If the same setup as above but the distance between the charges is now three times greater, how does the force compare to its original value?

• A. It becomes nine times smaller than the original
• B. It will still be four times the force, but divided by 9 since the distance is squared.
• C. It doubles the force
• D. It remains the same

Answer: (B)

The main idea is the inverse-square law: the electrostatic force between two charges goes as F ∝ 1/r^2. If you increase the separation by a factor of 3, the distance becomes 3r, so the denominator r^2 becomes (3r)^2 = 9r^2. That makes the force 1/9 of its original value. In other words, the new force is divided by 9. This directly reflects why increasing distance reduces force, not by a constant amount but by the square of the distance change. The correct description is that the force becomes one ninth of what it was originally (divided by 9 due to the 1/r^2 relationship).