Question

Two charges are placed between the plates of a parallel plate capacitor. One charge is +q1 and the other is q2 = +4.56C. The charge per unit area on each plate has a magnitude of = 8.63 × 10-3 C/m2. 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?

Answer 1

electric field between charge plates

E = sigma / epsilon_not

E = 8.63*10^-3 / 8.85*10^-12

E = 9.75*10^8 N/C

now,

force on q1 due to electroc field = force between q1 and q2

q1 E = k q1 q2 / r^2

9.75*10^8 = 9*10^9* 4.56 / r^2

r = 6.487 m

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