Answer
The minimum magnitude of the net force on particle 3 is $~~9.20\times 10^{-24}~N$
Work Step by Step
If particle 3 is placed between particle 1 and particle 2, then the force on particle 3 due to particle 1 is in the +x direction, and the force on particle 3 due to particle 2 is in the +x direction.
The net force on particle 3 is the sum of these two forces.
In part (a), we found that the magnitude of the net force on particle 3 is a minimum when $x = 2.00~cm$
We can find the minimum magnitude of the net force:
$F = \frac{4e^2}{4\pi~\epsilon_0~x^2}+\frac{108e^2}{4\pi~\epsilon_0~(0.0800-x)^2}$
$F = \frac{(4)(1.6\times 10^{-19}~C)^2}{(4\pi)~(8.854\times 10^{-12}~F/m)~(0.0200~m)^2}+\frac{(108)(1.6\times 10^{-19}~C)^2}{(4\pi)~(8.854\times 10^{-12}~F/m)~(0.0800~m-0.0200~m)^2}$
$F = 9.20\times 10^{-24}~N$
The minimum magnitude of the net force on particle 3 is $~~9.20\times 10^{-24}~N$
