 Now we're ready to tackle part D of this problem, to express the two forces as vectors. So what we just did on part C was to calculate the magnitude of force between the two charges. And if you refer back up to our diagram, we saw that because it was a positive and negative charges, they attracted each other, we can see the direction of those two forces. So we can start by putting this into words. The force on charge one points to the right, while the force on charge two points to the left. Now the way we keep track of these two forces in terms of them being vector signs is we represent the force on charge one due to charge two as F12. And the force on charge two due to charge one is F21. Now remember both of these have to have a magnitude of the force that we calculated up on the top. Because it points to the right, that's in the plus i hat direction, to the left is in the minus i hat direction. And there's no y component at all. There's no up and down aspect to this. So in terms of our standard vector notation, that means we have plus 0.36 Newton's i hat and 0 j hat for that force on charge one which points to the right. And then we've got minus 0.36 Newton's i hat plus 0 j hat for the force which points to the left. And this is consistent with the ij notation that we've been using starting from physics one.