Q110 / 150 · MHT CET 2026 (20 April Shift 1) · Physics
Two point charges $q_1$ and $q_2$ are '$l$' distance apart. If one of the charges is doubled and the distance between them is halved. The magnitude of the force becomes 'n' times, where 'n' is
- A$2$
- B$4$
- Correct answerC$8$
- D$16$
Solution
Initial force between the charges is given by Coulomb's law:
$F = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{l^2}$
When one of the charges is doubled, let $q_1' = 2q_1$ and $q_2' = q_2$.
When the distance between them is halved, let $l' = \dfrac{l}{2}$.
The new force is:
$F' = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1' q_2'}{l'^2}$
$F' = \dfrac{1}{4\pi\epsilon_0} \dfrac{2q_1 q_2}{\left(\dfrac{l}{2}\right)^2}$
$F' = \dfrac{1}{4\pi\epsilon_0} \dfrac{2q_1 q_2}{\dfrac{l^2}{4}}$
$F' = 8 \left( \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{l^2} \right)$
$F' = 8F$
Thus, the magnitude of the force becomes $8$ times the initial force, so $n = 8$.
Answer: $8$
$F = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{l^2}$
When one of the charges is doubled, let $q_1' = 2q_1$ and $q_2' = q_2$.
When the distance between them is halved, let $l' = \dfrac{l}{2}$.
The new force is:
$F' = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1' q_2'}{l'^2}$
$F' = \dfrac{1}{4\pi\epsilon_0} \dfrac{2q_1 q_2}{\left(\dfrac{l}{2}\right)^2}$
$F' = \dfrac{1}{4\pi\epsilon_0} \dfrac{2q_1 q_2}{\dfrac{l^2}{4}}$
$F' = 8 \left( \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{l^2} \right)$
$F' = 8F$
Thus, the magnitude of the force becomes $8$ times the initial force, so $n = 8$.
Answer: $8$
Hence, the correct answer is option: 'C'.