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

We can use Coulomb's law to calculate the distance between two point charges that would produce a force of 1.00 N between them.

Coulomb's law states that the force between two point charges is given by:

$F=k\frac{q_1{q}_2}{r^2}$

Where:
- $F$ is the force between the charges
- $k$ is Coulomb's constant ($8.99x{10}^{9}N{m}^2/C^2$)
- $q_1$ and $q_2$ are the magnitudes of the two point charges
- $r$ is the distance between the two charges

Given:
$q_1=q_2=75.0nC=75.0x{10}^{-9}C$
$F=1.00N$
$k=8.99x{10}^{9}N{m}^2/C^2$

Substitute these values into Coulomb's law and solve for $r$:

$1.00=(8.99x{10}^9)\frac{(75.0x{10}^{-9})(75.0x{10}^{-9})}{r^2}$

$1.00=(8.99x{10}^9)\frac{5625x{10}^{-18}}{r^2}$

Simplify:

$1.00=\frac{8.99x{10}^{9}x5625x{10}^{-18}}{r^2}$

$1.00=\frac{50,656.25x{10}^{-9}}{r^2}$

$r^2=\frac{50,656.25x{10}^{-9}}{1.00}$

$r^2=50,656.25x{10}^{-9}$

$r=\sqrt{50,656.25}x{10}^{-4}=71.171x{10}^{-4}=0.71171m$

Therefore, the two point charges must be approximately 0.71171 meters apart to have a force of 1.00 N between them.