A small sphere of radius `r_(1)` and charge `q_(1)` is enclosed by a spherical shell of radius r and charge `q_(2)` Show that if `q_(1)` is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge `q_(2)` on the shell is.

Explanation

A small sphere of radius `r_(1)` and charge `q_(1)` is enclosed by a spherical shell of radius `r_(2)` and charge `q_(2)`. Their centres are coincide to each other.

Total potential on outer spherical surface = potential due to charge `q_(2)` + potential due to charge `q_(1)`.
`:. V_(r_(2))=(kq_(2))/(r_(2))+(kq_(1))/(r_(2))`
`:. V_(r_(2))=K[(q_(2))/(r_(2))+(q_(1))/(r_(2))]`
Potential due to charge `q_(1)` on inside sphere
`V_(1) = (Kq)/(r_(1))`
Potential due to charge `q_(2)` at every point inside shell of radius `r_(2)` is same as the potential on its surface. Hence potential due to charge `q_(2)` on the sphere of radius `r_(1)`.
`V_(2) = (Kq_(2))/(r_(2))`
Total potential on sphere of radius `r_(1)`.
`V_(r_(1))=K[(q_(1))/(r_(1))+(q_(2))/(r_(2))]`
[ From equation (2) and (3) ]
The potential difference of inner and outer shell ,
`:. V_(r_(1))-V_(r_(2))=K[(q_(1))/(r_(1))+(q_(2))/(r_(2))-(q_(2))/(r_(2))-(q_(1))/(r_(2))]`
`=K [(q_(1))/(r_(1))-(q_(1))/(r_(2))]`
`:. V_(r_(2))-V_(r_(2))=Kq_(1)[(1)/(r_(1))-(1)/(r_(2))]`
`q_(1)` is positive charge and potential of inner sphere is always higher and potential outside the shell is lower.
If the two are connected by a wire, the charge `q_(1)` will always flow from the inner sphere to the outer spherical shell although `q_(2)` is present.
This fact is true for any shape of sphere .