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

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 :

$F=k\dfrac{q_1q_2}{r^2}$

i.e.

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

i.e

$\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2$

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

$F_2=\dfrac{F_1\times r_1^2}{r_2^2}$

$F_2=\dfrac{1\ N\times (0.02\ m)^2}{(0.08\ m)^2}$

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.