Respuesta :
Given data:
* The electrostatic force of repulsion between the charged bodies in the initial state is,
[tex]F=3.6\times10^{-5}\text{ N}[/tex]* The distance between the charged bodies in the initial state is,
[tex]d=0.12\text{ m}[/tex]* The distance between the charged bodies in the final state is,
[tex]\begin{gathered} d^{\prime}=0.24\text{ m} \\ d=2\times0.12 \\ d^{\prime}=2d \end{gathered}[/tex]Solution:
According to Coulomb's law, the electrostatic force of repulsion between the charged bodies in the initial state is,
[tex]F=\frac{kq_1q_2}{d^2}[/tex]where k is the electrostatic force constant, q_1 is the charge on the first charged body and q_2 is the charge on the second charged body,
The electrostatic force of repulsion between the charged bodies in the final state is,
[tex]\begin{gathered} F^{\prime}=\frac{kq_1q_2}{(2d)^2} \\ F^{\prime}=\frac{kq_1q_2}{4d^2} \\ F^{\prime}=\frac{1}{4}\times\frac{kq_1q_2}{d^2} \\ F^{\prime}=\frac{F}{4} \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} F^{\prime}=\frac{3.6\times10^{-5}}{4} \\ F^{\prime}=0.9\times10^{-5}\text{ N} \end{gathered}[/tex]Thus, the electrostatic force of repulsion between the charged bodies in the final state is,
[tex]\text{0}.9\times10^{-5}\text{ N}[/tex]