Two identical teflon rods are 10 centimeters long and rubbed with fur so that they each have a total negative charge of 20 microCoulombs that is uniformly distributed along their length. They are arranged along the same axis, with their ends 5 centimeters apart. What is the magnitude of the electrostatic force felt by each rod in Newton?

dq = 20^10^-6 dx/.1

dF = k (dq)^2/(2x)^2
with x from .025 to .125

so do double integral
F on one due to dq on the other
dF = dqleft (integral k dq/(2x)^2) from x = .025 to .125
let INT = (integral k dq/(2x)^2) from x = .025 to .125
or
dF = INT * integral dqleft from x = -.125 to x = -.025

To find the magnitude of the electrostatic force felt by each rod, we can use Coulomb's Law. Coulomb's Law states that the magnitude of the electrostatic force between two charged objects is given by:

F = (k * |q1 * q2|) / r^2

Where:
F is the electrostatic force
k is the electrostatic constant (9 × 10^9 Newton meter squared per coulomb squared)
q1 and q2 are the magnitudes of the charges on the two objects
r is the distance between the centers of the two objects

In this case, both rods have the same charge of -20 microCoulombs (-20 × 10^-6 C) and the distance between their centers is 5 centimeters (0.05 meters).

Plugging in the values:

F = (9 × 10^9 Nm^2/C^2) * |-20 × 10^-6 C * -20 × 10^-6 C| / (0.05 m)^2

Simplifying:

F = (9 × 10^9 Nm^2/C^2) * (20 × 10^-6 C)^2 / (0.05 m)^2

Calculating:

F = (9 × 10^9 Nm^2/C^2) * (400 × 10^-12 C^2) / (0.0025 m^2)

F = (9 × 10^9 Nm^2/C^2) * (4 × 10^-10 C^2) / (2.5 × 10^-3 m^2)

F = 36 × 10^-1 N

F = 3.6 N

Therefore, the magnitude of the electrostatic force felt by each rod is 3.6 Newtons.