FIGURE P23.44 shows a thin rod of length L with total charge Q...

Question

FIGURE P23.44 shows a thin rod of length L with total charge Q. Evaluate E at r=3.0 cm if L=5.0 cm and Q=3.0 nC.

Diagram of a charged rod of length L with charge Q, showing point P at distance r from the rod.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with electric fields. In this problem, we have a thin bar of length of eight centimeters and it is uniformly charged with a total charge of negative five nano cooling. The question was to determine the magnitude of the electric field at point A which is located at a distance of four centimeters from the top face of the bar. Below the question, we're given a diagram of what was described in the problem. We're also given four choices as are possible answers. Choice A is 1.4 multiplied by 10 to the two newtons per Coolum B is 2.5 multiplied by 10 to the three newtons per column. Choice C is 9.4 multiplied by 10 to the three newtons per column. Choice D is 5.5 multiplied by 10 to 4 newtons per column. So we're gonna start off by labeling a couple of these quantities. Let's label the total length of the bar here, the eight centimeters. Let's label that as L and let's label the distance from the top face of the bar to point A that four centimeters. We're gonna label that as D we're also going to need to have a coordinate system later on. So let's go ahead and make the very top of the bar our origin. So that means that my Y equal to zero point will be there. Let's have the positive Y direction going downward. So we're gonna start off by calculating the electric field. And here, um since the charge is negative, I know at point A that the electric field vector electric field is going to be pointing downwards towards the bar. That's because the charge is negative for the bar. And that's why we set the positive Y direction to go downward. That way our electric field is going in the positive Y direction we're interested in is the mag uh magnitude of our electric field. We just really need to calculate the Y component. So we call your definition for an electric field due to a continuous charge distribution. So here I'll have ey, that's gonna be equal to 1/4 pi epsilon knot multiplied by the integral of BQ divided by R squared. So the trick to these problems is figuring out what is the DQ? And how do we represent R in terms of the coordinate that um we're using, let's start off with the DQ here, I'm looking for a small piece of charge. And so that's just gonna be whatever the linear charge density is of that bar multiplied by some small link which is gonna be Dy. Now, the linear charge density Lamda, it's going to be equal to the total charge, which is um given as capital Q divided by the total length L. And so we'll have for DQ that's gonna be equal to capital Q divided by L multiplied by DY. That means we're gonna be integrating over the Y coordinate. The other quantity we need to look at is R and R is the distance from the point of interest in this case, point A to whatever piece of charge I'm looking at. So that's actually quite easy. Given the coordinate system that we set up, we're always gonna be a distance D away from the charge, then we'll have plus whatever Y coordinate we're at. So if we start off at zero, then it'll just be D. And when we're all the way at the other end, we'll have D plus L. And so our, our vector or our distance here is gonna be equal to D plus Y. So we now plug those back into our expression for the electric field and then begin to do the integration. We will have ey is equal to one over or pi epsilon knot. It's gonna multiply the interval. And for DQ, we'll have the quantity of capital Q divided by L and then we'll have by, and then that's gonna get divided by R squared, which in this case would be plus why that quantity is squared? Now, we need to look at our limits of integration. That's gonna actually go from zero to L. Now that we have the um integral set up, we can actually look at integrating this. So I can actually pull out some additional constants. So my EY is one equal, I still have the 1/4 pi absolutely not out. I can factor out the capital Q divided by L because those are both constants. And what I'm going to be left with is the integral of dy divided by D plus Y quantity squared. That's actually pretty easy to integrate. When I integrate that I get a negative one divided by B plus Y, that's gonna be evaluated from zero to L. Now, we just need to plug in our limits of integration. We'll have EY is equal to 1/4 pi absolutely not that gets multiplied by you divided by L. And then that's going to get multiplied by the quantity for the upper limit, I'll have negative one divided by the plus L. Then for the lower limit, I'll have plus one divided by D. At this point, I can plug in values for all my quantities. I'll have ey equal to the quantity of 1/4 pi epsilon knot. That is a constant, that is nine times 10 to the nine Newton meters squared per column squared. Then we'll have the total charge. In this case is five nano columns. I need to convert that column So it'll be five multiplied by 10 to the negative nine columns. We don't need to worry about the negative sign here. We've already taken care of the negative sign with the direction of our electric field. And I'll need to divide that charge by the length which is the eight centimeters. That's gonna be the 0.08 m. So we need to convert centimeters to meters by dividing by 100. And then that multiplies the quantity of negative one divided by you will have D which is the four centimeters which is 0.04 m plus the L is 0.08 m. And then we'll have less the one divided by D that's gonna be one divided by 0.04 m. So we can calculate um that value on the right hand side of that equation. So we'll have ey as equal to 9.4 multiplied by 10 to the three newtons for cool on here. Just kept two sign, two significant figure and the units do come out to be newtons per column. And so you notice that one of the columns cancels out with the columns from charge. And then um you have meter squared on the denominator there. Um One from the L and then one from the integration, you were left with units of nude for Coolum, which is the units for electric field. So that's gonna be our final answer. And if we look at what answer choice that corresponds to, that corresponds to answer C So just a quick little recap, um anytime that you're um looking at dealing with a continuous charge distribution for finding the electric field, the important quantity that you need to find is your DQ. And you typically need to write that as a charge per unit length or charge per unit area and multiplied by um a distance or an area and then use that as your limits of integration. So I hope that this has been useful and I'll see you in the next video.

FIGURE P23.44 shows a thin rod of length L with total charge Q. Evaluate E at r=3.0 cm if L=5.0 cm and Q=3.0 nC.

Key Concepts

Electric Field (E)

Linear Charge Density (λ)

Coulomb's Law

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