Hey, there is only one of me :(
You need to know the electric field E is constant between the plates and has value of E = sigma/eo = 1.3*10^-4/8.85*10^-12
= .147 *10^8
Two charges are placed between the plates of a parallel plate capacitor. One charge is +q1 and the other is q2= +5.00x10^-6 C . The charge per unit area on each plate has a magnitude of sigma=1.30 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?
2 answers
I will call q1 = q Now the force on q1 dues to the plates is:
F = q E = 0.147*10^8 q
The force on q due to the q2 is
F = 9*10^9 q (5*10^-6)/r^2
set those equal
q cancels
solve for r
F = q E = 0.147*10^8 q
The force on q due to the q2 is
F = 9*10^9 q (5*10^-6)/r^2
set those equal
q cancels
solve for r