Spring constant versus electric repulsion

[chopnhack]

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.

 

[(*steve*)]

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

 

[chopnhack]

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]

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

 

[Arouse1973]

Ok just realised C-2 is part of the units.
Adam

 

[(*steve*)]

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]

You have 2 issues.

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

 

[Laplace]

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

 

[chopnhack]

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.

 

[Laplace]

I took it as the spring is 20cm long when unstretched, then the charges are added.

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

 

[chopnhack]

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]

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]

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

 

[Laplace]

x=0.569715m

That's the answer I calculated.

 

[chopnhack]

That's the answer I calculated.

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

 

[Ratch]

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

He should write 1/C^2 to avoid confusion and bemusement.

Ratch

 

[(*steve*)]

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