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?
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:
\(F = k\frac{q_1q_2}{r^2}\)
Where:
- \(F\) is the force between the charges
- \(k\) is Coulomb's constant (\(8.99 x 10^9 N m^2 / C^2\))
- \(q_1\) and \(q_2\) are the magnitudes of the two point charges
- \(r\) is the distance between the two charges
Given:
\(q_1 = q_2 = 75.0 nC = 75.0 x 10^{-9} C\)
\(F = 1.00 N\)
\(k = 8.99 x 10^9 N m^2 / C^2\)
Substitute these values into Coulomb's law and solve for \(r\):
\(1.00 = (8.99 x 10^9) \frac{(75.0 x 10^{-9})(75.0 x 10^{-9})}{r^2}\)
\(1.00 = (8.99 x 10^9) \frac{5625 x 10^{-18}}{r^2}\)
Simplify:
\(1.00 = \frac{8.99 x 10^9 x 5625 x 10^{-18}}{r^2}\)
\(1.00 = \frac{50,656.25 x 10^{-9}}{r^2}\)
\(r^2 = \frac{50,656.25 x 10^{-9}}{1.00}\)
\(r^2 = 50,656.25 x 10^{-9}\)
\(r = \sqrt{50,656.25} x 10^{-4} = 71.171 x 10^{-4} = 0.71171 m\)
Therefore, the two point charges must be approximately 0.71171 meters apart to have a force of 1.00 N between them.