Question

Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the plates. What is the magnitude of the electric force on q1?

Answer 1

Answer:

The force exerted on the $q_1$ is  $F = 2.25*10^{3} \ N$

Explanation:

From the question we are told that

     The area is  $A = 2.34*10^{-3} \ m^2$

     The magnitude of charge placed on them is  $q = 7.07 * 10^{-7} C$

     The charge placed between the plate is $q_1 = 6.62 *10^{-5} C$

   

The electric field generated around the plate  is mathematically represented as

           $E = \frac{q}{A \epsilon_o}$

Substituting values

          $E = \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}$

         $E = 34*10^{6} \ V/m$

The force exerted the charge $q_1$ is  mathematically represented as

        $F = q_1 * E$

Substituting values  

        $F = 6.62 *10^{-5} * 34*10^{6}$

        $F = 2.25*10^{3} \ N$