Sunday
March 26, 2017

Post a New Question

Posted by on .

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 - ,

    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 - ,

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question