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
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?
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