To solve this problem, we will use Coulomb's law to calculate the electric force between the two charges. Coulomb's law is given by:

$F = k \frac{|q_1 \cdot q_2|}{r^2}$

Where:

• $F$ is the magnitude of the electric force.
• $k = 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2$ is Coulomb's constant.
• $q_1 = 6.96 \, \text{nC} = 6.96 \times 10^{-9} \, \text{C}$
• $q_2 = 4.54 \, \text{nC} = 4.54 \times 10^{-9} \, \text{C}$
• $r = 1.82 \, \text{m}$ is the distance between the charges.

(a) Calculating the force:

$F = \frac{(8.99 \times 10^9) \cdot (6.96 \times 10^{-9}) \cdot (4.54 \times 10^{-9})}{(1.82)^2}$

Let me compute this.

$F = \frac{(8.99 \times 10^9) \cdot (3.16144 \times 10^{-17})}{3.3124}$

$F = \frac{2.84185 \times 10^{-7}}{3.3124} \approx 8.58 \times 10^{-8} \, \text{N}$

(b) Is the force attractive or repulsive?

Since both charges are positive, the force is repulsive (like charges repel).

Final Answers:

(a) The magnitude of the force is approximately $8.58 \times 10^{-8} \, \text{N}$.
(b) The force is repulsive.

Let me know if you need further clarification or have additional questions.

Here are 5 related questions to enhance your understanding:

1. What would happen to the force if the distance between the charges was halved?
2. How would the force change if one of the charges was negative?
3. How does Coulomb's law relate to Newton's law of gravitation?
4. What happens to the force if one charge is doubled while keeping the distance the same?
5. Why is Coulomb's law inversely proportional to the square of the distance?

Tip: Always convert nanoCoulombs ($nC$) to Coulombs ($C$) by multiplying by ( 10