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) → 2d
By Coulomb's law, the force is proportional to the product of the charges and inversely proportional to the square of the distance.
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{d^{2}} \)
Doubling each charge makes the product four times. For the force to stay the same, \( d^{2} \) must also become four times.
\( \displaystyle \frac{(2q_1)(2q_2)}{d'^{2}} = \frac{q_1 q_2}{d^{2}} \;\Rightarrow\; d'^{2} = 4d^{2} \;\Rightarrow\; d' = 2d \)