[Music] so I'm going to go through another example of kul's law here and how to use it with a set of charges and it's going to be a little complicated to walk through so take your time on this but I want to show you a couple of the tricks of the trade here some of the shortcuts that you can use to make this problem solving a little more simple and uh ways of looking at these that I I like you learn so let me imagine for a second what I've got here is is five protons five blue charges there and I've glued them to the tabletop so they're stuck in position they form this sort of weird wedge shape I'm going to number them one to five and set them out so that they're in a 30° angle to each other and each of them is spaced by about a micron from from its neighbor so 1 to two is one micron two to four is a micron 1 to three is a micron and 3 to five is a micron uh a micron is a millionth of a meter 1 * 10us 6 what AM I do now is put another proton right in the middle between four and five right there and this proton I'm going to make loose he can he can start going as soon as I let go of him and in fact because all these guys repel each other what I would expect is he's going to shoot off to the top right of the screen as soon as I I were to let go of him and the question I want to ask is just exactly what will the acceleration of that light blue proton B when I let go well what I need to do here is calculate use the kol's force law to calculate what the force on that light blue proton is and then determine the acceleration from that force and as it sits right now it would be a rather complicated problem but one of the things I really want you to learn here is how to take advantage of symmetry in these situations so what I'm going to do is take this set of charges and just rotate it by 15° so now when I've rotated it what happens is all those charges are lined up so that I expect that the net force on the light blue guy is going to be straight out in the X Direction you'll see this work through in the math in just a moment and it's kind of neat how it happens but for now it's a way of making things a little simpler just by changing my Axis system I can set the axes however I want to I'm in charge of this so I'm going to go ahead and turn them by 15° and and choose it this way the other thing that's nice about this is if I set it up this way I can actually ignore four and five four pushes up on the light blue charge just as much as five pushes down because they're the same distance away and they're in line with each other so that means that there's no Point even calculating anything having to do with four and five at this point because the forces on them are just going to be equal and opposite to each other and in principle result in no force on the light blue charge it's only one two and three that do anything to start with well now let me start breaking it down and looking at one two and three and the first thing I'd like to do is figure out just how far 1 two and three are from zero and in particular what the displacement Vector from two to 0 3 to 0 and 1 to Z is well 2 to0 I can look at in the displacement vector there's that Pink Arrow I just drew there and it has a length of D where D was the spacing between these guys if I look at the symmetry of this obviously the distance from 1 to two and 2 to four is the same as the distance from 2 to zero so the length of that guy is D and since I've rotated my Axis system I now have this axis of symmetry that's 15° from the vshape so therefore I can pull out the the the displacement Vector from 2 to zero is just going to be D * the cosine of 15° in the X Direction and D * the S of 15 in the y direction if you can't quite figure that out go back and do some trig on it and make sure you you you see what's happening here the nice thing about this is it makes R30 easy to find because that's just a flip over the axis now R30 is also a length of D and it is D cosine 15 in the X Direction and just negative D sin 15 in the y direction it's just a flip of R20 r10 is a little trickier to find out but not that terribly difficult what I'll notice here is the distance from 2 and 3 in the X direction to zero is the same as the distance to one in other words there's symmetry about 2 and three if I draw a vertical line through there so that D cosine 15 it takes me to get back to 32 is just doubled up to get back to one so 1 is at 2 * D cosine uh D cosine 15 in the X Direction now what I can do is solve for the force that one causes on zero so just pull out my kum's law kqq over r^ 2 and the r in this case is 2D cosine 15 from that last slide and at this point all I can can do is is put a direction on it that is in the xat direction and I'm going to stop there instead of plugging in numbers on this and doing a lot of calculator jocking I'm going to notice that K and E and D are both going to all three of them going to show up in my next two calculations of the force due to charge two and charge three so let's just kind of put those in our pockets and hold on to it I'm going to let this equation sit as it is for a moment and go on to calculate what the force due to number two is the force of number two on charge zero is kqq over r^ 2 for it and the r in this case is just D plugging in the Q is the charge of a proton e so I have K e^2 Over d^2 Now what I'll have is the force in the in the R20 direction is k^2 over d^2 and the R20 is just the unit Vector that points from 2 to Zer if I go back and look at R20 and make it a unit Vector what I have is cosine 15 x hat plus sin15 y hat now the neat thing is the same thing is going to happen for f30 all of the calculation is the same as F20 except that last step where the Y component gets flipped and what this means is when I put these three forces together and add them up the Y component of F20 and the X Y component of f30 are are equal and opposite to each other and this is where I see mathematically that my idea of lining this thing up with the x-axis paid off there is no calculation of a y component because there is no y component due to symmetry so now all I have to do is grab those X components and put them all together and I'll keep the k^2 over d^2 out front for now and what I've got is k^ 2 over d^2 times this 1 4 cos^2 15 + 2 cosine 15 what I want to point out is everything in the parth there is just related to the geometry it's related to this vshape that I made in other words if I had done this with charges of 10 Kum Space by 5 km or something like that everything about what I've got is still the same the K the E and the D is the only thing that relates to the charge sizes and their distances what's in parenthesis is related to where they're located in relation to each other so if I grab the cosine of 15 and and do the the calculation here what I find is that it all adds up to be 2.2 as a factor multiplying k^2 over d^2 and then I get K is 9 8.99 * 10 9th e the charge of a a proton is the electron charge 1.62 10us 19 and D is 10us 6 one micron from the the given problem and plug all those guys in and what I get is a force of 5.08 * 10- 16 Newtons which seems very very small until I consider the fact that I'm being asked what the acceleration is and the acceleration is the force divided by the mass FAL ma after all and the mass of a proton is a very very small thing so 5.88 * 10- 16 when I divide by that 1.67 10-27 the mass of a proton I'm going to get a very large acceleration here 3 * 10 11 m/s squared so that proton is going to get shot out quite rapidly from this little vshape because it really is repelled by 1 two 3 and as soon as it is shot it will start being repelled by four and five as well that's a point to make here this acceleration that I've calculated is the acceleration at this exact instant as soon as that light blue circle moves as soon as the proton is shot out of the vshape the acceleration will be different because all the distances will have changed and also because now I will have components due to four and five pointing in the X direction as well so I can't just grab my equations for constant acceleration and calculate how fast a will be the proton be going a few seconds later or something like that because the acceleration is not going to be constant it's going to be very location dependent and that means this is going to be a tricky thing to calculate and we won't in fact get into that right now right now all I want to be able to do is calculate what that force is and therefore what the acceleration is because of the force