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.

Explanation

To find the potential of the hollow conducting sphere after it is completely filled with water, we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the Charge on a Single Drop**:
The potential \( V \) of a drop of radius \( r \) is given by the formula:
\[ V = \frac{kQ}{r} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \) is the Coulomb's constant and \( Q \) is the charge on the drop. Rearranging this gives:
\[ Q = \frac{Vr}{k} \]

2. **Determine the Number of Drops**:
The volume of the hollow conducting sphere of radius \( R \) is:
\[ V_{sphere} = \frac{4}{3} \pi R^3 \]
The volume of a single drop of radius \( r \) is:
\[ V_{drop} = \frac{4}{3} \pi r^3 \]
The number of drops \( n \) that can fit into the sphere is given by:
\[ n = \frac{V_{sphere}}{V_{drop}} = \frac{\frac{4}{3} \pi R^3}{\frac{4}{3} \pi r^3} = \left(\frac{R}{r}\right)^3 \]

3. **Calculate the Total Charge Inside the Sphere**:
The total charge \( Q' \) inside the sphere when filled with \( n \) drops is:
\[ Q' = nQ = n \left(\frac{Vr}{k}\right) = \left(\frac{R}{r}\right)^3 \cdot \frac{Vr}{k} \]
Simplifying this gives:
\[ Q' = \frac{VR^3}{kr^2} \]

4. **Calculate the Potential of the Hollow Sphere**:
The potential \( V_{big} \) of the hollow conducting sphere is given by:
\[ V_{big} = \frac{kQ'}{R} \]
Substituting \( Q' \) into this equation gives:
\[ V_{big} = \frac{k \cdot \frac{VR^3}{kr^2}}{R} = \frac{VR^2}{r^2} \]

### Final Result:
Thus, the potential of the hollow conductor after it is completely filled with water is:
\[ V_{big} = V \left(\frac{R}{r}\right)^2 \]