Charge Sharing between two metal spheres

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SUMMARY

The discussion focuses on the charge sharing process between two metal spheres, one with a radius of 0.48 cm and a charge of 0.15 µC, and the other with a radius of 0.95 cm and a charge of 0.090 µC. When connected by a thin wire, the total charge of 0.24 µC redistributes until both spheres reach the same electric potential. The relationship between charge and potential is expressed as (kQ1)/R1 = (kQ2)/R2, leading to the conclusion that the final charge on each sphere can be calculated to be 6.94 x 10-2 µC.

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  • Familiarity with the concept of electric potential (V)
  • Knowledge of Coulomb's law and its application
  • Basic algebra for solving equations
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This discussion is beneficial for physics students, educators, and anyone interested in understanding electrostatics and charge distribution in conductive materials.

diesel55
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Two small metal spheres are located 3.0 m apart. One has radius 0.48 cm and carries charge 0.15 uC. The other has radius 0.95 cm and carries charge 0.090 uC.

If the two spheres were suddenly connected by a thin wire, how much charge would move along it?

This is a question that I have been trying to solve for over a week and still can't get. Please Help. I need a detailed explanation because I know this concept will be used in later chapters.

I know I will add Q1 + Q2 = .24 uC
I was also told that V1 = V2
So (kQ1)/R1 = (kQ2)/R2

I was told to solve for one and plug it into the other, but I don't see the logic.
 
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If someone could please help me with this problem. I know the answer is 6.94*10-2 uC but I have no idea how to get that answer. I need to know this process so I can apply it in my current homework.
 
you want the final charges, it means they should come to equilibrium,it means that there shouldn't be any current between them, it happens when they have the same potential.
So assuming Q1 over first one and Q2 over second one when thay have radiuses R1 and R2
(kQ1)/R1 = (kQ2)/R2
and using Q1+Q2=Qtotal you can find charge on each one.
 

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