Question

two charged particles repel each other with a force of 10N when they are 30cm apart if they are repel each other by a force of 4N what would be the separation distance

Answer 1

Given data:

* The initial distance between the charged particles is,

`\begin{gathered} d_i=30\text{ cm} \\ d_i=0.3\text{ m} \end{gathered}`

* The initial force acting between the charged particles is F_i = 10 N.

* The final force acting between the charged particles is F_f = 4 N.

Solution:

By Coulomb's law, the electrostatic force between the charged particles in the initial case is,

`F_i=(kq_1q_2)/((d_i)^2)\ldots\ldots\text{.}(1)`

where k is the electrostatic force constant, q_1 is the charge on the first particle and q_2 is the charge on the second particle,

By Coulomb's law, the electrostatic force between the charged particles in the final case is,

`F_f=(kq_1q_2)/((d_f)^2)\ldots\ldots\ldots(2)`

Dividing (1) equation by (2) equation,

`\begin{gathered} (F_i)/(F_f)=((kq_1q_2)/((d_i)^2))/((kq_1q_2)/((d_f)^2)) \\ (F_i)/(F_f)=((d_f)^2)/((d_i)^2) \end{gathered}`

Substituting the known values,

`\begin{gathered} (10)/(4)=(d^2_f)/((0.3^()2)) \\ d^2_f=(10)/(4)*(0.3)^2 \\ d^2_f=0.225 \\ d_f=0.474\text{ m} \\ d_f=47.4\text{ cm} \end{gathered}`

Thus, the distance between the charges in the final state is 47.4 cm or approximately 0.5 m.