512 identical drops of mercury are charged to a potential of 2V each. The drops are joined to form a single drop. The potential of this drop is ___ V.

Answer

To solve the problem of finding the potential of a single drop formed by combining 512 identical drops of mercury, each charged to a potential of 2V, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Volume of the Drops**:
Each small drop has a volume given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of a small drop. Therefore, the total volume of 512 drops is:
\[ V_{\text{total}} = 512 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (512 r^3) \]

2. **Volume of the Larger Drop**:
The volume of the larger drop (formed by combining the smaller drops) can also be expressed as:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
where \( R \) is the radius of the larger drop.

3. **Equating the Volumes**:
Since the total volume remains constant when the drops are combined, we can set the two volume equations equal to each other:
\[ \frac{4}{3} \pi (512 r^3) = \frac{4}{3} \pi R^3 \]
Cancelling \(\frac{4}{3} \pi\) from both sides gives:
\[ 512 r^3 = R^3 \]

4. **Finding the Radius of the Larger Drop**:
Taking the cube root of both sides:
\[ R = \sqrt[3]{512} \cdot r = 8r \]

5. **Calculating the Charge**:
The charge \( q \) on each small drop can be related to its potential. The potential \( V \) of a small drop is given by:
\[ V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r} \]
Given that \( V = 2V \), we can express the charge \( q \) as:
\[ 2 = \frac{1}{4 \pi \epsilon_0} \frac{q}{r} \implies q = 2 \cdot 4 \pi \epsilon_0 \cdot r \]

6. **Total Charge in the Larger Drop**:
The total charge \( Q \) in the larger drop is:
\[ Q = 512q = 512 \cdot (2 \cdot 4 \pi \epsilon_0 \cdot r) = 1024 \cdot 4 \pi \epsilon_0 \cdot r \]

7. **Potential of the Larger Drop**:
The potential \( P \) of the larger drop can be expressed as:
\[ P = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R} \]
Substituting \( Q \) and \( R \):
\[ P = \frac{1}{4 \pi \epsilon_0} \frac{1024 \cdot 4 \pi \epsilon_0 \cdot r}{8r} \]
Simplifying this gives:
\[ P = \frac{1024 \cdot 4 \pi \epsilon_0 \cdot r}{32 \pi \epsilon_0 \cdot r} = 64 \cdot 2 = 128V \]

### Final Answer:
The potential of the larger drop is **128 V**.