Water from a tap, maintained at a costant potential V, is allolwed to fall by drops of radius r through a small hole into a hollow conducting sphere of radius R standing on an insulating stand until it fills the entire sphere. Find the potential of the hollow conductor after it is completely filled with water.

Answer

`v' = (R/v)^(2) v`

Explanation

Potential of a drop
`V = (1)/(4 pi epsilon_0 epsilon_r) q/( r)`
or `q = 4pi epsilon_0 epsilon_r r V`
Therefore, total charge
`Q = nq = 4 pi epsilon_0 epsilon_r r Vn`
(n = total number of drops)
Also, `(4)/(3) pi R^3 = n xx (4)/(3) pi r^3 or n = (R^3)/(r^3)`
Hence, potential of the hollow conductor
`V' = (1)/(4 pi epsilon_0 epsilon_r) Q/(R) = (1)/(4 pi epsilon_0 epsilon_r) xx (4 pi epsilon_0 epsilon_r r V n)/(R)`
=`r/R xx V xx ((R^3)/(r^3))`
=`((R)/( r))^2 V`.