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: Read the change.
Two identical charges feel a force $F$ when they are a distance $Y$ apart. We bring them closer so the gap becomes $Y/2$. What is the new force?

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

Step 3: Write the ratio.
Comparing the two situations, $\frac{F_2}{F_1} = \left(\frac{r_1}{r_2}\right)^2$.

Step 4: Put in the distances.
Here $r_1 = Y$ and $r_2 = Y/2$, so $\frac{r_1}{r_2} = \frac{Y}{Y/2} = 2$.

Step 5: Square it.
$\frac{F_2}{F_1} = 2^2 = 4$.

Step 6: Find the new force.
So $F_2 = 4F$. Bringing the charges to half the distance makes the force four times bigger.
\[ \boxed{F_2 = 4F} \]