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?

Explanation

To solve the problem of determining the number \( n \) of the last charged mercury drop that can still enter a spherical metal vessel, we can follow these steps:

### Step 1: Understand the Forces Acting on the Drop
When a charged drop of mercury falls into the spherical metal vessel, it experiences two main forces:
1. The gravitational force acting downwards, given by \( F_g = mg \), where \( m \) is the mass of the drop and \( g \) is the acceleration due to gravity.
2. The electrostatic force acting upwards due to the electric field created by the charge accumulated on the sphere.

### Step 2: Determine the Electric Field
When \( n \) drops have already entered the sphere, they create a total charge of \( nQ \) on the sphere. The electric field \( E \) at a distance \( r \) from the center of a charged sphere is given by:
\[ E = \frac{k \cdot nQ}{(h - R)^2} \]
where \( k \) is Coulomb's constant, \( n \) is the number of drops already inside, \( Q \) is the charge on each drop, and \( (h - R) \) is the distance from the drop to the center of the sphere.

### Step 3: Calculate the Electrostatic Force
The electrostatic force \( F_e \) acting on the drop due to the electric field is given by:
\[ F_e = Q \cdot E = Q \cdot \frac{k \cdot nQ}{(h - R)^2} = \frac{k \cdot nQ^2}{(h - R)^2} \]

### Step 4: Set Up the Equilibrium Condition
For the last drop (the \( n+1 \)th drop) to just enter the sphere, the gravitational force must equal the electrostatic force:
\[ mg = \frac{k \cdot nQ^2}{(h - R)^2} \]

### Step 5: Solve for \( n \)
Rearranging the above equation to solve for \( n \):
\[ n = \frac{mg(h - R)^2}{kQ^2} \]

### Step 6: Substitute Coulomb's Constant
Substituting \( k = \frac{1}{4\pi \epsilon_0} \):
\[ n = \frac{mg(h - R)^2}{\frac{1}{4\pi \epsilon_0} Q^2} = 4\pi \epsilon_0 mg(h - R)^2 \cdot \frac{1}{Q^2} \]

### Final Expression
Thus, the number \( n \) of the last drop that can still enter the sphere is given by:
\[ n = \frac{4\pi \epsilon_0 mg(h - R)^2}{Q^2} \]