Question

If the distance between two charges is doubled, what happens to the Coulomb force?

• A. It becomes double
• B. It becomes four times
• C. It becomes half
• D. It becomes one-fourth

Hint:

According to Coulomb’s law, electrostatic force varies inversely with the square of distance. If distance becomes \(2r\), force becomes \(\frac{1}{4}\). If distance becomes \(3r\), force becomes \( \frac{1}{9} \).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question explores the relationship between the electrostatic force and the distance separating two point charges.
Step 2: Key Formula or Approach:
According to Coulomb's Law, the force \( F \) between two charges is:
\[ F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \]
This implies that \( F \propto \frac{1}{r^2} \) (Inverse Square Law).
Step 3: Detailed Explanation:
Let the initial force be \( F_1 \) at distance \( r_1 \).
When the distance is doubled, the new distance \( r_2 = 2r_1 \).
The new force \( F_2 \) is:
\[ F_2 \propto \frac{1}{(r_2)^2} \]
\[ F_2 \propto \frac{1}{(2r_1)^2} \]
\[ F_2 \propto \frac{1}{4(r_1)^2} \]
Comparing this with the initial force:
\[ F_2 = \frac{1}{4} F_1 \]
Step 4: Final Answer:
If the distance is doubled, the Coulomb force becomes one-fourth of the original force.