Question

Two parallel plates of area 7.34 x 10^-4 m^2 have 5.83 x 10^-8 C of charge placed on them. A 6.62 x 10^-6 C charge q1 is placed between the plates. What is the magnitude of the electric force on q1? Hint: How is force related to the field? (Unit = N)

Answer 1

Answer:

* if the two plates have the same charge sign       F_net = 0

*if one plate is positive and the other is negative   F_net = 59.4 N

Explanation:

The electric field created by a parallel plate is

           E = $\frac { \sigma }{2 \epsilon_o }$

where sigma is the charge density

          σ = Q / A

we substitute

           E = $\frac{Q}{A \ 2 \epsilon_o }$

           E = $\frac{ 5.83 \ 10^{-8} }{7.34 \ 10^{-4}\ 2 \ 8.5 \ 10^{-12} }$

           E = 4.487 10⁶ N / C

the electric force is

          F = E / q

in this force it is a vector, so if the charges are of the same sign they repel and if they are of the opposite sign they attract. In this case, the test load is between the two plates that have the same load sign, so the forces are in the opposite direction.

* if the two plates have the same charge sign

          F_net = F₁ - F₂

          F_net = q (E₁ -E₂)

since the electric field does not depend on the distance

           E₁ = E₂

in consecuense  

           F_net = 0

In a more interesting case

*if one plate is positive and the other is negative

           F_net = F₁ + F₂

           F_net = q (E₁ + E₂) = 2 q E₁

           F_net = 2 6.62 10⁻⁶  4.487 10⁶

           F_net = 59.4 N

net force goes from positive to negative plate