Question

The force of electrostatic repulsion between two small positively charged objects, A and B, is 3.6 x 10⁻⁵ N when AB = 0.12m. What is the force of repulsion if AB is increased to 0.24 m

Answer 1

Given data:

* The electrostatic force of repulsion between the charged bodies in the initial state is,

`F=3.6*10^(-5)\text{ N}`

* The distance between the charged bodies in the initial state is,

`d=0.12\text{ m}`

* The distance between the charged bodies in the final state is,

`\begin{gathered} d^(\prime)=0.24\text{ m} \\ d=2*0.12 \\ d^(\prime)=2d \end{gathered}`

Solution:

According to Coulomb's law, the electrostatic force of repulsion between the charged bodies in the initial state is,

`F=(kq_1q_2)/(d^2)`

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,

`\begin{gathered} F^(\prime)=(kq_1q_2)/((2d)^2) \\ F^(\prime)=(kq_1q_2)/(4d^2) \\ F^(\prime)=(1)/(4)*(kq_1q_2)/(d^2) \\ F^(\prime)=(F)/(4) \end{gathered}`

Substituting the known values,

`\begin{gathered} F^(\prime)=(3.6*10^(-5))/(4) \\ F^(\prime)=0.9*10^(-5)\text{ N} \end{gathered}`

Thus, the electrostatic force of repulsion between the charged bodies in the final state is,

`\text{0}.9*10^(-5)\text{ N}`