To calculate the electrical force of attraction between two charges, we use Coulomb's Law:

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

Where:

• $F$ is the electrical force,
• $k = 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2$ (Coulomb's constant),
• $q_1$ and $q_2$ are the charges in coulombs,
• $r$ is the distance between the charges in meters.

Given:

• $q_1 = 0.0005 \, \text{C}$,
• $q_2 = -0.0008 \, \text{C}$ (magnitude only since force is scalar for attraction),
• $r = 2.5 \, \text{m}$,
• $k = 8.99 \times 10^9$.

Substitution:

$F = (8.99 \times 10^9) \cdot \frac{|0.0005 \cdot -0.0008|}{2.5^2}$

1. Multiply $q_1$ and $q_2$: $0.0005 \cdot 0.0008 = 4 \times 10^{-7}$

2. Calculate $r^2$: $2.5^2 = 6.25$

3. Plug values into Coulomb’s Law: $F = (8.99 \times 10^9) \cdot \frac{4 \times 10^{-7}}{6.25}$

4. Simplify the fraction: $\frac{4 \times 10^{-7}}{6.25} = 6.4 \times 10^{-8}$

5. Final calculation: $F = (8.99 \times 10^9) \cdot 6.4 \times 10^{-8}$

$F = 575.36 \, \text{N}$

Final Answer:

The electrical force of attraction between the two balloons is approximately 575.36 N.

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Would you like further clarification or detailed breakdowns of any part?

Relative Questions:

1. How does the distance between charges affect the force of attraction?
2. What would happen to the force if one charge were doubled?
3. Why does Coulomb's Law use the absolute value of charges?
4. What is the significance of the sign of the charges in attraction vs. repulsion?
5. How would the force change if the medium were not air but water?

Tip:

Always square the distance in Coulomb’s Law, as this significantly impacts the magnitude of the force!