Homework Help: Physics

Posted by Trevor on Wednesday, February 6, 2008 at 7:29pm.

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?

Physics - Damon, Wednesday, February 6, 2008 at 7:36pm
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
Physics - Damon, Wednesday, February 6, 2008 at 7:42pm
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

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