Two point charges, initially 2 cm apart, are moved to a distance of 10 cm apart. By what factor does the resulting electric force between them change?

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This question involves the law of interaction between the two point electric charges, also known as Coulomb's Law. It states that the force between the two point charges is inversely proportional to the square of the distance between them and directly proportional to the product of the magnitude of the charges. It also states that the force directed along the line connecting the two charges, and it is attractive when the charges are different in sign (unlike) and repulsive when they are the same in sign (like).

Mathematically, the magnitude of the force between the two charges with the magnitudes `q_1` and `q_2` is expressed as

`F_(12) = k(q_1*q_2)/r^2`

Here, r is the distance between the charges and k is the universal constant that relates units of charges (Coulombs) to the units of force (Newtons).

`k = 8.99*10^9 N*m^2/C^2`

Since the force is inversely proportional to the distance, when the distance increases by a factor of 5 (from 2 cm to 10 cm), the force decreases by a factor of 25.

This could be shown by direct calculation:

`F_(2cm) = k(q_1*q_2)/(0.02)^2`

`F_(5cm) = k (q_1*q_2)/(0.1)^2`

`F_(5cm)/F_(2cm) = k(q_1*q_2)/(0.1)^2 * (0.02)^2/(kq_1*q_2) = 0.0004/0.01 = 1/25 `

`F_(5cm) = 1/25 F_(2cm)`

The resulting electric force between the charges decreases by a factor of 25.

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