Question

How far apart (in mm) must two point charges of 70.0 nC (typical of static electricity) be to have a force of 1.30 N between them?

Answer 1

Answer:r=5.824 mm

Explanation:

Given

Charge $q_1=70 nC$

$q_2=70 nC$

Force between them $F=1.30 N$

Electrostatic Force is given by

$F=\frac{kq_1q_2}{r^2}$

where $q_1$ and $q_2$ are the charge on the Particles

r=distance between them

$k=Coulomb\ constant =9\times 10^9 N-m^2/C^2$

$1.3=\frac{9\times 10^9\times 70\times 70\times 10^{-18}}{r^2}$

$r^2=\frac{9\times 10^9\times 70\times 70\times 10^{-18}}{1.3}$

$r=\sqrt{33.923\times 10^{-6}}$

$r=5.824\times 10^{-3}$

$r=5.824 mm$