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Two initially uncharged identical metal spheres, 1 and 2, are connected by an insulating spring (unstretched length L0 = 1.39 m, spring constant ks = 20.7 N/m), as shown in the figure. Charges +q and –q are then placed on the spheres, and the spring contracts to length L = 0.51 m. Recall that the force exerted by a spring is Fs = -ks Δx, where Δx is the change in the spring’s length from its equilibrium length.

a) Determine the charge q.

b) If the spring is coated with metal to make it conducting, what is the new length of the spring?

For a) I got 2.29*10^-5 C which is correct, I just have no idea how to figure out b.

Bailey
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    This seems to be a physics question, not a mathematical one. Unfortunately I seem to recall that [Physics.SE] takes a dimmer view of homework questions than this site does. – hmakholm left over Monica Sep 28 '16 at 12:17
  • You may want to consider asking over at the physics forum http://physics.stackexchange.com/ – Carser Sep 28 '16 at 12:17
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    Also, $2.29\times 10^{-6}$ cannot possibly be a correct answer to part (a), since the question asked for a charge rather than a dimensionless number. – hmakholm left over Monica Sep 28 '16 at 12:20
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    What will happen to the charges when the spring becomes conducting? What will this mean for the forces? – user121049 Sep 28 '16 at 12:22
  • I know that the spring will stretch, I am just not sure how to calculate how far. – Bailey Sep 28 '16 at 12:26
  • What happens in a conductor placed between two charges? – Neal Sep 28 '16 at 12:29
  • Will charges spread evenly across it? I'm not sure if that makes sense, I just know that all points on a conductor have the same potential so that's kind of where my train of thought was going. If that is true, will the spring just return to its original length? – Bailey Sep 28 '16 at 12:34
  • The charges will cancel out so there is no longer any force acting on the spring. – user121049 Sep 28 '16 at 17:01

1 Answers1

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a.

We use the equation $$ F = \frac{kq1q2}{d^2} $$

$$ F = \Delta X spring constant $$ q1 = q2 so it becomes q^2
spring constant = 20.7
d = .51
delta x = 1.39 - .51
k = coulombs constant = 8.99x10^9

so $$ .88 * 20.7= \frac{8.99*10^9q^2}{.51^2}$$ so $$ sqrt(\frac{(.88 * 20.7).51^2}{8.99*10^9}) = q$$

Q = -2.2295x10^-5

b. The net charge = 0 if charges are connected. The answer is the original length of the spring 1.39

user484926
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