Answer
The ratio of the electrostatic force between A and B at the end of experiment 2 to that at the end of experiment 1 is $~~\frac{3}{8}$
Work Step by Step
Experiment 1:
After sphere C touches sphere A, both sphere A and sphere C have a charge $2Q$
After sphere C touches sphere B, both sphere B and sphere C have a charge $\frac{2Q-6Q}{2} = -2Q$
We can find the electrostatic force between sphere A and sphere B:
$F_1 = \frac{(2Q)(-2Q)}{4\pi~\epsilon_0~r^2} = -4~\frac{Q^2}{4\pi~\epsilon_0~r^2}$
Experiment 2:
After sphere C touches sphere B, both sphere B and sphere C have a charge $-3Q$
After sphere C touches sphere A, both sphere A and sphere C have a charge $\frac{4Q-3Q}{2} = \frac{Q}{2}$
We can find the electrostatic force between sphere A and sphere B:
$F_2 = \frac{(Q/2)(-3Q)}{4\pi~\epsilon_0~r^2} = -\frac{3}{2}~\frac{Q^2}{4\pi~\epsilon_0~r^2}$
We can find $\frac{F_2}{F_1}$:
$\frac{F_2}{F_1} = \frac{-\frac{3}{2}~\frac{Q^2}{4\pi~\epsilon_0~r^2}}{-4~\frac{Q^2}{4\pi~\epsilon_0~r^2}} = \frac{3}{8}$
The ratio of the electrostatic force between A and B at the end of experiment 2 to that at the end of experiment 1 is $~~\frac{3}{8}$
