Question

Two point charges each experience a 1-N electrostatic force when they are 2 cm apart. If they are moved to a new separation of 8 cm, what is the magnitude of the electric force on each of them?

Answer 1

Step-by-step explanation:

Force between two point changes, F₁ = 1 N

Distance between them, r₁ = 2 cm = 0.02 m

We know that the electrostatic force is given by :

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

i.e.

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

i.e

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

Let F₂ is the force when the distance between the charges is 8 cm, r₂ = 0.08 m

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

`F_2=(1\ N* (0.02\ m)^2)/((0.08\ m)^2)`

F₂ = 0.0625 N

So, the distance between the sphere is 8 N, the new force is equal to 0.0625 N. Hence, this is the required solution.