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------------------------------------------------ At what separation is the electrostatic force between a +12.6 C point charge and a +27.7 μC point charge equal in magnitude to 1.64 N?

Express your answer in meters.

The separation at which the electrostatic force between two point charges is equal to 1.64 N is approximately 0.84 meters by using Coulomb's law .

Explanation:

The electrostatic force between two point charges is given by Coulomb's law:

F = k * (|q1 * q2| / r^2)

Where F is the force, k is the Coulomb constant, q1 and q2 are the magnitudes of the charges, and r is the separation between the charges.

Given that the force magnitude is 1.64 N, q1 = 12.6 C, q2 = 27.7 xC, and they are equal, we can solve for r:

1.64 N = (9 x 10^9 N*m^2/C^2) * (12.6 C) * (27.7 x 10^-6 C) / r^2

Simplifying the equation and solving for r gives us:

r = sqrt((9 x 10^9 N*m^2/C^2) * (12.6 C) * (27.7 x 10^-6 C) / 1.64 N)

r = 0.84 meters

Learn more about Coulomb's law here:

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