Two point charges exert a 8.00 N force on each other. What will the force become if the distance between them is increased by a factor of six?

The force between two point charges is governed by **Coulomb's Law**:

\[
F = k \frac{{|q_1 q_2|}}{{r^2}}
\]

where:
- \( F \) is the force between the charges,
- \( k \) is Coulomb's constant (\(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\)),
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.

When the distance is increased by a factor of 6, the new distance becomes \( 6r \). Since the force is inversely proportional to the square of the distance, the new force \( F' \) is related to the original force \( F \) by:

\[
F' = F \left( \frac{r}{6r} \right)^2 = F \left( \frac{1}{6} \right)^2 = F \frac{1}{36}
\]

Given that the original force is 8.00 N, the new force will be:

\[
F' = \frac{8.00 \, \text{N}}{36} = 0.222 \, \text{N}
\]

Thus, the force will decrease to **0.222 N** when the distance between the charges is increased by a factor of 6.

Would you like any further details or clarification?

Here are 5 related questions:
1. How does the force between two charges change if the charge magnitudes are doubled?
2. What happens to the force if the distance is halved instead of increased?
3. How is Coulomb's constant derived, and what does it represent?
4. Can Coulomb's law apply to charges in mediums other than a vacuum?
5. What are the similarities and differences between gravitational and electrostatic forces?

**Tip**: Always remember that forces involving charges follow an inverse square law, which means doubling the distance decreases the force by a factor of four.

Two point charges exert an 8.00 N force on each other. What will the force become if the distance between them is increased by a factor of six?

Learn how the electrostatic force between two point charges changes when the distance is increased by a factor of six. Using Coulomb's Law, we explain the relationship between force and distance and calculate the new force, which becomes 0.222 N from an initial 8.00 N.

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