Posted by Phil on Friday, January 20, 2012 at 3:38pm.
Figure (a) shows charged particles 1 and 2 that are fixed in place on an x axis. Particle 1 has a charge with a magnitude of q1 = 12.0e. Particle 3 of charge q3 = +13.0e is initially on the x axis near particle 2. Then particle 3 is gradually moved in the positive direction of the x axis. As a result, the magnitude of the net electrostatic force on particle 2 due to particles 1 and 3 changes.
Figure (b) gives the x component of that net force as a function of the position x of particle 3. The scale of the x axis is set by xs = 1.70 m. The plot has an asymptote of F2,net = 1.370 × 1025 N as x → ∞.
As a multiple of e and including the sign, what is the charge q2 of particle 2?
Here are the two figures:
h t t p : / / i . i m g u r . c o m / I U C 3 6 . p n g

Physics  drwls, Friday, January 20, 2012 at 3:58pm
As x> infinity, it no longer affects the force on particle 2, F2.
Use the value of F2 when x = infinity, the fixed distance separation of 1 and 2, and Coulomb's equation to determine the charge of particle 2.

Physics  Bob, Sunday, February 5, 2017 at 12:22pm
As above, as the xvalue for particle 3 approaches infinity the only force acting on particle 2 is from particle 1. The asymptote is the value of that force. Set the equation for (kq1q2)/(r^2) = 1.370e25N and solve for q2.
Note that you can solve for r with the knowledge that the Fnet provided is the constant force from q1 onto q2 and that the net force is 0 at the xvalue where the graph crosses the x axis.
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