as 10 to the -5. Fulham's for me or swear, plug that into our calculator and come up with A Charge of 2.6 times 10 to the -13 times. So that is our answer here And we go up to our potential answers and that is answer C2. times 10 to the -13 pull ups. All I have for this problem. See you in the next video

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