Calculate the electric potential due to a dipole whose dipole moment is $4.2 \times 10^{-30} \textrm{ C}\cdot\textrm{m}$ at a point $2.4 \times 10^{-9} \textrm{ m}$ away if this point is

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The electric potential at this point here on the axis connecting the two ends of the dipole is going to be the potential due to the positive charge plus the potential due to the negative charge, the negative charge resulting in a minus here because its charge is going to be of magnitude Q with a minus because it's negative and the potential due to the negative charge is k Q over the distance r separating the point from the positive charge plus an additional l distance between the two charges. And basically I'm deriving the potential of a dipole formula here. And so we have k Q factored out and that's going to be one over r minus one over r plus l and then multiply this by r plus l on the top and on the bottom. And then multiply this fraction by r over r. And what you get is r plus l times one here and then minus the r times one there. And then on the bottom we have a common factor, r times r plus l and the top, the r minus r makes just l on the top and in the bottom. We're gonna call it r squared because the idea is that this l separation between the charges is insignificant and negligible in comparison to this distance r from the positive charge to the point of interest. So the bottom just becomes r squared. Because this is essentially zero. Now dipole moment is, is defined as the charge on each end of the dipole times the distance separating them l and so we can rewrite this formula as k times P instead of Q l over r squared and we're given the dipole moment is 4.2 times ten to the minus 30 coulomb metres. Multiply that by columb’s constant and divide by the distance between the charges, of 2.4 times ten to the minus nine meters squared. And that gives 6.6 times ten to the minus three volts is the potential at this point here. Now when the point of interest is 45 degrees closer, with respect to the line connecting the dipoles or the line connecting the charges closer to the positive charge. I mean it must be over here. I mean if it was 45 degrees on this side then it would be closer to the negative charge. So, the distance from this point to each of the charges is different. Different by this much Delta r. And so we have the potential due to the positive charge minus the potential due to the negative charge and negative charge distance is r plus delta r. And we're doing the same algebra here and ending up with k Q times the difference between their distances over r squared which is the same thing that happened up here. We have k Q times the difference in the distances from the point to the charges and the differences l, in that case. But in this case it's not as simple as this delta r but the delta r is the adjacent leg of this right triangle here and so we use cosine 45 times l to get this delta r and we assume that this is approximately 45 degrees just as this r because this angle is here because since l is so small compared to r, these lines are essentially parallel. I mean on my drawing they obviously are not but the assumption is that this r is much much greater than l. So we have delta r is l times Cos theta. And so there we have it. We've derived that formula in the textbook k Q l cos theta over r squared is approximately what the potential due to the dipole is and substituting P for Q l for the dipole moment and we get that coulomb’s constant times 4.2 times ten to the minus 30 coulomb meters, dipole moment, times cosine of 45 divided by the distance between the charges of 2.4 times ten to the minus nine meters squared, we get 4.6 times ten to the minus three volts. For this scenario, same idea except our angle is 135 degrees instead of 45. So it's the straight line angle of a 180 minus the 45 in here to make an angle of 135 that would we plug into our formula, this angle being the angle from the the axis of the dipole to the line connecting the positive charge to the point of interest and so it's a 135 degrees in this case and so same calculation as before except we got Cos 135 into the cos 45 and we get negative 4.6 times ten to the three volts.