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

Explanation:

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

$F\propto \dfrac{1}{r^2}$

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,

$\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2\\\\F_2=\dfrac{F_1r_1^2}{r_2^2}\\\\F_2=\dfrac{1\times 2^2}{(0.25)^2}\\\\F_2=64\ N$

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