Question

What is the force of repulsion between the charges below?+0.001+0.0025m

Answer 1

The distance between the charges is given as,

`d=5\text{ m}`

The value of the charges are given as,

`\begin{gathered} q_1=0.001\text{ C} \\ q_2=0.002\text{ C} \end{gathered}`

According to the Coulomb's Law, the force of repulsion between these charges is,

`F=k(q_1q_2)/(d^2)`

where k is the electrostatic constant.

The value of k is,

`k=9*10^9Nm^2C^(-2)`

Substituting the known values,

`\begin{gathered} F=9*10^9*(0.001*0.002)/(5^2) \\ F=7.2*10^2\text{ N} \\ F=720\text{ N} \end{gathered}`

Thus, the repulsive force acting between the charges is 720 N.