A very long, straight wire has charge per unit length 3.20 * 10^-10 C/m. At what distance from the wire is the electricfield magnitude equal to 2.50 N/C?

Video transcript

Hey, everyone. Welcome back in this problem. We have an extremely long linear bar charged by friction to 1.6 Nanako gallons per meter. The electric field has a magnitude of 6.4 newtons per cologne at a point are measured from the bar and were asked to determine the value of our. Alright, so let's write out what we're given. Now, we're given the charge 1.6 Nana columns per meter. Now this is a linear bar. So this is going to be a linear charge density. And so Lambda, the linear charge density is equal to 1. nana Coolum per meter. Well, let's convert this into our standard unit of just columns per meter. Okay to do that to get from Nano to just cool. Um we're gonna multiply by 10 to the 9 -9. So we get 1.6 times 10 to the negative nine times per meter. And we also know that the magnitude of our electric field denoted by E is equal to 6.4 newtons per column. All right. Now we're told that this is an extremely long linear bar and because its extremely long recall that we can use the following equation E the magnitude of the electric field Is equal to Lambda. The linear charge density divided by two pi absolutely not. Are, they were absolutely not as an electric constant and R is the distance. Alright, so we know e we know lambda, we were given those in the problem. We've written them out above and we know epsilon not the electric constant. The only thing we don't know is this value of R, which is what we're looking for that distance from the point to the bar. So let's just fill in what we know and all we have left to do is solve. So we have 6.4 Newtons per column is equal to 1.6 times 10 to the -9 columns per meter Divided by two pi the electric constant is given by 8.8, 5 times 10 To the negative cool M squared per Newton meter squared times big R because that distance little R we're told is going to be called R and that's what we're looking for is valuable. All right. So we can multiply everything from this right hand denominator up to the left hand side, we're gonna have our times two pi Times 8.8, 5 times 10 to the -12 Coolum squared per Newton meter square Times 6.4 newtons per Coolum. All of that is equal to 1.6 times 10 to the - columns per meter. And again, we're looking for our, so let's go ahead and isolate for our by dividing by all of this other stuff we're gonna have that R is equal to 1.6 times 10 to the negative nine columns per meter divided by two pi times 8.85 times 10 to the negative 12 Coolum squared per Newton meter squared times 6.4 newtons per cool. Alright, let's start with the units here in the denominator. We have cool um squared divided by Newton meter squared times newtons per cool. Um So the unit of Newton is going to divide out and one of the cool um terms will divide out. So we're left with cool um per meter squared. Now we have colon per meter divided by Colin per meter squared. That's gonna leave us with a unit of meter. Okay? And we know that our is a distance. So that makes sense the units work out. And if we do this on our calculator, we're gonna get a value for our approximately 4.496 m. Okay. The answer choices have two significant digits. So if we round to two significant digits, we get 4.5 m and that is the value of our we were looking for, we go back up to our answer choices. We see that that corresponds with answer choice B 4. m. Thanks everyone for watching. I hope this video helped see you in the next one.