A ring of radius R has total charge Q. At what distance along ...

Question

A ring of radius R has total charge Q. At what distance along the z-axis is the electric field strength a maximum?

Accepted Answer

Everyone in this problem. A ring with diameter D has a total charge Q distributed uniformly along its circumference. The ring is centered at the origin of the XZ plane. And we're asked to determine the distance from the origin along the Y axis at which the magnitude of the electric field is maximum. We're given four answer choices here. Option A plus or minus D divided by two. Option B plus or minus D divided by three. Option C plus or minus D divided by two, route two and option D plus or minus D divided by two, route three. So let's start and let's think about what we're looking for. OK. We wanna find a distance from the origin. OK. Where the electric field is maximum? Well, what is the electric field in this case when we have a ring charge? Ok. Well, recall that the electric field at a point along the Y axis due to a ring in the XZ plane like we have that's centered at the origin is gonna be given by the phone. We're gonna say that er Y is equal to one divided by four pi epsilon not multiplied by Y multiplied by Q divided by Y squared R squared to the exponent three HS. Now, looking at this equation, we know epsilon, not that constant Y is going to be that position along the y axis that we're looking for. He was going to be our charge, which we know R is the radius. Hey, we're told the diameter so we can find the radius just by dividing by two. And so let's go ahead and substitute everything we know into this equation. So we're gonna just simplify this and write it as E and in this case, E is gonna be one divided by four pi absolutely not multiplied by A Y multiplied by little Q divided by Y squared plus D divided by two all squared and all to the exponent the HS Yeah, this is the electric field out of position one. We want to maximize the electric field. What do we do when we want to maximize something? Well, we're gonna take the derivative, we're gonna set it equal to zero and we're gonna find the corresponding Y position. OK. So we are going to find the derivative de ID Y, OK. The derivative of the electric field with respect to the position. And we have one divided by four pi epsilon knot. That's all a constant open front. So we can just leave it open front. Now we're gonna take the derivative of the numerator of our function. OK. We have Y multiplied by Q the derivative is just going to be Q. Now, we're gonna multiply by the denominator. OK. We're using our quotient rule Y squared plus D divided by two squared all to the exponent three halves. And we're gonna subtract the numerator, Y multiplied by Q multiplied by the derivative of the denominator. Now, the derivative of the denominator is gonna be a few steps. So we need to use our exponent rule. We're gonna bring that power down three halves. We're gonna multiply by the original Y squared plus D divided by two squared all to the exponent of three halves minus one, which is one half. And then chain roll tells us we need to take the derivative of what is inside that function as well. The derivative of Y squared plus D divided by two squared is just going to be two Y. OK. So that's the numerator of our big messy derivative. And the denominator is just gonna be the original denominator squared. So we have Y squared plus D divided by two squared. And all of this is gonna be to the exponent three. It originally it was to the three halves when we square that we're left with an exponent of three, all right. So this looks really, really messy, but we're gonna go ahead and simplify, OK. So let's start by simplifying these um kind of messy terms. There's this Y squared plus D divided by two squared term. OK. It shows up in every single part of this equation or this function, we have an exponent of three halves, an exponent of half an exponent of three. OK. So the lowest exponent is one half. So we're gonna take a factor of one half out, divide it out and see what's left. OK. So we have one divided by, or pi epsilon in the numerator. If we take out a factor of one half. And right now, we have three halves, we're gonna be left with just an exponent of one. So we get Q multiplied by Y squared D divided by two squared minus. We're doing the same thing here. Y squared plus D divided by two squared to the one half. We divide out that exponent one half and this entire factor to the exponent one half. And what we're left with is why multiplied by Q multiplied by three halves multiplied by two, Y. Well, the three halves multiplied by two is gonna give us three. Then we have Y multiplied by Y multiplied by Q. So we're gonna have three Y squared multiplied by Q in the denominator. Remember we're dividing this term out from the numerator and denominator. So we have to do the same in the denominator. We originally had a fact of an exponent, sorry A three, we're dividing by an exponent of a half. So we're gonna subtract those exponents and we get Y squared plus D divided by two squared all to the exponent by paths. Mm All right. So we're getting there, we've simplified a little bit. It looks a little better. Let's keep going. And what we're gonna do is we're gonna expand in the numerator. So we get de divided by DY is equal to one, divided by four pi epsilon not and multiplied by, well, we have Q multiplied by Y squared and then we have negative three Y squared. Q OK. So one minus three, that's gonna give us negative two. And then we have Q multiplied by D divided by two squared. OK. D divided by two squared gives us D squared divided by four. And so we end up with Q multiplied by D squared divided by four minus two Y squared Q. And this is all divided by that same denominator. Nothing to simplify there Y squared plus. And actually, we can, we can simplify our D divided by two squared. So we'll do that. So we have D squared divided by four all to the exponent by paths. All right. Now, remember what we're trying to do. Don't get hung up in all of these simplifying steps. We are trying to find the magnitude of the electric field that maximizes this. OK. Sorry, we're trying to find a maximum electric field. We're trying to find the position that maximizes the electric field. OK. So we've taken the derivative. Now we need to set it equal to zero. And we said we were gonna do that from the start. So we're gonna set this thing equal to 01, divided by four pi epsilon multiplied by QD squared, divided by four minus two Y squared, Q all divided by Y squared velocity squared divided by four to the five paths. OK. Our four pi epsilon, not that is a constant. So we can multiply both sides by four pi epsilon. Not no problem. That's gonna go away. Our denominator Y squared plus D squared divided by four to the exponent five halves. OK. D is nonzero. So this denominator is never going to be zero. OK. So we have no issues here. That will never be zero will never be dividing by zero. No issues. OK? So in order for this equation on the right hand side to be equal to zero, the numerator must be equal to zero. That's the only way that this is gonna work. And so what we have is that zero must be equal to Q multiplied by D squared divided by four minus two Y squared Q OK. Again, that numerator must be equal to zero. Now, we want to solve for why we wanna find the position that maximizes this electric field. So we're gonna move that to the left hand side by adding, we have two Y squared, Q is equal to Q multiplied by D squared divided by four. OK. The Q value is going to divide out. That's just a constant. We could have actually taken it out at the very beginning of this and left it with this one divided by four pi ups or not constant. Um Either way you are going to get the same answer. We divided it out here. Now we're gonna divide both sides by two. We get Y squared is equal to D squared divided by eight. And finally, we take the square root, we get that Y is gonna be plus or minus the square root of D squared divided by eight. OK. We can split this square root into the numerator and denominator. The square root of D squared is just gonna be D in a square root of eight. We can write as to route two. And so we get our final answer here that the position Y along the Y axis that is going to maximize our electric field for this ring charge is going to be plus or minus D divided by two root two. If we compare this to our answer choices, we can see that this corresponds with answer choice. C thanks everyone for watching. I hope this video helped see you in the next one.

A ring of radius R has total charge Q. At what distance along the z-axis is the electric field strength a maximum?

Key Concepts

Electric Field

Symmetry in Charge Distribution

Maximization Techniques

Watch next