Question

3 ) find the electrical force between the two charges Q1=3mc ,Q2=-6mc when they are 0.3 m parted ?

Answer 1

Answer:

  F = 3.6 10⁶ N

Explanation:

The expression for the electric force is

         F =  $k \ \frac{q_1 q_2}{r^2}$

in this case it indicates that the charge q1 is doubled

       q₁ = 2   3 10⁻³ C

       q₁ = 6 10⁻³ C

   

let's reduce the magnitudes to the SI system

        q₂ = 6 10⁻³ C

        r = 0.3 m

let's calculate

       F = 9 10⁹ 6 10⁻³ 6 10⁻³ / 0.3²

       F = 3.6 10⁶ N