Answer:
The electric force between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by Coulomb's law:
\[ F = \frac{{k \cdot |q_1 \cdot q_2|}}{{r^2}} \]
Where:
- \( F \) is the magnitude of the electric force,
- \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.
Given that the electric force between the charges A and B is \( -560 \, \text{N} \), and the charges are \( +6 \, \mu \text{C} \) and \( -8 \, \mu \text{C} \) respectively, we have:
\[ 560 = \frac{{8.99 \times 10^9 \cdot |6 \times 10^{-6} \cdot (-8 \times 10^{-6})|}}{{r^2}} \]
\[ 560 = \frac{{8.99 \times 10^9 \cdot 48 \times 10^{-12}}}{{r^2}} \]
\[ r^2 = \frac{{8.99 \times 10^9 \cdot 48 \times 10^{-12}}}{{560}} \]
\[ r^2 = \frac{{431520 \times 10^{-3}}}{{560}} \]
\[ r^2 = 770 \times 10^{-3} \]
\[ r = \sqrt{770} \, \text{m} \approx 27.74 \, \text{m} \]
Therefore, the distance between the two charged particles is approximately \( 27.74 \, \text{m} \).
