Capacitance of two spherical conductors

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Two spherical conductors connected by a wire and charged with 7 µC will have their charge distributed to ensure equal electric potential across both spheres. The charge distribution can be determined using the relationship that the potential difference between the spheres is zero, leading to the conclusion that charge will move until equilibrium is reached. Although the potential of the system can be considered additive (V1 + V2), it is not necessary for solving the charge distribution problem. The use of total charge and the sum of the radii helped derive the correct potential difference, demonstrating a practical approach despite initial confusion. Understanding that the charges equalize potential is crucial for solving such problems.
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Given that two spherical conductors of radii r1 and r2 are separated by
x>>(both r), connected by a thin wire, and charged with 7 uC - how is the charge distributed between the spheres (no charge on wire)?

I tried using charge density but:

Qtot/A1+A2 = Q1/A1 + Q/A2 doesn't work by definition

then maybe:

C(dV) = Q for either sphere - but how do you find dV without knowing the charge on each first? However, I got the correct answers when I flukily got dV by using the total charge of the system (7uC) and the sum of the radii (r1+r2) and multiplying by C of each sphere. I don't understand why this worked though.
 
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Well,i don't follow your "proof".I would advise you to use the fact that the charges on the 2 spheres are distributed in such a way,as to insure equal electric potential for each sphere...

Daniel.
 
why do they have to have the same electric potential?
 
Because if they didn't charge would be moving from one to another until they did.So better assume that they already do.

Daniel.
 
ok, I see. This means that the potential difference between them is 0 right? Also, would the potential of the system be V1 + V2?
 
Yes to the first question...As for the second,you don't need that piece of information.

Daniel.
 
Ok, but what if I wanted to know the potential of the system - would it be V1 + V2?
 
Yes,potential (electric) is additive.But,again,you don't need this info.

Daniel.
 

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