In the figure three Identical conducting spheres initially h

In the figure three Identical conducting spheres initially have the following charges: sphere A, 2Q; sphere B, -7Q; and sphere C, 0. Spheres A and B are fixed in place, with a center-to-center separation that is mu C h larger than the spheres. Two experiments are conducted. In experiment 1, sphere C is touched to sphere A and then (separately) to sphere B, and then it is removed. In experiment 2, starting with the same Initial states, the procedure is reversed: Sphere C is touched to sphere B and then (separately) to sphere A, and then it is removed. What is the ratio of the electrostatic force between A and B at the end of experiment 2 to that at the end of experiment 1?

Solution

when two spheres let be 1 and 2 touch each other ; the sphere shares total charge between them Q₁+Q₂ / 2

in experiment 1 sphere C is touched to sphere A ; there by reducing the charge on sphere A to half the initial value

before touching charge on C = 0 charge on A = +2Q

after touching charge on C = +Q charge on A = +Q

now sphere C is touched with B

after touching charge on C = +Q-7Q /2 = -3Q charge on B = -3Q

Force F1 between A and B after experiment 1

                                   F = k Qa * Qb / d^2        with k - coloumb constant

                                      = k Q* - 3 Q / d^2

                                     = - k 3Q^2 /d^2   attractive force

in experiment 2 sphere C is first trouched to sphere B ; there by reducing the charge on sphere B to half the initial value

before touching charge on C = 0 charge on B = -7Q

after touching charge on C = -3.5Q charge on B = -3.5Q

now sphere C is touched with A

before touching charge on C = -3.5Q charge on A = +2Q

after touching charge on C = -3.5Q +2Q /2 = -0.75Q charge on A = -0.75Q

Force F2 between A and B after experiment 2

                                   F₂ = k Qa * Qb / d^2        with k - coloumb constant

                                      = k * -0.75Q *-3.5 Q / d^2

                                     = k * 2.625Q^2 /d^2   repulsive force

Ratio of the force F2 / F1 =( k * 2.625Q^2 /d^2 ) / ( k 3Q^2 /d^2 ) = 0.875