Question

how does the electric force between two charged particles change if the distance between them is decreased by a factor of 3?

Answer 1

it is increased by a factor of 9

Step-by-step explanation:

just took the test

Answer 2

F' = 9F

Step-by-step explanation:

The electric force between charges is given by :

`F=(kq_1q_2)/(r^2)`

It means,

`F\propto (1)/(r^2)`

If distance decreased by a factor of 3, r' = r/3

Net force will become,

`F'=(kq_1q_2)/(r'^2)\\\\=(kq_1q_2)/(((r)/(3))^2)\\\\=9(kq_1q_2)/(r^2)\\\\=9* F`

It means new force will become 9 times that of the initial force.