At a certain distance from a point charge, the poten-tial and electric-field magnitude due to that charge are 4.98 V and 16.2 V/m, respectively. (Take V = 0 at infinity.) (b) What is the magnitude of the charge?

Video transcript

Welcome back everybody. We are taking a look at two particles here. One was already previously placed at point B. And we are told that a group of students places a charged particle at point A. Now this makes the magnitude of the electric field at point B. 84.6. And we are told that the potential there is 50 volts and we are tasked with finding what is the magnitude of charge on point A. Well, we know that the magnitude of the electric field is equal to coolants, constant times the absolute the magnitude of the charge divided by R squared. And we know that the potential at a point is equal to Cullen's constant times Q. All over are. Now, here's the thing in both of these equations, we have two unknowns, both Q and R. So what I'm gonna do is I'm going to to manipulate these equations to where we can solve for one variable and then the other. So here's what I'll do. I will divide the potential equation by the magnetic field equation. We are going to get K. Q over R times the reciprocal of this equation right here. So we have R squared over K. Q. Now, these terms are going to cancel out and this term is going to cancel out leaving us with the fact that our radius R. R. R is just the potential divided by the electric field. So let's go ahead and solve for that first that R is equal to V over E. Which for us is equal to 50 divided by 84.6, Plugging this into our calculator. We get that our radius here is 0.591. Or our distance between the two is 0.591. Right now with that in mind we can just use one of the original equations. I'm going to use our potential equations here and we can go ahead and solve for R. Q. In order to isolate Q. I'm actually gonna multiply by our over K. On both sides and you'll see That these terms cancel out leaving us with that. Our charge is equal to our potential times our distance between them. All over columns constant. Let's go ahead and plug those values in. We have our potential of 50V Our distance between them of 0.591. All divided by columns constant of 9.0 times 10 to the ninth. When you plug this into our calculator we get 3.28 times 10 to the negative ninth columns or 3.28 nano columns corresponding to our answer choice of C. Thank you all so much for watching. Hope this video helped. We will see you all in the next one