Two identical like-charged conductive spheres are placed very close

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When two identical like-charged conductive spheres are placed close together, their charges redistribute due to repulsion, moving to the far ends of each sphere. This redistribution results in an effective center-to-center distance, rc, that is larger than the physical center-to-center separation, r, between the spheres. The charges on the surfaces of the spheres create a scenario where the forces between them can be described by Coulomb's law. The discussion highlights the importance of understanding charge distribution in conductive materials. Ultimately, the effective distance between the charges increases as a result of the repulsive forces acting on them.
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Homework Statement


Two identical like-charged conductive spheres are placed very close to each other, the charges willl redistribute themselves on the sphere's surface with an effective center-to-center distance rc so as to approximately satisfy the Coulomb's law, F = ke q1 q2 / (rc)^2. Is the value of rc larger or smaller than the physical center-to-center separation r between the two spheres? Explain

Homework Equations


The Attempt at a Solution


I guess the surfaces between the spheres will become neutral, and the charges will redistribute on the other side in each sphere, then i have no idea what to do next with the question.
I have not study for five years, forget most of them already, please help, thx
 
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You're right. Due to repulsion, the charges will end up on the far ends of the spheres. Hence the distance between the charges will be greater than the center to center distance between the spheres.
 
Okay, thanks very much
 
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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