Question

A charged particle A exerts a force of 2.45 μN to the right on charged particle B when the particles are 12.2 mm apart. Particle B moves straight away from A to make the distance between them 18.2 mm. What vector force does particle B then exert on A?

Answer 1

`F_2 = 1.10 \mu N`

Step-by-step explanation:

As we know that the electrostatic force is a based upon inverse square law

so we have

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

now since it depends inverse on the square of the distance so we can say

`(F_1)/(F_2) = (r_2^2)/(r_1^2)`

now we know that

`r_2 = 18.2 mm`

`r_1 = 12.2 mm`

also we know that

`F_1 = 2.45 \mu N`

now from above equation we have

`F_2 = (r_1^2)/(r_2^2) F_1`

`F_2 = (12.2^2)/(18.2^2)(2.45\mu N)`

`F_2 = 1.10 \mu N`