. The separation d between A and B is the same as the separation between B and C. One of the charges is positive and two are negative, as the drawing shows. Consider the net electrostatic force that each charge experiences due to the other two charges. Rank the net forces in descending order (greatest first) according to magnitude. (a) A, B, C (b) B, C, A (c) A, C, B (d) C, A, B (e) B, A, C

the descending order of the forces is

          B A C

Therefore the correct result is e

Explanation:

To answer this exercise we must use that the forces are vector magnitudes and the value is given by Coulomb's law

                   F = k q₁q₂ / x²

Consider the following scheme                          

              A    B     C

             -q  + q   + q

                  x      x

The forces are

           F_AB = k q² / x²

           F_AC = k q² / 4x²

           Fo = k q² / x²

           F_AB = Fo

           F_AC = ¼ Fo

           F_BC= F_AB

The forces on the charge A are

          F_A = F_AB + F_AC

          F_A = Fo + ¼ Fo

          F_A= 5/4 Fo

The forces on the charge B

        F_B = -F_AB - F_BC

        F_B = - Fo - Fo

        F_B = -2 Fo

The forces on the load C

        F_C = F_BC –F_AB

        F_C = Fo - ¼ Fo

        F_C = ¾ Fo

For this distribution of charges the descending order of the forces is

          B A C

Therefore the correct result is e

If the charge distribution has the scheme          

              A      B     C

             + q   -q    +q

                    x     x

Force on A

           F_A = FA_B - F_BC

           F_A = Fo - ¼ Fo = ¾ Fo

Force on B

         F_B = F_AB –F_BC

        F_B = 0

Force on  C

          F_C = F_AC - F_BC

          F_C = ¼ Fo - Fo

          F_C = - ¾ Fo

For this distribution of charges the descending order of the forces is

          A B C

result is  a

moving ball hits still ball. mover stops, hit ball moves. momentum transfer.