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?

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).

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).