Question

Constants Part A wo charged dust particles exert a force of 7.5x102 N n each other What will be the force if they are moved so they are only one-eighth as far apart? Express your answer using two significant figures.

Answer 1

New force,
`F'=48* 10^3\ N`

Step-by-step explanation:

It is given that,

Force acting between two charged particles,
`F=7.5* 10^2\ N`

We need to find the force if they are moved so they are only one-eighth as far apart.

The force between two charged particles separated at a distance of r is given by :

`F=k(q_1q_2)/(r^2)`............(1)

If the charges are one-eighth as far apart then, r' =(1/8)r and new force is given by :

`F'=k(q_1q_2)/(((r)/(8))^2)`..........(2)

Dividing equation (1) and (2) :

`(F)/(F')=(1)/(64)`

`F'=7.5* 10^2\ N* 64`

F' = 48000 N

or

`F'=48* 10^3\ N`

Hence, this is the required solution.