Spring constant versus electric repulsion

Discussion in 'Electronics Homework Help' started by chopnhack, Jul 9, 2017.

1. chopnhack

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Apr 28, 2014
Is this question this simple to answer? I feel like some of these questions are too easy....Makes me uneasy like I am missing something!

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?

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

When in equilibrium, the springs are stretched to 3 meters.

2. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
What is the force between the two charges when they are 3 metres apart?

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3. chopnhack

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Apr 28, 2014
F = k ×(q1q2/r^2)
9x10^9 N×m2×C-2 × [(40μC^2)/3m^2]
F=1.6x10^-6N this is the force of the two particles acting against each other
the force of the spring at 3m stretch would be 360N

I need to rethink this. Thanks for pointing that out! Knew it was too easy.

Arouse1973 likes this.

5,164
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Dec 18, 2013
Hi John
What is the C-2? Also is it micro or milli-Coulombs. You say mC in your post and then your calculations show uC.
Cheers

5,164
1,087
Dec 18, 2013
Ok just realised C-2 is part of the units.

6. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
You have 2 issues. The force between the two charges falls as the distance between them increases, and the spring tension (which is a force acting in the other direction) increases as the distance between it's ends increases

You need to find a distance where these forces balance.

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7. chopnhack

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Apr 28, 2014
I think I have a lot more than two issues right now

This seems like it needs to be solved by a system of equations.

I was thinking of graphing the two against each other to see where they cross.

first equation boils down to y = 14.4/x^2 the second equation is y = 120x.

x = 0.493242m

Last edited: Jul 10, 2017
8. Laplace

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Apr 4, 2010
How do you account for the fact that when the charges are 20 cm apart the force in the spring is zero?

9. chopnhack

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Apr 28, 2014
I didn't assume that when I read the question. I took it as the spring is 20cm long when unstretched, then the charges are added.

10. Laplace

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Apr 4, 2010
So was the question, "How far does the spring stretch?" or "How far apart are the charges in equilibrium?"

11. chopnhack

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Apr 28, 2014
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?

12. Laplace

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Apr 4, 2010
So the spring does not exert any force until the length of the spring exceeds 20 cm. Is that what your equation for the spring force indicates? (the second equation is y = 120x.)

13. chopnhack

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Apr 28, 2014
I believe so, since the original length is 0.2m. It doesn't start exerting force until it is stretched beyond 0.2m. So it should be amended to y=120*(x-0.2m)?

14.4/x^2 = 120(x-0.2)

x=0.569715m

14. Laplace

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Apr 4, 2010

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15. chopnhack

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Apr 28, 2014
Thank you for helping me through that! I don't think I would have spotted that on my own.... physics is tricky!!

16. Ratch

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Mar 10, 2013
He should write 1/C^2 to avoid confusion and bemusement.

Ratch

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17. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
C^-2 or C-2 would both be clearer AND consistent with what he is required to use.