Question

when the distance between the charged particle is half the force between them becomes​

Answer 1

Answer:

the force becomes 4 times.

Explanation:

according to the formula of the force between charged particles

$f = k \times \frac{q1 \times q2}{ {r}^{2} }$

where k is proportionality constant

q1 & q2 are the two charges

and r is the distance between them.

If the distance gets half then bew force, F' will be

${f}^{.} = k \times \frac{q1 \times q2}{ ({ \frac{r}{2} })^{2} } = k \times \frac{4q1 \times q2}{ {r}^{2} }$

${f}^{.} = 4 \times f$