To answer this question, we need to understand Coulomb's Law, which tells us the force between two point charges. Coulomb's Law states:
[tex]\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \][/tex]
Here:
- [tex]\( F \)[/tex] is the magnitude of the force between the charges,
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\(8.9875 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\)[/tex]),
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the charges, and
- [tex]\( r \)[/tex] is the distance between the centers of the two charges.
According to the question, initially, two charges produce a force of 12 N. Without loss of generality, assume that this force is attractive, meaning the charges have opposite signs, but the same could be argued if the force was repulsive (charges having the same sign). This is important only for direction, not magnitude.
When one of the charges changes its sign (from positive to negative or from negative to positive), there are a few important observations to make:
1. The magnitudes [tex]\( |q_1| \)[/tex] and [tex]\( |q_2| \)[/tex] do not change; only the sign of one charge changes.
2. The distance [tex]\( r \)[/tex] between the charges remains the same.
3. Coulomb's constant [tex]\( k \)[/tex] is a universal constant and does not change.
Since the magnitude of the force depends only on the magnitudes of the charges, the distance between them, and Coulomb's constant, changing the sign of one charge does not affect the magnitude of the force at all. It will, however, change the direction of the force. If the initial force was attractive, switching the sign on one charge will make it repulsive and vice versa.
Therefore, the new force after changing the sign of one charge will still be 12 N in magnitude, but its direction will be the opposite of what it originally was.
