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At what separation is the electrostatic force between a 112.2 μC point charge and a 29.1 μC point charge equal in magnitude to 1.57 N?

Answer :

The separation at which the electrostatic force between a 112.2 μC point charge and a 29.1 μC point charge equals 1.57 N is approximately 5.62 meters.

To find the separation, we can use Coulomb's law, [tex]\(F = k \frac{{|q_1 q_2|}}{{r^2}}\)[/tex], where F is the electrostatic force, k is Coulomb's constant 8.99xN [tex]m^2[/tex][tex]C^2[/tex] [tex]q_1\)[/tex] and [tex]q_2\)[/tex] are the charges, and r is the separation distance. We're given F = 1.57N, ([tex]q_1[/tex] = 112.2 x[tex]10^{-6}[/tex]C, and [tex]\(q_2[/tex] = 29.1x [tex]10^{-6}[/tex]C.

We rearrange Coulomb's law to solve for r: r = [tex]\sqrt{\frac{{k |q_1 q_2|}}{F}}\)[/tex].

Plugging in the given values, we get r = [tex]\sqrt{\frac{{(8.99 \times 10^9) \times (112.2 \times 10^{-6}) \times (29.1 \times 10^{-6})}}{{1.57}}\)[/tex]

Calculating this expression gives us r [tex]\approx[/tex]5.62 m.

In this problem, we found the separation between the two charges at which the electrostatic force equals a given value using Coulomb's law. By rearranging the equation and substituting the given values, we solved for the separation, obtaining approximately 5.62 meters. This means that at a distance of approximately 5.62 meters, the electrostatic force between the two charges is equal to 1.57 N, as required.

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