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.
Answer & explanation
Correct answer: option 2
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,
$\frac{kq_1q_2}{r^2}=\frac{k2q_12q_2}{(r')^2}$
$⇒(r')^2=4r^2$
$⇒r'=\sqrt{4r^2}=2r≡2d$