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Spring constant versus electric repulsion

chopnhack

Apr 28, 2014
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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.
 

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What is the force between the two charges when they are 3 metres apart?
 

chopnhack

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What is the force between the two charges when they are 3 metres apart?
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

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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
Adam
 

(*steve*)

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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.
 

chopnhack

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You have 2 issues.

I think I have a lot more than two issues right now :mad::confused::p:D

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
 
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Laplace

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

chopnhack

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

chopnhack

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

Laplace

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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.)
 

chopnhack

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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
 

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