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: Recall Coulomb's law.
The force between two charges goes as $F \propto \tfrac{1}{r^2}$. So halving the distance has a strong effect.

Step 2: Set up the ratio.
\[ \frac{F_2}{F_1} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{Y}{Y/2}\right)^2 = 2^2. \]

Step 3: Evaluate.
$2^2 = 4$, so $F_2 = 4F$.

Step 4: State it.
\[ \boxed{4F} \]