Two charged particles, placed at a distance d apart in vacuum, exert a force F on each other. Now, each of the charges is doubled. To keep the force unchanged, the distance between the charges should be changed to ________.

Fill in the blank with the correct answer from the options given below.

$2 d$

The correct answer is Option (2) → $2 d$

According to Coloumbs law,

Force, $F = \frac{kq_1q_2}{r^2}$  $[r=d]$

[$r$ = Distance between the charge particle]

Now, let the new distance become $r'$

According to Question,

$\frac{kq_1q_2}{r^2}=\frac{k2q_12q_2}{(r')^2}$

$⇒(r')^2=4r^2$

$⇒r'=\sqrt{4r^2}=2r≡2d$