The magnitude of the force between two charges can be calculated using Coulomb's Law:

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

Where:

• $F$ is the force between the charges.
• $k$ is Coulomb's constant, approximately $8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2$.
• $q_1$ and $q_2$ are the magnitudes of the charges.
• $r$ is the distance between the charges.

Given:

• $q_1 = 1.5 \times 10^6 \, \text{C}$
• $q_2 = 3.2 \times 10^4 \, \text{C}$
• $r = 1.5 \, \text{m}$

Step 1: Substitute the values into Coulomb's Law:

$F = (8.99 \times 10^9) \cdot \frac{(1.5 \times 10^6)(3.2 \times 10^4)}{(1.5)^2}$

Step 2: Simplify the terms:

• $(1.5 \times 10^6)(3.2 \times 10^4) = 4.8 \times 10^{10}$
• $(1.5)^2 = 2.25$

So: $F = (8.99 \times 10^9) \cdot \frac{4.8 \times 10^{10}}{2.25}$

Step 3: Divide $4.8 \times 10^{10}$ by $2.25$:

$\frac{4.8 \times 10^{10}}{2.25} = 2.1333 \times 10^{10}$

Step 4: Multiply by $8.99 \times 10^9$:

$F = (8.99 \times 10^9) \cdot (2.1333 \times 10^{10}) = 1.917 \times 10^{20} \, \text{N}$

Final Answer:

$F \approx 1.92 \times 10^{20} \, \text{N}$

This is the magnitude of the force between the two charges.

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Do you want details about the calculation steps or any clarification?
Here are some related questions to explore further:

1. What is Coulomb's law and how does it relate to electric force?
2. How does the distance between two charges affect the magnitude of the force?
3. What happens if one of the charges is negative?
4. How is Coulomb's constant derived, and what does it represent?
5. How would the force change if the charges are placed in a medium with a dielectric constant?

Tip: Remember, the force between charges increases significantly as the distance decreases because it is inversely proportional to the square of the distance.