Question

What separation distance do two point charges of +1.0µC and −1.0µC exerts a force of attraction on each that is 440N? 

Answer 1

According to Coulomb's Law,

F = $\frac{k.q1.q2}{r^2}$

Here,

F = 440 N

k = 9 x 10⁹

q₁ = 1 x 10^-6 C

q₂ = -1 x 10^-6 C

Hence,

r² = $\frac{k.q1.q2}{F}$

= (9 x 10⁹ x 1 x 10^-6 x 1 x 10^-6) / 440

r² = $\frac{9}{440000}$

Hence,

r = 0.0045 m

= 4.5 mm