Two positive point charges q are placed on the x-axis, one at x =...

Question

Two positive point charges q are placed on the x-axis, one at x = a and one at x = -a. (a) Find the magnitude and direction of the electric field at x = 0.

Accepted Answer

Hey, everyone in this problem, we're told that two small spheres separated by a distance to D lie on the X axis. Each sphere carries a positive charge. Q, the origin of the X axis lies at the midpoint of the two charges. And were asked to find the electric field magnitude and direction at the origin. We're told to treat the spheres like point charges. Alright. So let's just draw a little diagram of what's going on here. So we have our X axis here, then we have our first point charge, okay, positive Q. And we know that the origin is the midpoint. These lie on the X axis source. Second charge is going to be over here in the negative X axis. And that's also a charge of plus que. Now this one is going to be located at negative D and the one in the positive X axis is located at positive D. Okay, because the total distance between them We're told is two D which tells us that the distance from each point to the origin is going to be D. Alright. So there's a little picture representing what we have going on here. And we're asked to find the electric field magnitude and direction at the origin. Okay. So right here at the origin, no recall that the magnitude of the electric field is given by the following E is equal to Q divided by four pi epsilon, not R squared where Q is the charge epsilon naught is the electric constant and R is the distance from the charge to the point of interest. Alright, so let's start by looking at the charge that's located at X equals positive de. So The electric field magnitude E is going to be given by Q divided by four pi absolutely not times positively squared, which gives us an electric field magnitude of Q divided by four pi epsilon not D squared. Alright. And again, this is for the charge at X equals positive de. Now we need to figure out what direction this is going okay. This is the magnitude of the electric field at the origin due to this point. But what's the direction? All right. Well, let's think about this in terms of our diagram, this X equals plus D charge, this is a positive charge. So the electric field is going to point away from the positive charge. Okay. So if we're looking at the electric charge along the X axis, it's going to be pointing away from that charge. And so at the origin that electric field that is caused by that charge is going to be pointing to the left or in the negative X direction. And so the vector representing the electric field at the origin due to the charge up plus D is going to be Q over four pi epsilon not D squared times negative I okay. So it's that magnitude we found times the direction which is in the negative extraction. So the negative I direction, alright, let's do the same for the X equals negative D case. Okay. So the charge located at X equals negative D, when we look at the magnitude of the electric field, we have Q okay. The charge divided by four pi epsilon not now this time we have negative D squared. Okay because we're going from the origin to the left in order to get to this charge. So that's the negative X direction to have negative D squared. And this is gonna give us a magnitude of the charge of Q divided by four pi upsilon not D squared. And so the magnitude is the same for both of these charges, which makes sense because they have, they are the same charge and they are located at the same distance away when we talk about the direction of this one. Okay, let's go back to our diagram and do the same thing we did before. Okay, we're going to have the electric field pointing away from the positive church. So it's going to be pointing like this towards the origin where we're measuring our electric field. And so when this electric field caused by that charge gets to the origin, it's going to be pointing in the, to the right or in the positive extraction. And so when we write the vector representing the electric field at the origin due to the charge at X equals negative D, we have the magnitude Q over four pi epsilon naught D squared times of direction which is the positive X direction. So positive I now we have information about the electric field caused by the charge and the positive X axis, the one from the negative X axis. Now we need to figure out that total electric field caused by both of those charges at the origin. And so the electric field at the origin is just going to be equal to the sum of the two. Okay. It's going to be the sum of the electric field from the X equals positive D charge. And from the X equals negative D charge. If we add these two together, we have Q divided by four pi epsilon, not D squared times negative I plus Q divided by four pi epsilon, not D squared times positive I and this is going to equal zero and that is going to be the magnitude of our electric field there. Now this makes sense because let's look back at her picture, we have two charges, they have the same sign, same magnitude we're measuring the electric field at the midpoint of both of them. Okay. So the electric field has the same magnitude, it's in the opposite direction. And so that's going to cancel each other out. And so we have a, an electric field of zero at the origin. Alright. If we look at our answer choices, we see that this corresponds with answer choice B the electric field is zero. Thanks everyone for watching. I hope this video helped see you in the next one.

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Two positive point charges q are placed on the x-axis, one at x = a and one at x = -a. (a) Find the magnitude and direction of the electric field at x = 0.

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