Question

What is the magnitude of the force a 1.5 x 10^-7 C charge exerts on a 7.2 x 10 ^-5 C charge located 2.8 m away?

Answer 1

Given:

`\begin{gathered} Q_1=1.5*10^(-7)\text{ C} \\ Q_2=7.2*10^(-5)\text{ C} \\ r=2.8\text{ m} \end{gathered}`

The electrostatic force is given as,

`F=(KQ_1Q_2)/(r^2)`

Here, K is the electrostatic force constant.

Putting all values,

`\begin{gathered} F=\frac{(9*10^9*1.5*10^(-7)*7.2*10^(-5))^{}}{(2.8)^2} \\ =0.0124\text{ N} \end{gathered}`

Therefore, the magnitude of the force between the charges is 0.0124 N.