Two conducting spheres of radii `6 km` and `12 km` each, having the same charge `3xx10^(-8)C`, are kept very far apart. If the spheres are connected to each other by a conductiong wire, find
i. the direction and amount of charge transferred and
ii. final potential each sphere.

Explanation

i. As equal charge is given to both the spheres, the potential `V(propq//R)` of the smaller sphere will be higher and, hence, charge will flow from the smaller to the larger sphere when they are connected. Now, as charge is shared in proportion to capacity, and the capacity of the spherical conductor is proportional to its radius,
`q'_(1)=R_(1)/(R_(1)+R_(2))q=6/(6+12)(3+3)xx10^(-8)=2xx10^(-8)C`
And
`q'_(2)=R_(2)/(R_(1)+R_(2))q=(12)/(6+12)(3+3)xx10^(-8)=4xx10^(-8)C`
So charge transferred is
`(q_(1)-q'_(1))=(q'_(2)-q_(2))=1xx10^(-8)C`
ii. After sharing the final potential of each sphere,
`V'_(1)=(q'_(1))/C_(1)=(q'_(2))/(4piepsilon_(0)R_(0))=(9xx10^(9)xx2xx10^(-8))/(6xx10^(-2)) = 3kV`
and
`V'_(2)=(q'_(2))/C_(2)=(q'_(2))/(4piepsilon_(0)R_(2))=(9xx10^(9)xx4xx19^(-8))/(12xx10^(-2))=3kV`
So after sharing, both the conductors are at the same potential, i.e., `V'_(1)=V'_(2)=V=3kV`.