You measure an electric field of 1.25×106 N/C at a distance of 0.150 m from a point charge. There is no other source of electric field in the region other than this point charge. (a) What is the electric flux through the surface of a sphere that has this charge at its center and that has radius 0.150 m? (b) What is the magnitude of this charge?

Video transcript

Everyone. In this problem, we are asked to first find the electric flux or five E two, a caution sphere of radius R centered at the location of the object. And then second find the charge of the object itself. Where in this particular example, we have a student measuring the strength of an electric field created by a small positively charged object at a distance of R equals 0.2 m from it And comes up with a value of two times 10 to the power of six Newton per cool. Um so I am just going to start us off with just listing off things that we actually know. So we know that the R is 0.2 m. It is already an S I and E which is the electric field, which is the value that is being measured here is equals 22 times 10 to the power of six newton per cool on. So next, you want to recall that To solve for the electric flux or five E 5 E can be calculated by multiplying E&A. So the electric flux through an area is defined as the product of the component of the electric field perpendicular to the area times the area. So in this case, we have the A to B because we have the caution of a sphere, so the area of the sphere is going to Ashley V four pi R squared. So that is just the surface area of the entire sphere. So we actually have all the value needed. We have to E N D R. So we can solve 45 E. So five is then going to be two times 10 to the power of six Newton per Kula with four pi the R is going to be 0.2 m squared squared. So far E will then corresponds to one times 10 to the power of six Newton times meter squared per cool launch just like. So, so that is part one of our problem. So next, we want to move on to the second part which is asking for the charge of the object itself. In this case, we want to apply Gaza's law Where the five e can be calculated by defining the charge by the free permitted city of space or epsilon. Not. So we asked the charge here and the epsilon not is just going to be a constant which is going to be 8. times 10 to the power of minus 12 Coolum squared over Newton times meter squared. So because the epsilon is constant, we can actually solve for Q which is just five E multiplied by absolute or not. The fight is going to be the same as what we have found previously in the first part and multiplying that with our constant here, Peter squared. And therefore, we can actually determine that the charge is then going to be 8.85 times 10 to the power of -6 cool. Um we can cross out all the units and leave out one cool um here. So the charge is going to be this and that will be all for the second part. So we found that the electric flux by E corresponds to one times 10 to the power of six Newton meter square per cool. Um And the Q corresponds to 8.85 times 10 to the power of minus six cool um which will build corresponds to option C here and the O B O for this particular problem. If you guys have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that'll be all for this video. Thank you.