Eight small drops, each of radius `r` and having same charge q are combined to form a big drop. The ratio between the potentials of the bigger drop and the smaller drop is

Options

A. `8:1`
B. `4:1`
C. `2:1`
D. `1:8`

Answer

To solve the problem of finding the ratio between the potentials of a bigger drop and smaller drops, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We have 8 small drops, each with a radius \( r \) and charge \( q \). When combined, they form a larger drop. We need to find the ratio of the electric potential of the larger drop to that of a smaller drop.

2. **Volume of the Drops**: The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
For 8 small drops, the total volume is:
\[ V_{\text{small}} = 8 \times \frac{4}{3} \pi r^3 = \frac{32}{3} \pi r^3 \]
This volume equals the volume of the larger drop, which we denote as \( R \):
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]

3. **Equating Volumes**: Setting the volumes equal gives us:
\[ \frac{32}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ 8r^3 = R^3 \]

4. **Finding the Radius of the Larger Drop**: Taking the cube root of both sides, we find:
\[ R = 2r \]
This means the radius of the larger drop is twice that of the smaller drops.

5. **Charge on the Larger Drop**: The total charge on the larger drop is the sum of the charges of the 8 small drops:
\[ Q_{\text{large}} = 8q \]

6. **Calculating the Potentials**: The electric potential \( V \) at the surface of a charged sphere is given by:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant.

- **Potential of the Larger Drop**:
\[ V_{\text{large}} = \frac{k \cdot 8q}{R} = \frac{k \cdot 8q}{2r} = \frac{4kq}{r} \]

- **Potential of a Smaller Drop**:
\[ V_{\text{small}} = \frac{kq}{r} \]

7. **Finding the Ratio of Potentials**: Now, we can find the ratio of the potentials:
\[ \text{Ratio} = \frac{V_{\text{large}}}{V_{\text{small}}} = \frac{\frac{4kq}{r}}{\frac{kq}{r}} = \frac{4kq}{r} \cdot \frac{r}{kq} = 4 \]

### Final Answer:
The ratio between the potentials of the bigger drop and the smaller drops is:
\[ \text{Ratio} = 4:1 \]