How Do Charges on Beads Affect Spring Stretching?

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The discussion revolves around the mechanics of a spring system with charged beads. A lightweight spring stretches when a mass is hung from it, and further stretches when charged beads are attached. The key point is the equilibrium condition where the restoring force of the spring equals the electric force exerted by one bead on the other. There is confusion about whether the restoring force should be considered as equal to two electric forces due to the interaction between the beads. Ultimately, the consensus is that only one electric force is needed for calculations, as the forces acting on each bead are balanced in the system.
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The question states:

"You have a lightweight spring whose unstretched length is 3.28 cm.
You're curious to see if you can use this spring to measure charge.
First, you attach one end of the spring to the ceiling and hang a 2.57 g
mass from it. This stretches the spring to a length of 4.37 cm. You then
attach two small plastic beads to the opposite ends of the spring, lay
the spring on a frictionless table, and give each plastic bead the same
charge. This stretches the spring to a length of 3.78 cm. What is the
magnitude of the charge on each bead?"

So, after finding the spring constant using the mass hung from the
ceiling, we can calculate the restoring force of the spring in the
spring-2bead system. This restoring force must be equal to the electric
force exerted by the beads since the spring is at equilibrium. This is
the part that I'm having trouble with. Shouldn't the restoring force be
equal to 2x the electric force? Each bead will exert a force on the
other and cause each to stretch the spring in opposite directions. In other words, the electric force of each bead on the other (so there are two forces) is what actually causes the spring to stretch to its equilibrium length. If we calculate using only one electric force, aren't we making the assumption that one bead is held still in position?

The answer as it stands is found by equating the restoring force to only one electric force so where have I gone wrong?

Thanks.
 
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nothing123 said:
If we calculate using only one electric force, aren't we making the assumption that one bead is held still in position

Hi nothing123! :smile:

But if one bead was held still in position, wouldn't the length of the spring be the same? :smile:
 
Yes, you're right and I have reasoned it this way. Nevertheless, if one wasn't held in place, why would the result be the same?
 
Why shouldn't it be? :confused:

How could any other part of the spring "know" what was happening at one end? :smile:
 
This is
the part that I'm having trouble with. Shouldn't the restoring force be
equal to 2x the electric force?

We are applying Newton's second law on the Bead. There are only two forces working on it-electrostatic and spring force. So F_s=F_q not F_s=2F_q
 

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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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