Question

Two point charges, initially 2.0 cm apart, experience a 1.0 N force. If they are moved to a new separation of 0.25 cm, what is the electric force between them (in N)?

Answer 1

Step-by-step explanation:

Th electric force between charges is inversely proportional to the square of distance between them. It means,

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

Initial distance, r₁ = 2 cm

Final distance, r₂ = 0.25 cm

Initial force, F₁ = 1 N

We need to find the electric force between charges if the new separation of 0.25 cm. So,

`(F_1)/(F_2)=((r_2)/(r_1))^2\\\\F_2=(F_1r_1^2)/(r_2^2)\\\\F_2=(1* 2^2)/((0.25)^2)\\\\F_2=64\ N`

So, the new force is 64 N if the separation between charges is 64 N.