Answer
$\frac{q_C}{q_B} = 1.333$
Work Step by Step
After particle B is moved to the opposite side, the net force on particle A decreases. This shows that initially, the forces on particle A due to the other particles were in the same direction, which shows that particle B and particle C have the same sign.
Since particle B has been moved to a new location that is the same distance from particle A, the magnitude of the force on particle A due to particle B is the same.
We can find the magnitude of the force on particle A due to particle B:
$F_B+F_C = 2.014\times 10^{-23}~N$
$F_C-F_B = 2.877\times 10^{-24}~N$
$2F_B = 1.7263\times 10^{-23}~N$
$F_B = 8.6315\times 10^{-24}~N$
We can find the magnitude of the force on particle A due to particle C:
$F_B+F_C = 2.014\times 10^{-23}~N$
$F_C = 2.014\times 10^{-23}~N-F_B$
$F_C = 2.014\times 10^{-23}~N-8.6315\times 10^{-24}~N$
$F_C = 1.15085\times 10^{-23}~N$
We can find $\frac{q_C}{q_B}$:
$\frac{q_C}{q_B} = \frac{F_C}{F_B}$
$\frac{q_C}{q_B} = \frac{1.15085\times 10^{-23}~N}{8.6315\times 10^{-24}~N}$
$\frac{q_C}{q_B} = 1.333$
