Electrically charged drops of nercury fall from an altitude h into a sperical metal vessel of radius R. There is a small opening in the upperapart of the vessel. The mass of eanc drop is m, and the charge on the drop is Q. What will be the number n of the last drop taht can still enter the sphere?

Answer

`(4 pi epsilon_(0)mg(h-R)^(2))/(Q^(2))`

Explanation

Each charged drop that falls into the conducting vessel increases the charge of the vessel by Q. The accumulated charge on the conducting vessel is uniformly distributed on the surface of the vessel. The charged spherical vessel creates its own electric field that can be calculated by assuming the entire charge of the sphere to be concentrated at its center. Let n drops have accumulated in vessel and (n +1) th be in the state of equilibrium at a height h. The equilibirium exists under the influence of repulsive Coulomb force (directed upward) and force of gravity (downward) thus, we have
`((nQ)Q)/(4 pi epsilon_(0)(h-R)^(2))=mg`
Hence `n=(4 pi epsilon_(0)mg(h-R)^(2))/(Q^2)`.