Question

How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them?

Answer 1

Answer:

0.71 cm

Explanation:

$q_1=75.0nC=75\times 10^{-9}C$

$1nC=10^{-9}C$

$q_2=75.0nC=75\times 10^{-9}C$

Force between two charges=1 N

Coulomb's law of force

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

Where k=$9\times 10^9Nm^2/C^2$

Using the formula

$1=\frac{9\times 10^9\times 75\times 10^{-9}\times 75\times 10^{-9}}{r^2}$

$r^2=0.50625\times 10^{-4}$m

$r=\sqrt{0.50625\times 10^{-4}}=0.71\times 10^{-2} m$

$r=0.71\times 10^{-2}\times 10^{2}=0.71\times 10^{-2+2}=0.71\times 10^0=0.71 cm$

Using formula

$1 m=10^2cm, a^x\cdot a^y=a^{x+y}, a^0=1$

Hence, the distance between two charges =r=0.71 cm