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

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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.

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