CALC An electric dipole at the origin consists of two charges ±q ...

Question

CALC An electric dipole at the origin consists of two charges ±q spaced distance s apart along the y-axis.a. Find an expression for the potential V(x, y) at an arbitrary point in the xy-plane. Your answer will be in terms of q, s, x, and y.

Accepted Answer

Hello everyone. Let's go through this practice problem. Find the expression for the electric potential V of X sub A and Y sub A at point A in the xy plane where an electric dipole is positioned at the origin with two charges positive and multiplied by E and negative and multiplied by E separated by a distance of D along the X axis. Your response should be in terms of N multiplied by edX sub A and Y sub A. This problem isn't a multiple choice problem but it is a problem that while simple still requires a bit of work because this is the sort of thing where it's extremely important to first draw a diagram to make sure you understand visually what's going on. So I'm gonna drive crude sketch myself. Here is an xy plane. So there is X and there is Y. And what the problem is describing is that at around the origin, there is an electric dipole with both a positive and a negative end. So I'll say the negative end is right about here and there's a positive end over here. The problem says that there is a distance D between them which means that each point, each pole is positioned at a distance D divided by two away from the origin. And there's a third point a at some other arbitrary position on the xy plane, we aren't told where it is. It doesn't matter, we can pick any place. I'm just gonna draw it right here. Call point A with the position X sub A why? And the problem is simply asking us to find an expression for the electric potential at point A. So what we can see from the diagram is that this point has an electric potential from both the negative and positive points. So we need to account for both factors. The first thing I'm going to do is represent the distance visually between point A and the two different point charges. So I'm gonna call this distance our sub negative because this is the, the distance between point A and the negative pole. Then I have another point leading to the positive poll our sub plus. Yeah. So recall that the formula for electric potential V is equal to the charge of the point that is the point producing the potential divided by four multiplied by pi multiplied by epsilon knot multiplied by the distance between the charge creating the potential and the point we're analyzing. So in this case that R could represent either the R positive or the R minus since there are two different components. So what I'm going to do is represent the two different potentials that the point A is going to be experiencing from both of the poles. First, the potential from the positive pole V sub plus. So referring back to the formula, the charge of that particle goes in the numerator. And the problem tells us that each charge can be represented as N multiplied by E presumably the number of electrons or the number of elementary charges that make up the full charge of those point charges. So for the positive case, it's just going to be N multiplied by E no need to include a negative sign or anything like that. And this is being divided by four high epsilon knots. But it's also being divided by R sub positives the distance from point A to the positive pole. And this is where things can start to get a little bit mathematically messy because we're not given that distance, we're not allowed to just use the variable R sub plus. So we'll have to use some geometry such as the Pythagorean theorem in order to get an expression for that distance in terms of D NX and Y. So if you look back at our diagram, we can try and break up these RS into their components because point A is of course located at a position X sub A and Y sub A applying the Pythagorean theorem with the positive component. For example, we can see that the X component is going to be equal to the D divided by two part right here plus the additional distance, the additional horizontal distance between point A and the Y axis and the vertical component is of course just the Y sub A, just the vertical positioning of point A. So what I'm going to do is use the Pythagorean theorem to represent this. So I'm going to write our sub plus as the square root of X sub A plus D divided by two squared plus Y sub A squared. Now that's under a square root. And so all of this represents the electric potential from the positive pole of the dipole. Now let's do the exact same process before the negative potential that it's experiencing from the negative dipole. So once again, the magnitude of the charge is ne but because it's negative, it's going to be a negative ne the rest of the process is the same four multiplied by five, multiplied by epsilon knot. And then we're once again going to have to do a Pythagorean theorem type thing for the R sub minus for the distance between point A and the negative dipole of the negative pole. This time, the distance that we're using for the horizontal component is going to be the horizontal positioning minus the D divided by two because that's the, that's being excluded from the horizontal position in this triangle here. So instead it's going to be the square root of X sub A minus D divided by two squared plus waa A squared and all of that is under the square root. So now we have determined expressions for the individual potentials. But the problem is asking us for a formula representing the complete potential. And for, for us, this isn't very hard to do because electric potential is a scaler and not a vector. We don't need to worry about fancy things like adding up components or, or doing tricky math. All we have to do to find the total potential is simply add the positive and negative components together. And that's all we gotta do. So I'm gonna switch my black ink and I'm going to simply add these together. So, so V positive, that's and E divided by four pi epsilon not multiplied by the square root of X sub A plus D divided by two squared plus Y sub A squared and then plus the negative component which has a minus in it. So I'm gonna say minus NE divided by four pi epsilon knot multiplied by the square roots of X sub A minus D divided by two squared plus Y sub A squared. So that's kind of messy, but we can simplify it slightly by factoring out the NE and the four pi epsilon knot. So instead I'm going to write this as NE divided by four pi epsilon knot then multiply in big brackets, one divided by the square root of X sub A plus D divided by two squared plus Y sub A squared minus a big denominator. Again, one divided by the square root of X sub A minus D divided by two squared plus Y sub A squared, make that a little clear squared. And that's it. That's about as much as we can simplify it. So that means that this is our answer to the problem. We don't have any multiple choice options to, to compare this to, but this is our answer to the problem and that is it for this problem. I hope this video helped you out if you need more help and you think you want more practice. Please check out some of our other tutoring videos which will give you more experience with these types of problems, but that's all for now. And I hope you all have a lovely day. Bye bye.

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CALC An electric dipole at the origin consists of two charges ±q spaced distance s apart along the y-axis.a. Find an expression for the potential V(x, y) at an arbitrary point in the xy-plane. Your answer will be in terms of q, s, x, and y.

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