Question

'F' is the force between the two identical charged particles placed at a distance 'Y' from each other. If the distance between the charges is reduced to half the previous distance then force between them becomes

• A. \(\frac{F}{4}\)
• B. \(4F\)
• C. \(2F\)
• D. \(\frac{F}{2}\)

Hint:

Remember: Force is inversely proportional to the square of separation. Halving distance multiplies force by 4; doubling distance divides force by 4.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Two identical charges experience force \(F\) at distance \(Y\). Distance is reduced to \(Y/2\). New force?

Step 2: Key Formula or Approach:
Coulomb’s law: \(F \propto \frac{1}{r^2}\).

Step 3: Detailed Explanation:
\[ \frac{F_2}{F_1} = \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{Y}{Y/2} \right)^2 = (2)^2 = 4. \] Thus \(F_2 = 4F\).

Step 4: Final Answer:
Force becomes \(4F\), option (B).