how far apart are two point charges of 5.0x10^-6 C and 6.0x10^-6 C if they produce an electric force on each other of 0.62 N?
Coulomb's law is discussed here:
http://www.physicsclassroom.com/Class/estatics/U8L3b.cfm
just plug in your numbers.
To find the distance between two point charges, we can use Coulomb's law, which states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's law is:
F = k * (q1 * q2) / r^2
where:
F is the electric force between the charges,
k is the electrostatic constant (9.0 x 10^9 N m^2/C^2),
q1 and q2 are the magnitudes of the charges,
and r is the distance between them.
In this case:
F = 0.62 N,
q1 = 5.0 x 10^-6 C,
q2 = 6.0 x 10^-6 C,
and we need to find r.
Rearrange the formula to solve for r:
r^2 = k * (q1 * q2) / F
Plugging in the given values:
r^2 = (9.0 x 10^9 N m^2/C^2) * (5.0 x 10^-6 C * 6.0 x 10^-6 C) / 0.62 N
Calculate the right side of the equation:
r^2 = (9.0 x 10^9 N m^2/C^2) * (30.0 x 10^-12 C^2) / 0.62 N
r^2 = (9.0 x 10^9 N m^2/C^2) * (0.30 x 10^-11 C^2) / 0.62 N
r^2 = 13.7 x 10^-2 m^2
Take the square root of both sides to find r:
r = √(13.7 x 10^-2 m^2)
Simplify the expression:
r ≈ 0.117 m
Therefore, the distance between the two charges is approximately 0.117 meters.