At what distance would the repulsive force between two electrons have a magnitude of 8.50 N? I have no idea how to do this problem... can someone help me please... 3x10^-4/8.50?
To determine the distance at which the repulsive force between two electrons has a magnitude of 8.50 N, we need to use Coulomb's law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Coulomb's law formula is given as:
F = k * (|q1| * |q2|) / r^2
Where:
F is the magnitude of the force between the particles,
k is the electrostatic constant (k = 9 x 10^9 N m^2/C^2),
|q1| and |q2| are the magnitudes of the charges on the particles, and
r is the distance between the charges.
In this case, we have two electrons, and since they both have the same charge, we can use the same absolute value, which is 1.6 x 10^-19 C.
Substituting the given values into the equation, we have:
8.50 N = (9 x 10^9 N m^2/C^2) * ((1.6 x 10^-19 C)^2) / r^2
To solve for r^2, we rearrange the equation:
r^2 = ((9 x 10^9 N m^2/C^2) * ((1.6 x 10^-19 C)^2)) / 8.50 N
Now, let's calculate it:
r^2 = ((9 x 10^9 N m^2/C^2) * (2.56 x 10^-38 C^2)) / 8.50 N
= (23.04 x 10^-29 N m^2/C^2) / 8.50 N
= 2.71 x 10^-29 m^2
To find the distance r, take the square root of both sides:
r = sqrt(2.71 x 10^-29 m^2)
r ≈ 1.65 x 10^-15 m
Therefore, the distance at which the repulsive force between two electrons has a magnitude of 8.50 N is approximately 1.65 x 10^-15 meters.
Use Coulomb's law.
F = k Q^2/R^2
In this case, Q is the electron charge and R is the separation distance.
Set F = 8.50 N and solve for the separation R. Your course material should tell you the value of k, if you do not know it. You can also find it many places online, such as http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l3b.html