Question

If the distance between two charges is doubled, the Coulomb force becomes:

• A. Twice
• B. Half
• C. One-fourth
• D. Four times

Hint:

In Coulomb's law, force varies inversely with the square of distance. If distance doubles, force becomes \(1/4\); if distance triples, force becomes \(1/9\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question tests the relationship between electrostatic force and the distance separating two point charges.
Step 2: Key Formula or Approach:
According to Coulomb's Law:
\[ F = k \frac{q_1 q_2}{r^2} \]
This shows an inverse-square law relationship, where \(F \propto \frac{1}{r^2}\).
Step 3: Detailed Explanation:
Let the initial force be \(F_1 = \frac{k q_1 q_2}{r^2}\).
If the new distance \(r_2 = 2r\), the new force \(F_2\) is:
\[ F_2 = \frac{k q_1 q_2}{(2r)^2} \]
\[ F_2 = \frac{k q_1 q_2}{4r^2} \]
Substituting \(F_1\) into the equation:
\[ F_2 = \frac{1}{4} F_1 \]
Step 4: Final Answer:
The force becomes one-fourth (\(1/4\)) of the original value.