Physics: Electric fields and parallel plate capacitor?

Question

I'm stuck on this. My approach would be to place Coulombs law on one side and the electric plate capacitor equation on the other and solve for r, but I am also missing the value for q1 so I have 2 unknowns. What step am I missing?

Here's the question:

2 charges are placed between the plates of a parallel plate capacitor. One charge is +q1 and the other is q2=+5.00 x 10^-6 C. The charge per unit area on each plate ha a magnitude of sigma=1.3 x 10^-4 C/m^2. The force on q1 due to q2 equals the force on q1 due to the electric field of the parallel plate capacitor. What is the distance r between the two charges?

Here's the equations I want to use (just try to imagine o and E are Greek. ;)

k|q1||q2| / r^2 = o / Eo

Answer 1

I came across your question when I was looking how to do mine.  I figured it out, so I figured I'd help you out.

The key statement in the question is "The force on q1 due to q2 equals the force on q1 due to the electric field of the parallel plate capacitor".

So basically, "the force on q1 due to q2" is represented by F = k*|q1|*|q2| / r^2

And "the force on q1 due to the electric field" is represented by the equation F = E * q1 .

So , what you have is  k*|q1|*|q2| / r^2 = E * q1 .  The q1 's cancel out and you can Find E by using the E = sigma / E_o

IT's all plug and chug from there.

hope that helps!