Electric potential and distribution of charge

[Lanza52]

Homework Statement

A conducting sphere of radius r1 = 10 cm and charge q1 = 2 µC is placed far apart from a second conducting sphere of radius r2 = 30 cm and charge q2 = 3 µC. The two spheres are then connected by a thin conducting wire. What is the potential on the surface of the first sphere after the two spheres are connected by the wire? Use the reference V = 0 for r at infinity.

Homework Equations

Meh...

The Attempt at a Solution

A total of 5mC(using m for micro, not not milli) is in this system. When the wire is connected, the charge would distrbute evenly according given a surface charge density of 5mC/(area of sphere one + area of sphere two). So the charge on one would be (5mC/(area of 1+area of 2))(area of 1). And the voltage would be the line integral from infinity to 10x10^-2 of q/(4pie0r) dl)

I end up with 44,937.8V after all is said and done. But the "correct" answer is ~112,000V.

Going backwards from 112,100V, you get that V*4*pi*e0*r = Q from the line integral to find potential, or that the Q on the first sphere is 1micro coloumb. Which would require that the first sphere is 1/4 the area of the second sphere given a uniform charge distribution. Where as the area ratio is 1/9.5 or something close.

Any thoughts?

 

[ehild]

The Attempt at a Solution

A total of 5mC(using m for micro, not not milli) is in this system. When the wire is connected, the charge would distrbute evenly according given a surface charge density of 5mC/(area of sphere one + area of sphere two).

No. The charge will distribute between the spheres so that the potential of both spheres is the same. Otherwise the potential difference causes a driving force for the charges from one sphere to the other one.

ehild