Question

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

find the amount of the force when Q1 is doubled ?​

Answer 1

F = 3.6 10⁶ N

Step-by-step explanation:

The expression for the electric force is

F =
`k \ (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