Two point charges repelling, stretching a spring

1. Jul 9, 2017

chopnhack

1. The problem statement, all variables and given/known data

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?

2. Relevant equations
1) F = k*(q^2)/r^2 where k is the constant of proportionality
2) F = kx where k is the spring constant

3. 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.

2. Jul 9, 2017

person123

Remember that the spring is originally 20 cm.

3. Jul 9, 2017

chopnhack

Yes, but they don't ask for the change in length but the final length. Unless I am missing something else you are hinting at?

4. Jul 9, 2017

person123

What does x stand for in the equation for the force of a spring?

5. Jul 9, 2017

chopnhack

distance when stretched

6. Jul 9, 2017

person123

It's the change in length when stretched.

7. Jul 9, 2017

chopnhack

so the force of the spring when stretched 0.4932m * 120N/m = 59N
the force of the particles apart from each other at 0.4932m --> F = 14.4Nm^2/(0.24324) = 59N

8. Jul 9, 2017

person123

Is 0.4932 m the change in length or the new length?

9. Jul 9, 2017

chopnhack

I took it to be the total new distance between the particles, hence the current length of the spring.

10. Jul 9, 2017

person123

And isn't x supposed to be the change in the length of the spring and not the new length of the spring?

11. Jul 9, 2017

chopnhack

that means that the force on the spring will only be 35N! 120n/m * deltaX. That means there is a problem with this approach. sigh....

My first approach at the problem was to equate the two directly together and the problem there was I got an answer of 3 meters that made no sense. As someone pointed out to me - check the value of force at 3m for each equation and you will see the problem. I did indeed see that the force at 3m for the particles was extremely small where as with the spring it was relatively high.

k = constant of proportionality

F = k ×(q1q2/r^2)

9x10^9 N×m2×C-2 × [(40μC^2)/0.2m^2]

F = 360N

k = spring constant = 120 N×m-1

F = kx
360N = 120 N×m-1 × x
x = 3m

so then if we change the equation to:

14.4/x^2 = 120*(x-0.2)
x = 0.569715m

12. Jul 9, 2017

person123

Why are you finding the original electrostatic force instead of the final?

13. Jul 9, 2017

chopnhack

In that particular attempt I thought I just needed to set the equation to each other to find their equivalency. But the two apparently weren't linear. The next attempt was only flawed in what you pointed out, the improper use of the x in hooke's law. Now that I corrected that I believe I go it. The values of force work out also.

14.4/x^2 = 120*(x-0.2)
x = 0.569715m

14. Jul 9, 2017

person123

My apologies. I agree with what you did and it matches up with my results so I think it should be correct.

15. Jul 9, 2017

chopnhack

I appreciate your efforts to help me! Thanks mate :-)