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,

$F_i=\frac{kq_1q_2}{(d_i)^2}\ldots\ldots\text{.}(1)$

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,

$F_f=\frac{kq_1q_2}{(d_f)^2}\ldots\ldots\ldots(2)$

Dividing (1) equation by (2) equation,

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

Substituting the known values,

$\begin{gathered} \frac{10}{4}=\frac{d^2_f}{(0.3^{)2}} \\ d^2_f=\frac{10}{4}\times(0.3)^2 \\ d^2_f=0.225 \\ d_f=0.474\text{ m} \\ d_f=47.4\text{ cm} \end{gathered}$

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