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

0.71 cm

Step-by-step explanation:

`q_1=75.0nC=75* 10^(-9)C`

`1nC=10^(-9)C`

`q_2=75.0nC=75* 10^(-9)C`

Force between two charges=1 N

Coulomb's law of force

`F=(kq_1q_2)/(r^2)`

Where k=
`9* 10^9Nm^2/C^2`

Using the formula

`1=(9* 10^9* 75* 10^(-9)* 75* 10^(-9))/(r^2)`

`r^2=0.50625* 10^(-4)`m

`r=\sqrt{0.50625* 10^(-4)}=0.71* 10^(-2) m`

`r=0.71* 10^(-2)* 10^(2)=0.71* 10^(-2+2)=0.71* 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