Question

The electrostatic force between two small charged particles is 5.2 x 10^-7 N at a distance of 2 meters. What will be the new force if the distance is

increased to 4 meters apart?

Answer 1

`F_2=1.3* 10^(-7)\ N`

Step-by-step explanation:

Given that,

Initial force,
`F=5.2* 10^(-7)\ N`

distance, d = 2 m

New distance, d' = 4 m

We need to find the new force. The electrostatic force between two charges is given by :

`F=k(q_1q_2)/(d^2)`

So,

`(F_1)/(F_2)=((d_2)/(d_1))^2\\\\(F_1)/(F_2)=((4)/(2))^2\\\\(F_1)/(F_2)=4\\\\F_2=(F_1)/(4)\\\\F_2=(5.2* 10^(-7))/(4)\\\\F_2=1.3* 10^(-7)\ N`

So, the new force is equal to
`1.3* 10^(-7)\ N`.