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Two point charges repelling, stretching a spring

  • Thread starter chopnhack
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  • #1
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Homework Statement



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?

Homework Equations


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!
 

Answers and Replies

  • #2
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Remember that the spring is originally 20 cm.
 
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  • #3
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Remember that the spring is originally 20 cm.
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
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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?
What does x stand for in the equation for the force of a spring?
 
  • #5
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What does x stand for in the equation for the force of a spring?
distance when stretched
 
  • #7
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It's the change in length when stretched.
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
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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
Is 0.4932 m the change in length or the new length?
 
  • #9
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Is 0.4932 m the change in length or the new length?
I took it to be the total new distance between the particles, hence the current length of the spring.
 
  • #10
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I took it to be the total new distance between the particles, hence the current length of the spring.
And isn't x supposed to be the change in the length of the spring and not the new length of the spring?
 
  • #11
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And isn't x supposed to be the change in the length of the spring and not the new length of the spring?
:sorry::H
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
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9x10^9 N×m2×C-2 × [(40μC^2)/0.2m^2]
Why are you finding the original electrostatic force instead of the final?
 
  • #13
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Why are you finding the original electrostatic force instead of the final?
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
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My apologies. I agree with what you did and it matches up with my results so I think it should be correct.
 
  • #15
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I appreciate your efforts to help me! Thanks mate :-)
 

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