A very long insulating cylinder of charge of radius 2.50 cm carries a uniform linear density of 15.0 nC/m. If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads 175 V?

Video transcript

Welcome back, everyone. We are making observations about an electric fence and we are told that it's constructed with insulating rods that have a radius of two cm. You're told that each Rod holds a homogeneous linear charged density of 10 nano Kums per meter. And we wish to achieve a potential between some point off the rod and the rod itself of 100 and 50 volts. We are tasked with finding what is the location of that point P measured from the surface of the rod. Let's think about this for a second. Say we have the center of our rod and it has our radius right here of two centimeters. Here is our point P but say we made a second circle around R R center point. Well, I would just deem this point our R two or our second radius. Therefore, R P is just R two minus R one and that will come into play a little bit later. All right. So before diving into this problem here, let's read out our answer choices. We have that the potential difference of 1 50 volts is achieved at a 500.4 point 61 centimeters away. From the surface. Answer. Choice B 2.61 centimeters away from the surface. Answer choice C 6.61 centimeters away from the surface or answer choice D two centimeters away from the surface. All right. So how are we gonna do this? Well, we know that the potential difference which we have right here is just equal to our linear charge density divided by two pi times epsilon, not times the natural log of our outer radius minus our or divided by our inner inner radius. What I'm gonna do here is I'm actually gonna multiply both sides by two pi epsilon knot over our linear charge density here. And what you'll see happen on the right hand side is that this term and this term cancel each other out. What we are then left with is that the natural log of our outer radius divided by our inner radius is equal to this formula right here. So let's go ahead and plug in our values that we know potential difference is 150 times two pi times our constant epsilon knot, which is 8.85 times 10 to the negative all divided by our linear charge density of 10 nano Kums per meter. But we want Kums per meter. So I'm gonna multiply that by 10 to the negative ninth. And what this gives us in our calculator is 100.834. Now, there's a property of natural logs that states, if we raise E, which is the base of a natural log to the power of our solution, we can set that equal to what's ever inside the natural log. So let's go ahead and do that. We have the R two over R one is equal to E to the power of 10.834. And what I'm gonna do is I'm gonna multiply both sides by R one. And you'll see on the left hand side, R one cancels out and we finally have a formula for our desired R two. Let's go ahead and figure out what that is. We have E to the power of 20.834 times two giving us 4.61 centimeters. Now we can go ahead and calculate our point P here by this formula. We have our R two is 4.61 centimeters minus our inner radius of two, giving us a final answer of 2.61 centimeters to achieve a potential difference of 150 volts away from the surface of the rod which corresponds to our final answer. Choice of B Thank you all so much for watching. I hope this video helped. We will see you all in the next one.