Please explain the highlighted portions of the solution. How can we use the zero net charge and the size of the first 2 charges to determine the relative location of the third charge?

25. Point charges of 5.00 uC and - 3.00 uC are placed 0.250 m apart. (a) Where can a third charge be placed so that the net force on it is zero? (b) What if both charges are positive? Solution (a) We know that since the negative charge is smaller, the third charge should be placed to the right of the negative charge if the net force on it is to be zero. So if we want Fret = F, + F2 = 0, we can use F = k 4142 to write the forces in terms of distances: kq,92 k929 = kq 42 =0, or since r, = 0.250m + d andr2 = d, 5x10- C 3x10-6 (0.250 m + d} d2 -, or d = (0.250m)+ 3d so that OpenStax College Physics Instructor Solutions Manual Chapter 18 3 (0.250m) and finally, d = (0.250 m) = 0.859m 1 - V3 5 The charge must be placed at a distance of 0.859 m to the far side of the negative charge. (b) This time we know that the charge must be placed between the two positive charges and closer to the 3 uC charge for the net force to be zero. So if we want Fnet = F, + F2 = 0, we can again use F = k 4192 to write the forces in terms of distances: kq,92 k929 = kq 91 + 92) =0 5x10- C 3x10-6 Or since r = 0.250m - 12 (0.250 -12) or 722 = 2(0.250 m - 72) , or 2(0.250m) 72 = =(0.250m -rz), and finally r; = = 0. 109 m 1 + 15 The charge must be placed at a distance of 0.109m from the 3 uC charge