Two spherical point charges each carrying a charge of 40 mC are attached to the two ends of a spring of length 20 cm. If its spring constant is 120 Nm-1, what is the length of the spring when the charges are in equilibrium?
1) F = k*(q^2)/r^2 where k is the constant of proportionality
2) F = kx where k is the spring constant
The Attempt at a Solution
Both equations can be graphed and see where they intersect. I had some trouble getting my calculator to do that, so I set both equations equal to each other and then solved for x.
1) F = 14.4Nm^2/r^2
2) F = 120N/m * r
14.4Nm^2/r^2 = 120N/m * r
x = 0.493242m
Is the logic sound? If you try to solve for r, the forces do not equate.
Thanks in advance!