To calculate the electrical force between charges \( q_2 \) and \( q_3 \), we can use Coulomb's Law, which is given by the formula:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where:
- \( F \) is the magnitude of the electrical force,
- \( k = 8.99 \times 10^9 , \text{N} \cdot \text{m}^2/\text{C}^2 \) (Coulomb's constant),
- \( q_1 \) and \( q_2 \) are the two charges,
- \( r \) is the distance between the charges.
In this case:
- \( q_2 = -4 , \text{C} \)
- \( q_3 = 3 , \text{C} \)
- The distance between \( q_2 \) and \( q_3 \) is \( 1 , \text{m} \) (as given).
Substituting the values into Coulomb's Law:
\[ F = k \frac{|(-4) \cdot 3|}{(1)^2} \] \[ F = 8.99 \times 10^9 \frac{|-12|}{1} \] \[ F = 8.99 \times 10^9 \times 12 \] \[ F = 1.0788 \times 10^{11} , \text{N} \]
Since \( q_2 \) is negative and \( q_3 \) is positive, the force will be attractive. Therefore, we will represent the force as negative.
Thus, the electrical force \( F_{23} \) between \( q_2 \) and \( q_3 \) is:
\[ F_{23} = -1.0788 \times 10^{11} , \text{N} \]
Rounded to two significant figures, this yields approximately:
\[ F_{23} \approx -1.1 \times 10^{11} , \text{N} \]
The answer is:
–1.1 × 10¹¹ N.