Two similar charges, when placed 2 cm apart, repel each other with a force of 44.1 N. Find the magnitude of either charge.

Hello!
This problem requires using Coulomb's law, one of the basic electrostatic laws. It states that for point charges q_1 and q_2, which are stationary relative to each other, the electrostatic force is:
F = k*(q_1*q_2)/r^2.
Here r is a distance between these points and k is a constant which depends on the electrostatic properties of a medium in which the charges are reside. For air it is about 9*10^9 N*m^2/C^2.
It is given that both charges are equal: q_1 = q_2 = q. Then:
q^2 = (F*r^2)/k  and  |q| = r sqrt(F/k).
Numerically it is:
0.02*sqrt(44.1/(9*10^9)) = 2*10^(-2)*sqrt(4.9/10^9) = 2*10^(-2)*sqrt(49/10^(10)) =
=14*10^(-2)*10^(-5) = 1.4*10^(-6) (C).
C is for the Coulomb, the Si unit for electric charge. Note that we transformed 2 cm into 0.02 m to use the proper units.
So the answer is: the magnitude of each charge is 1.4*10^(-6) C. The sign of this charge is still unknown, charges may be both positive or both negative.
https://www.physicsclassroom.com/class/estatics/Lesson-3/Coulomb-s-Law

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