Two point charges placed a distanced apart in vacuum exert a force of magnitude F on each other. One of the two charges is doubled. To keep the magnitude of force same the separation between the charges should be changed to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $sqrt{2}d$
Given:
Initial force between charges, $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$
When one charge is doubled, the new force becomes
$F' = \frac{1}{4\pi\varepsilon_0}\frac{(2q_1)q_2}{r'^2} = \frac{2}{4\pi\varepsilon_0}\frac{q_1 q_2}{r'^2}$
For the force to remain same, $F' = F$
$\Rightarrow \frac{2}{r'^2} = \frac{1}{r^2}$
$\Rightarrow r'^2 = 2r^2$
$\Rightarrow r' = r\sqrt{2}$
∴ The separation between the charges should be increased to $r\sqrt{2}$.