Question

The distance between the two identical charges is reduced by$\frac{1}{4}$. The force of repulsion between them is

• A. $\frac{1}{4}$of initial force
• B. $4$ times the initial force
• C. $16$ times the initial force
• D. $8$ times the initial force

Answer 1

The correct option is C

$16$ times the initial force

Step 1: Given and assumptions:

The magnitude of each charge$=q$

The distance between them$=r$

Step 2: Calculation of force:

$F=\frac{K.\left(q\right).\left(q\right)}{r^2}=\frac{K.q^2}{r^2}$ --------(i)

If the distance between them is reduced by $\frac{1}{4}$ then the force becomes $F\text{'}$

$$\begin{gathered} F\text{'}=\frac{K.\left(q\right).\left(q\right)}{{(r/4)}^2}=\frac{16.K.q^2}{r^2} \\ fromequation\left(i\right) \\ F\text{'}=16F \end{gathered}$$

Therefore force is 16 times the initial value.

Hence, the option $\left(c\right)$ is correct.