The electrostatic force between two charges q1 and q2 is given by
where
is the Coulomb's constant
is the distance between the two charges.
In our problem, the two charges are two electrons, so their charges are equal and equal to
By substituting these values, we find the intensity of the force between the two electrons:
This is the magnitude of the force each electron exerts to the other one. The direction is given by the sign of the charges: since the two electrons have same charge, they repel each other, so the force exerted by electron 1 is toward electron 2 and viceversa.
Answer:
0.306mm
Explanation:
The radius of the conductor is 3mm, or 0.003m
The area of the conductor is:
A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2
The current density is:
J = 130/2.8*10^-5 = 4.64*10^6 A/m
According to the listed reference:
Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s
The distance traveled is:
x = v*t = 0.34 * .90 = 0.306 mm