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Electrically charged drops of mercury fall from an altitude h into a spherical metal vessel of radius R. There is a small opening in the upper part of the vessel. The mass of each drop is m, and the charge on the drop is Q. What will be the number n of the last drop that can still enter the sphere?

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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 crates its own electric field that can be calculated by assuming the electric charge of the sphere to be concentrated at its center. Let n drops have accumulated in the vessel and (n + 1) th be in the state of equilibrium at a height h. The equilibrium exists under the influence of repulsive coulomb force (directed upward) and force of gravity (downward). Thus, we have
`((n Q)Q)/(4 pi epsilon_0(h - R)^2 ) = m g`
Hence
`n = (4 pi epsilon_0 m g(h - R)^2)/(Q^2)`.
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